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What Exactly Makes Probability A Requirement To Solving This Puzzle?

I'm no expert on these things, but do the various published simple solutions rely on Probability? Glkanter —Preceding unsigned comment added by Glkanter (talk • contribs) 23:22, 20 December 2009 (UTC)

Yes. Martin Hogbin (talk) 00:06, 21 December 2009 (UTC)

Maybe for the original 1/3, I can see that. And for the total of all outcomes equaling 1. Am I allowed to mix disciplines when solving a problem logically? Glkanter (talk) 01:03, 21 December 2009 (UTC)

The question, in pretty well any form, asks the question, 'What is the probability...'. That makes it a probability question in my book. 86.132.187.240 (talk) 15:28, 21 December 2009 (UTC)

The question is 'Should the contestant switch?' Glkanter (talk) 18:15, 23 December 2009 (UTC)

Yes you are right, but if this question is not to be answered on the basis of the probability ow winning by switching, on what other basis should it be answered? Martin Hogbin (talk) 17:14, 14 January 2010 (UTC)

I was only pointing out to 86.132.187.240 that the actual question does not include the word 'probability'. He needs a different argument. Oh, no. That's not you, is it? Besides, is 'What is the probability the sun will rise in the east tomorrow?' a probability question?

Yes. (86.132.187.240 was me, by the way, I forgot to sign) Martin Hogbin (talk) 17:40, 14 January 2010 (UTC)

Maybe to a Mathematician. And I suppose he can solve it. Would an Astronomer call it a probability question? Could the question reasonably go on an Astronomy test? Would it be an Astronomy problem or a Probability problem? Maybe a little of both? Glkanter (talk) 18:00, 14 January 2010 (UTC)

The real response to your question is 'Logic'. Here's part of it. There's 2 goats, and the contestant can only choose 1 of them at most. Hence, there is always a goat for Monty to show. That's not probability. It's logic, and a very big part of the simple solutions. The first sentence of the Article is wrong. The MHP is not strictly a probability puzzle. In fact, it's a story problem that begins, "Suppose you are on a game show...". Maybe that's where certain editors get confused. It's not a test question on a university exam. Glkanter (talk) 17:31, 14 January 2010 (UTC)

Because the result of what the player will get if they choose to switch is uncertain, it depends on the random position of the car, the problem must be about probability.

On the other hand, I completely agree that, "It's not a test question on a university exam", I have made this same point myself many times. Martin Hogbin (talk) 17:40, 14 January 2010 (UTC)

More detailed response to Rick's contrived scenario

This is an interesting but somewhat contrived scenario that indicates the errors of thinking by those who support the Morgan approach that deserves a detailed response.

Lets say Huckleberry is not a contestant but works for the FCC.

This is a bad start. I and others have explained many times that any probability problem must be answered from a defined state of knowledge (or, if you are using modern probability theory, defined distributions based on the information given). I do not think this statement is in serious doubt. It has always been assumed by everyone that we treat the problem from the SoK of a contestant on the show. Whitaker's questions starts, 'Suppose you are a contestant...'. But let is see how it looks from anothe angle.

His job is to make sure the show is fair. He doesn't know much about probability so looks up the problem on Simplepedia. The article has the "combining doors" solution and various other "simple" explanations that convincingly say that 1/3 of the time you should win by staying and 2/3 of the time you should win by switching. He watches the show. He keeps track of how many times players who switch win and how many times players who stay with the first choice win. He sees 3000 shows in which 2700 players stay with their original choice and only 300 switch (people think it doesn't matter so most stay with their original choice). Of the 2700 "stickers" 900 win, and of the 300 switchers 200 win. 1/3 of the stickers win and 2/3 of switchers win. Perfect. He sleeps soundly, convinced everything is OK.

And so he should, the show is indeed perfectly fair to the contestant in that the overall odds come out as would be expected. The players who lose might complain that it was unfair to then but that is just bad luck. Overall the show is fair.

But, it's a game show, so it's run by humans. It turns out what's really going on is they perfectly randomize the initial car placement but the host is told only "open one of the other doors if the player initially picks the car". The host figures it doesn't matter which one he picks in this case, so doesn't pay any particular attention to it.

This is bizarre in the extreme, they randomise the car placement but implement some peculiar rule to decide which door to open rule. I take your statement to mean that the host is told to implement some unspecified rule to decide which door to open when there is a choice. Still, I suppose it could happen.

Huckleberry's boss also doesn't know much about probability and looks up the problem on Wikipedia which has a reference to the Morgan et al. paper and says that how the host picks between two goats matters. The boss thinks about it and decides to see if Huckleberry is really doing his job. The boss looks at tapes of the show and keeps track of the 6 individual pairs of player pick and door the host opens separately, i.e. player picks door 1 host opens door 2, player picks door 1 host opens door 3, etc. The boss determines that even though 2/3 of the players who pick door 1 win by switching (and 1/3 win by staying) players who pick door 1 and see the host open door 2 would win by switching with probability about .625 (rather than .66) and players who pick door 1 and see the host open door 3 would win by switching with probability about .714 (rather than .66). Using the Morgan et al. formula he found on Wikipedia, the boss works backwards, and solves .625=1/(1+p) and .714=1/(1+q) and sees that instead of p=q=1/2 the host is exhibiting a p=.6 preference for the leftmost door and a q=.4 preference for the rightmost door. That's odd, he thinks. Exactly 2/3 of the players who switch win and exactly 1/3 of the players who stay win, but yet the host is exhibiting a slight preference for one goat door over another.

Huckleberry's boss then thinks a bit harder and realises that the host action is not important from the contestant's perspective unless the contestant is aware of what the host's policy is. He decides to consider the to possibilities separately. Firstly he imagines that the contestant has not watched tapes of the show and thus will have no idea of how the host decides which door to open. He concludes that, as the contestant has no knowledge of the host policy, and, as the overall chances of winning by switching are fair, there is no problem. He briefly worries about how this relates to an individual contestant but soon realizes that although from the Sok of the producer the odds are either 1 or 0 depending where the car was placed the odds from the SoK of the player are 2/3.

The boss then wonders about the player who may have studied the show and found out about the host's strange door policy. He soon realises that the player could use this information to gain an advantage over the TV station, although it looks as though no one has done this so far. Anyway that is the station's problem, if they want to give the player an advantage by using an non-random policy that is their problem.

Looking into it some more, the boss notices that the host seems to open door 2 more often on Mondays, Wednesdays, and Fridays, and less often on Tuesdays and Thursdays. The boss talks to the host and the host says to make it "random" he always picks the leftmost door (if he has a choice) on Mondays, Wednesdays and Fridays, and the rightmost door on Tuesdays and Thursdays.

The boss wonders why they choose to give the player an advantage in that way and buys Huckleberry a drink.

Martin Hogbin (talk) 11:38, 24 December 2009 (UTC)

Sorry lads, for all your theoretical considerations, but the shows only took place on Fridays (except on Good Friday of course). Nijdam (talk) 13:02, 24 December 2009 (UTC)

Of course, the shows were completely different from the MHP as it is generally known. Monty in fact never offered the swap on the real show. I was responding to Rick's rather contrived situation and erroneous conclusions. When are you going to tell me exactly what makes an event an condition in a probability problem. Yes, you have told me before but four different answers, none of which agrees with the answer given in the WP article. Martin Hogbin (talk) 13:36, 24 December 2009 (UTC)

Yes it's a contrived scenario. The point is only that the chance of winning for an individual player on the show or even of potentially identifiable subgroups of players on the show does not have to be the same as the overall average across all players. All players get the same odds (2/3 chance of winning by switching) if and only if all of the following conditions are met

You are quite right that if the host has a particular door opening policy (say, always opens rightmost door permitted by the rules) then the subgroup of players who themselves happen to have a (probably unwitting) appropriate 'policy' of deciding whether to switch or swap will do better than those who have the wrong policy (it might turn out that the player's initial door choice is important too). But, the players cannot know any of this as they do not know the host's door opening policy. A player, or a member of the the audience, or us (who only have the information given in the question) cannot know which subgroup a player is in thus from that particular state of knowledge they all have the same chance.

So you are right insofar that subgroups of players who are identifiable to someone who knows the host door choice policy may not all have the same chance of winning by switching, but who cares? Certainly not the player (or the audience or the FCC). For any given player it may turn out that they are in the 'lucky' subgroup or it may turn out that they are in the 'unlucky' group but they still have a 2/3 chance of winning by switching. To explain this again, overall the players who switch have a 2/3 chance of winning therefore from the player's state of knowledge she has a 2/3 probability of winning, that is what probability means. Sure, someone with greater knowledge, say the producer who is watching the show or even someone who knows only the host door opening policy, may have a better idea of which players will win by switching, but neither the player, nor members of the audience, nor those who only have the information presented in the question have any basis on which to assign any probability other than 2/3 to this event. Martin Hogbin (talk) 12:57, 30 December 2009 (UTC)

1) the car is uniformly placed

Yes.

2) the player's choice is independent of the initial car placement (it doesn't do any good to randomize the car placement if a stagehand tells the contestant where the car is)

Sure.

3) the host must open a door with knowledge of where the car is and must reveal a goat

Never disputed.

4) the host must choose which of two "goat doors" to open equally randomly

See my reply above.

5) the host must make the offer to switch

Of course. In fact he must be known to always do this.

6) We should add that the player has not studied the history of the show. Otherwise this becomes more of a game theory problem (in which the host can do no better to act randomly). Martin Hogbin (talk) 12:54, 30 December 2009 (UTC)

If conditions 1-5 are in effect, it doesn't matter if the player has studied the history of the show - the probability for each and every player on the show will be exactly the same as the probability for any other player. This is my exact point. And this is exactly what the FCC cares about, i.e. whether it is possible for a player to gain an advantage over other players. -- Rick Block (talk) 17:44, 30 December 2009 (UTC)

You are right in that 'if conditions 1-5 are in effect, it doesn't matter if the player has studied the history of the show ' but as conditions 1-5 may not always be the applied. I think it is wise to make clear whether we consider the possibility that the player might study form. I have to admit that I do not know exactly what the FCC cares about, I am not sure that they would worry about a player being able to gain some advantage over the expected odds by studying form, especially as this is does not reduce the odds for other players, but maybe they would object to this. If I were them I would not care if TV stations were silly enough to let viewers gain an unexpected advantage over them. Nobody would mind if a casino gave a 40:1 payout for a single number in roulette, in fact there is no legal objection to players of blackjack card counting to gain an advantage over the house. It is up to the casino to take whatever steps they feel necessary to avoid losing money. Martin Hogbin (talk) 12:59, 1 January 2010 (UTC)

The "unconditional" solutions ignore #3 through #5 and say the answer of winning by switching is 2/3 regardless of whether these are in effect. Another way to put it is that these solutions assume these conditions. Yet another way to put it is that these solutions are valid only if these conditions hold. All of them, except #4, were explicitly addressed by vos Savant in her followup columns, in particular the explicit experiment she suggested in her second followup. In my opinion, the solution is not complete unless each and every one of these conditions (including #4) are mentioned. -- Rick Block (talk) 17:10, 24 December 2009 (UTC)

That is exactly why I mention my point six. We can drop point 4 and, provided the player has no information on the hosts door opening policy, either from studying history of the show or in some other way, the odds of winning by switching from the players SoK are still always exactly 2/3, individually, overall, any way you like.

I am not sure why you continue to complicate this d1scussion with points 1,2,3,5. We have long since agree these standard game rules. The only issue being considered is whether the host door choice policy matters. To those who do not know what it is it does not. Martin Hogbin (talk) 12:54, 30 December 2009 (UTC)

See example in new section below. 86.132.187.240 (talk) 15:18, 30 December 2009 (UTC)

Note: If X (choice) and C (car) are independent: P(C=X)=SUM(c) P(X=c|C=c)P(C=c)=SUM(c) P(X=c)P(C=c). As a consequence if either X or C is uniformly distributed P(C=X)=1/3. Hence instead of #1 the choice may be uniformly made.Nijdam (talk) 13:15, 28 December 2009 (UTC)

I think not. If the host has a choice for a particular numbered door for example, the initial car placement and the player's choice should both be random. This is what I was trying to show in my analysis page. I would be interested in completing this page with you, using whatever notation you prefer, as I think it shows some interesting results. Martin Hogbin (talk) 13:07, 30 December 2009 (UTC)

Two urns

There are two identical urns. One contains 10 white balls and 20 black balls, the other contains 20 white balls and 10 black balls. A person picks a ball from one of the urns. Using only the information given, what is the probability that it is a black ball? Martin Hogbin (talk) 15:22, 30 December 2009 (UTC)

This seems the same as Nijdam's question just above. How about if there are 20 white balls (and no black balls) in one of them and 40 black balls (no white) in the other? Or 20 white balls (no black) in one and 20 each white/black in the other? I'm not sure where you're going with this, but my point is that these are all conditional probability questions. -- Rick Block (talk) 17:32, 30 December 2009 (UTC)

Perhaps you would care to answer my question. What is the probability that a black ball is chosen? Then I will answer yours. Martin Hogbin (talk) 17:45, 30 December 2009 (UTC)

I thought I did answer it by referring to the section above. More explicitly, there's a 2/3 chance of picking a black ball from one of them and a 1/3 chance from the other. Assuming the person is picking randomly between the urns, the overall probability of ending up with a black ball is therefore (1/2)(2/3 + 1/3) = 1/2.

And, ah yes, where you're clearly going is that this means that the host's preference between two goats doesn't matter since the player doesn't know it. In your urn problem what you're asking is the overall chance. No one is arguing that the overall chance of winning by switching in the MHP is anything other than 2/3. We all agree on average, across all players, 2/3 who switch will win regardless of any host preference between two goats. However, just like your urn problem where an individual player has a 1/3 or 2/3 chance depending on which urn they pick from which averages out to a 1/2 chance, in the MHP an individual's chance depends on the host preference and only averages out to a 2/3 chance.

That is what probability means. A single player on the show has a 2/3 probability of winning, that is it. There is no separate probability and overall probability, except in the unrealistic case that the host's door opening policy is known.

To make your urn problem more similar to the MHP rephrase it like this. There are two urns, A and B, containing only white and black balls. Randomly choosing between urns, a person's probability of picking a black ball is 1/2. If a person picks a ball from urn A what is this person's probability of picking a black ball?

I suspect your answer would be 1/2, and there is certainly a sense in which that is the right answer. But if we aren't given that the fraction of black balls is the same in both urns, another answer is "anything at all" since all we really know is that if the fraction of black/total balls in urn A is X it must be (1-X) in urn B. -- Rick Block (talk) 19:07, 30 December 2009 (UTC)

Your problem above is not a good model of the MHP unless the question is changed to, 'If a person picks a ball from one of the urns what is this person's probability of picking a black ball', or in the MHP the player is constrained to always initially choose door 1. Martin Hogbin (talk) 12:16, 31 December 2009 (UTC)

The point is that whatever host policy you decide upon, (host always pick door 1, picks leftmost door, picks centre door, in all cases, where allowed by the rules) the player's initial door choice can change the effect of the host's choice. The result is that, for an initial random choice by the player and a fixed host door opening policy, the players chance of winning by switching is 2/3. Martin Hogbin (talk) 20:41, 30 December 2009 (UTC)

The point Nijdam is making above is that there are two different events. The first is where the player chooses a door. This event can hardly be construed as anything other than a random choice by the player, even if the initial distribution is non-random. The second event is where the host opens a door. We all agree this event is done with knowledge of where the car is (or else how does the host always show a goat). The host is therefore not acting randomly. You're saying in the one case where the host picks between goats, the player must assume the host IS acting randomly or, perhaps more accurately, that we must average all possible host preferences. [Rick]

That is not quite what I meant but, if the player has no knowledge of the host policy, this would not be unreasonable. It is what Morgan attempt to do but get wrong, the correct answer in this case is 2/3 again. However even if we adopt a single fixed host policy, say host always opens the highest numbered door (within the agreed rules), if we average over all possible games you will see that the probability of winning by switching is still 2/3, try it for yourself. So even for a fixed host policy, the chances of a player winning by switching are 2/3 (unless of course the player knows the host policy, in which case she might be able to improve her odds of winning, but this is a different question and part of game theory). By the way, I made a mistake above which I have struck out. Martin Hogbin (talk) 12:16, 31 December 2009 (UTC)

You haven't responded to the point I made above about studying the history. I think we all agree that if the host chooses randomly between two goats each and every player has a 2/3 chance of winning by switching and studying the history cannot possibly reveal anything else. Do you disagree that if the host isn't constrained to choose randomly between two goats there is a possibility of obtaining relevant information from studying the history? [Rick]

Yes, of course. It then becomes a matter of game theory but has already been pointed out, if the host has a policy and the player finds this out then the player can do better than 2/3 overall so, in fact the best policy for the host is to act randomly (when there is a choice). Martin Hogbin (talk) 12:16, 31 December 2009 (UTC)

And, yes, it might be fairly complicated (e.g. if player picks door 1 host picks door 2 if possible, if player picks door 2 host picks door 1 if possible, if player picks door 3 host chooses randomly if the car is behind door 3). All I'm saying is that there's a difference between saying the player's chance is 2/3 on average and saying the player's chance is 2/3 in all cases. You seem to be denying that these are different, or perhaps saying the former implies the latter. -- Rick Block (talk) 22:18, 30 December 2009 (UTC)

Except in the case where the player knows, either by studying the history of the game or by some other means, the host door opening policy the odds of winning (from the player's SoK) by switching are 2/3 plain and simple. The overall odds are the odds, since there is nothing known to the player to distinguish different circumstances. Martin Hogbin (talk) 12:16, 31 December 2009 (UTC)

I cannot see your point about studying the history, perhaps you could remind me please. My opinion is that it is assumed by nearly everyone that the player has not studied the history of the game and that this possibility forms no part of the standard MHP. I would also note that anyone who studied the history of the game would not only attempt to gain information about the host's door opening policy but also about the initial car placement, which also might turn out to be non-random.

Studying history, along with trying to guess the show's strategy all belong to a variant section where game theory aspects are discussed. They only serve to confuse the standard MHP even further. Martin Hogbin (talk) 13:09, 31 December 2009 (UTC)

My point about studying the history is that there's a difference between saying the player's chance is 2/3 on average and saying the player's chance is 2/3 in all cases (or in some particular case, such as if the player picks Door 1 and host opens Door 3).

Sorry but you are just plain wrong. The probability of the player winning by switching is always 2/3 from her SoK (does not know the host door opening strategy). I would like to convince you of this, regardless of how the article progresses. Martin Hogbin (talk) 02:47, 1 January 2010 (UTC)

And I would like to convince you that the probability given a specific initial player pick and specific door the host has opened (from the perspective of the puzzle solver, regardless of what we consider the SoK of a contestant) does not have to be 2/3 and is indeed an undetermined value between 1/2 and 1 with an average of 2/3 unless the host is constrained to pick between two goats randomly. But, alas, I think this is not going to happen.

I take you now agree that from the SoK of the contestant the odds are simply 2/3 since you no longer want to treat the problem from this perspective. That is fine, let us treat the problem from the point of view of the puzzle solver. We have to solve the problem on the information given in the problem statement, this is what Morgan purport to do. Whitaker's problem statement gives us no information as to how the car was originally placed. What do we do? We could assign probabilities to each door and conclude rather uninterestingly that the player's odds of winning by switching are from 0 to 1. The only alternative is to apply the principle of indifference and take it that it is equally likely to initially be placed be behind any door. This may not, in reality, be true, perhaps it is more often placed behind door 2, perhaps it is always placed behind door 2, the problem statement does not tell us. We thus have two options, apply a standard principle of probability from which we hope to answer the question or give what is essentially a non-answer. Morgan do as I would do and chose the former.

Whitaker's problem statement gives us no information as to how the host chooses which door to open if the player has originally chosen the car. What do we do? We could assign probabilities to each door and conclude that the player's odds of winning by switching are from 1/2 to 1. The only alternative is to apply the principle of indifference and take it that it is equally likely that the host will open either door in the above case. This may not, in reality, be true, perhaps the host always opens the highest numbered door allowed by the rules. We thus have two options, apply a standard principle of probability from which we hope to answer the question or give what is essentially a non-answer. I would chose the former and thus take it that there is an equal chance that the host will choose either door.

However, it is not even necessary for us to apply the principle of indifference to the host's action. We could take a non informative (uniform) prior distribution of host door choice parameter (Morgan's q) and integrate this to get the answer of exactly 2/3 again.

In fact, it is not even necessary to assume anything at all about the hosts door opening policy, he could have a fixed and policy of always opening the highest numbered door possible for example and still the odds of an individual player winning by switching would be exactly 2/3, unless we restrict ourselves only to the specific case that the player initially chooses door 1 and the host reveals a goat behind door 3. This is shown on my analysis page (this is currently being rewritten to use standard notation, any help welcome). Alternatively you could just consider these two cases: the player initially chooses door 1 and the host reveals a goat behind door 3 and the player initially chooses door 1 and the host reveals a goat behind door 2. Both are always possible for any host strategy. If we do not restrict our problem to the specific case that the host opens door 3, the host door choice parameter is unimportant.

A sudden rush of blood to the head caused me to be a bit overenthusiastic above. Martin Hogbin (talk) 12:42, 2 January 2010 (UTC)

This is an interesting discussion which I would like to continue but it need not stop us reaching agreement on how to best write the article. Martin Hogbin (talk) 13:56, 1 January 2010 (UTC)

Perhaps one last try. If you worked for the FCC, would it be OK with you if the host said he always opens the rightmost door (if he can) on Mondays and Wednesdays, leftmost door on Tuesdays and Thursdays, and literally flips a coin to decide which of two goat doors to open on Fridays? Average chance of winning by switching is 2/3. [Rick]

That would be absolutely fine by me. I might wonder why the TV station would give away more cars than they need to, but that is their problem. Martin Hogbin (talk) 14:08, 1 January 2010 (UTC)

Average chance of winning by switching given the player picks door 1 and the host opens door 2 (or door 3) is 2/3. Average chance of winning for players who initially pick door 1 on Mondays and Wednesdays is 2/3. But on Mondays and Wednesdays if the player picks door 1 and the host opens door 3 the chance of winning (whether the player knows it or not) is 1/2, while on these same days players who pick door 1 and see the host open door 2 win with probability 1 (the latter case happens less often than the former case to make the average 2/3). Are you really saying in your view with this host the chance of winning by switching for a player who picks door 1 and sees the host open door 3 on a Monday is 2/3 unless the player knows the host's policy? And, if a different player does know the host's policy this player has a probability of 1/2?

Players who do not know the hosts policy have exactly a 2/3 chance of winning by switching, from their SoK, from your SoK, who somehow knows the host strategy the odds are 1/2 or 1, you will know which and thus your overall odds will be better than 2/3. Players who discover the host's strategy from studying previous shows will also have this advantage. You are forgetting a basic principle that probability is a state of knowledge. Martin Hogbin (talk) 14:08, 1 January 2010 (UTC)

I'm fine with deferring any discussion of anything other than the symmetric case to a "Variants" section, but I would really like to see a conditional probability analysis in the initial "Solution" section - again, much like talk:Monty Hall problem#Proposed unified solution section. -- Rick Block (talk) 06:11, 1 January 2010 (UTC)

You think it's confusing to approach the problem using conditional probability. I think the wording of the problem deliberately leads people to approach it conditionally, and they do - thinking about the specific case where the player has picked Door 1 and the host has opened Door 3. Furthermore, they typically do this incorrectly thinking 2 unknowns must have equal probability, and this is exactly why people arrive at the 1/2 answer and exactly why they so insistently believe 1/2 is correct.

I suggest we stop trying to convince each other and have the article present both approaches (to the "symmetric" problem). My goal here is not to knock the unconditional approach, but to also show how to correctly address the problem conditionally (since this is how I think people screw it up). I think this means explaining how the unconditional and conditional approaches differ. Can we find a way to do this that you don't find offensive? -- Rick Block (talk) 00:00, 1 January 2010 (UTC)

I am happy to present more than one approach but I think that initially the solution should be in what I will call a non-conditional form. This does not mean necessarily mean unconditional just that we do not mention the issue at the start. Many sources do not mention this issue. As you will see from my statement to the mediator I am happy to include Nijdam's proposed statement about the probability of having the car not changing when the host opens a door, or perhaps we could use footnotes. What I do not want to do at the start is mention that the host might have a choice of doors to open. I do not think that we are too far apart. Martin Hogbin (talk) 02:47, 1 January 2010 (UTC)

I think we have to mention the "issue" of why the probability in the given case (player picks door 1 and host opens door 3) is the same as the average probability across all cases. The probability in this case is a conditional probability. Asserting it is the same or simply ignoring it does not explain it, and (IMO) fails to address the core of the problem. -- Rick Block (talk) 06:11, 1 January 2010 (UTC)

Rick, let us make clear that it is not my and I guess not your intention to give a first solution in complete technical terms using the word conditional probability. I've got the idea, that's what some are afraid of, whether they understand the conditional nature or not. Like you, I want to make sure the "simple solution", not taking the conditional nature into account, is not taken for granted. I think you also will support a first formulation without mentioning the term conditional, but in some way indicating the incompleteness. Nijdam (talk) 11:36, 1 January 2010 (UTC)

I am not going to argue about your statement 'The probability in this case is a conditional probability'. although, I think the issue is not as clear cut as you and Nijdam think. I accept that there is an event, the host opens one of the doors, that might possibly affect the probability that we are trying to calculate and thus we might be wise to treat this as a condition, just to be on the safe side. To make this clearer, I am prepared to accept that the K&W formulation is, strictly speaking, conditional.

Therefore we need, as you say, to give some rationale as to why the host opening a does does not affect the probability that the player has chosen the car. There are two that I can see, the first is symmetry, there is nothing to suggest that opening of one door is different from opening the other when the player has originally chosen the car. The other argument is that, the host chooses chooses randomly so no information can be conveyed by this action, the player's state of knowledge is therefore unchanged.

The problem, as you have pointed out before, is that neither of these arguments is specifically mention in any reliable source, as far as I know. Maybe a footnote of some kind might help.

Alternatively, perhaps we could start with Whitaker's question, give the u/c solution then state that some sources consider this solution to be the solution of a particular interpretation (a bit of licence here) of Whitaker's statement and give Morgan's unconditional formulation, and that a better formulation (here we give the Morgan restatement of the problem) requires a conditional solution. This discussion should probably return to the main talk page now. Martin Hogbin (talk) 11:45, 1 January 2010 (UTC)

Issues about the Morgan paper

1. Method of numbering doors

In the Morgan paper, the assumption is made that the doors are numbered statically, i.e. the doors have a static identity through repeated experiment. An alternative assumption would be that the door initially picked by the contestant is always called No. 1, while the door opened by the host is always called No. 3. Based on this, I postulate:

a. The assumption should be made explicit.
b. "The elegant solution that assumes no additional information" is incorrect, in that it needs additional information.
Heptalogos (talk) 20:57, 31 December 2009 (UTC)

There is plenty wrong with the Morgan paper. Here is my comprehensive criticism of it. Martin Hogbin (talk) 02:50, 1 January 2010 (UTC)

Little is wrong with the paper, following science rules. Most of your critics is about the application of it, i.e. mismatch with reality. Some of it I agree with, some not. Anyway, I found an error in the paper which is an error following their own rules. You mention another relevant aspect not explained in the paper, which is the implicit law (?) that the problem must be treated as one of conditional probability. Heptalogos (talk) 11:49, 1 January 2010 (UTC)

No one has yet given me a clear description of what exactly makes it necessary to treat a given event as a condition in a probability question. Martin Hogbin (talk) 12:31, 1 January 2010 (UTC)

No one??Nijdam (talk) 13:42, 1 January 2010 (UTC)

2. The strategy of Morgan et al.

Morgan chooses not to write an article about the Monty Hall problem, but about several solutions offered to a specific question asked in Parade. They reject all solutions and then present the only possible elegant solution, which is very complex and of no obvious use. What is their strategy? Do they choose a problem of public interest every year? Are they requested by society? Is there any scientific relevance? Do they help anyone by providing their solution? Are they constantly spying for an opportunity to show their own quality, even if they have to misinterpret a public issue, from the public point of view? The relevance of their action is in their strategy, ironically similar to the subject of their solution. Heptalogos (talk) 16:29, 3 January 2010 (UTC)

3. The need for conditional solution

Morgan stated the problem to be conditional and start solving it using conditional formula. Also, they disapprove other unconditional solutions given, which they say would only be right as a solution to the unconditional problem. Thus, the assumption seems to be made implicitly that all conditional problems should be solved using conditional method, for which no scientific proof is given. Actually, it is scientifically valid that if two events A and B are statistically independent, P(A|B) = P(A), which is exactly the case in the 'conditional' problem they are solving. What we need next, is a scientific source which states that there is no need for conditional methodology when conditions are independent on the calculated probability. Simple as that. Heptalogos (talk) 16:39, 3 January 2010 (UTC)

4. The position of a single event

I was hesitating to add this fourth issue, because it's a very formal point of criticism to some probability statements in general. But then I reminded myself of the extreme formality of the arguments of Morgan to find a condition in a number. Moreover, I find the following formality much more inspiring. Morgan calculated some probabilities of winning by switching, considering the entire sample space of probabilities. However, they had to deal with one unique event within that space. The third (!) assumption implicitly made by Morgan is that the event of interest (suppose you're in...) is placed randomly within the sample space. Amen. Heptalogos (talk) 22:28, 3 January 2010 (UTC)

To put it differently: no assumption of this kind would have to be made if the question had been something like: "what would the switching chance of contestants be if they..." instead of describing a single situation. Heptalogos (talk) 22:36, 3 January 2010 (UTC)

5. Inconsistant formality

Morgan is being as formal as stating that some unconditional answers given are not formally answering the exact question in which a door number is given as a condition, even if the detail of the number has no statistical dependeny on the probability of winning by not switching. Next they present a solution that needs no assumptions. But at the end of the paper is mentioned that prior information of the player as to the location of the car, or the car not being placed randomly initially, "are unlikely to correspond to a real playing of this particular game show situation". So, we have aspects which, in relation to the requested probability, are irrelevant and those that are unlikely, of which the first are welcomed and the last ignored. Heptalogos (talk) 22:33, 5 January 2010 (UTC)

A conceptually equivalent urn problem

I have always maintained that there were two issues in Morgan's treatment of the MHP, the basic problem and Morgan's conditional treatment, and that each issue obfuscates the other. This is simple urn problem which shows the meaning of Morgan's conditional treatment. It is intentionally not mathematically equivalent to the Monty Hall problem. However the concepts of knowledge and probability involved are the same.

A host has ten white balls and ten black balls. He must put ten balls in to each of two urns, numbered 1 and 2. He may do this any way he chooses so long as there are ten balls in each urn. For example he may put all the black balls in one and all the white balls in the other, or he may put five of each colour in each urn, or any other combination that he chooses, so long as there are ten balls in each urn.

The player then picks one of the urns and takes a ball from it.

We assume that no one has studied the history of the game.

The unconditional formulation

This might be stated thus. The host will place the balls in the urns as described above and the player will then choose an urn and pick a ball from it. What is the probability that the ball will be black?

The answer is 1/2 because there is nothing which makes it more likely that the player will pick a black ball than a white ball.

The main point is: what is the mechanism of chance, that determines the probability? Only if we know this, we can answer the question. People easily use the term probability, without specifying its meaning. There are two mechanisms, the policy of the host and the policy of the player. If both are random, the answer is 1/2.Nijdam (talk) 12:43, 2 January 2010 (UTC)

What you say is, of course, true. For all we know, the host might always place all the black balls in urn 1 and the player might always pick urn 1. If you take that position the problem is clearly insoluble.

Rather than ask about mechanisms I think it is better to ask about initial probability distributions for the sample set. What is the probability that the host puts N black balls in urn 1 and what is the probability that the player picks urn 1? How should we answer such questions? As you imply, we could say that these distributions are unknown and thus the problem cannot be solved. This is exactly the same with the MHP, for all we know the producer might always place the car behind door 1 and the player might always pick door 1. In a real-world situation these questions are vitally important and might make the actual answer very different from that expected from assumed random action. For example ask one person to think of a digit from 1 to 9 then ask another to guess what it is. The probability of a correct guess is, in reality, likely to be better than 1/9.

The question we have to answer here and for the MHP is what initial distributions do we assume? We might start by admonishing the questioner for not defining the question with sufficient precision. Fine, I accept my admonishment, I cannot speak for Whitaker. So now what? We have, as I see it, only two alternatives. Insist that the problem cannot be solved, or make some reasonable assumptions, state what they are, and answer the question on that basis.

Before we make our decision we should perhaps consider the origin and purpose of the question. Is it a real-world question, in which case we maybe should think very hard about the likely starting distributions, based on our general knowledge, or is it a mathematical puzzle, in which case we might suppose that we should take unknown distributions to be uniform, especially as we cannot answer the question any other way. There are two things that I would add to this, firstly that we should be consistent in or decisions, and secondly that we should state the assumptions that we have made. I think that in this case application of the principle of indifference is justified for all unstated distributions. Martin Hogbin (talk) 13:50, 2 January 2010 (UTC)

Conditional formulation

The host has place the balls in the urns as described above and the player has chosen urn number 1 and taken a ball from it. What is the probability that the ball is black?

Some points to note

1. The host, by their choice of ball distribution, cannot change the overall odds of a player winning, just as the MHP host cannot, by means of their door opening policy, change the player's overall odds of winning by switching.
2. The host, by their choice of ball distribution, cannot change the odds of an individual player winning, because the host cannot know which urn the player will chose. In the MHP, the host cannot, by means of their door opening policy, change an individual player's odds of winning by switching.
3. The urn chosen is clearly identified, just as in the MHP the door opened by the host is identified.
4. The host knows what is in each urn, just as the host knows his door opening policy in the MHP.
5. The player does not know what is in each urn, just as the player does not know the host door opening policy in the MHP.
6. The problem statement does not tell us what is what is in each urn, just as it does not tell us the host's door opening policy in the MHP.

The conditional answer

So what is the probability the the player has drawn a black ball.

From the SoK of the host it is N/10 where N is the number of black balls in urn 1.

From the SoK of the player it is 1/2 as the player does not know what is in the urns.

From the SoK of the problem solver it is 1/2 as no information about the distribution of the ball in the urns is given in the problem statement.

Does anyone disagree with these answers? Let us leave for later discussion the best way to get them. Martin Hogbin (talk) 10:32, 2 January 2010 (UTC)

The strategy of science

Indeed, the distinction between conditional and unconditional is in the number of the urn. What you are demonstrating, is that this number (the event of the specific host choice) has no statistical dependency on the player's choice. When no statistical dependency exists between events A and B, P(A|B) = P(A). But the problem is still conditional simply because we ask P(A) given event B. So far most of us agree.

What we don't agree about is the need for using conditional formula. In math there appears to be no common rule or law which tells us exactly when to use conditional formula. There even seems to be no reliable definition of "conditional". Some say "any given event", others say "any event reducing the sample space". The use of such definitions is unclear. Both actually don't have any value regarding the method of solution. Heptalogos (talk) 14:06, 2 January 2010 (UTC)

I agree. I have seen five definitions of what is conditional as, as you say, even for conditional problems there may be a clear rationale for using an unconditional solution. Martin Hogbin (talk) 14:55, 2 January 2010 (UTC)

For the future, I foretell that the common definition of conditional probability will include the statistical dependency between probability and condition. However, the paradox will still not be solved! Because, how do we prove independency? Should conditional formula be used? There seems to be a logical loop. It's all in the question. What should we ask? Should we use all given information in the question, or does some of it have no relevance? How can we be sure?

The paradox is that of learning through mistakes, of evolution, iteration, efficiency. Should we try as much as possible, use our energy to think out our best chances, or don't take any chance at all? Good knowledge needs good questions, but good questions need good knowledge. When we finally reach the perfect question, we already have the answer!

Conditional methodology doesn't take any chances, but consequent use of it would kill us all, because of it's unefficiency. Most of the times there's no use or sense in it. So we have to consider when to use it, which means taking a chance! Heptalogos (talk) 14:06, 2 January 2010 (UTC)

I agree again. We cannot possible take every event into account for every problem. Martin Hogbin (talk) 14:55, 2 January 2010 (UTC)

The strategy of Morgan et al. has considered it quietly: they ripped the information which simply couldn't be used scientifically, but they used everything that could be calculated. However, if the question would hold multiple conditions which had no statistical relevance, but could all be calculated through, they would probably call it a useless question and don't even start at it. So the randomness of their science is in deciding which questions to aks and answer. Which is exactly what makes science an overall subjective process, namely the strategy of it. Heptalogos (talk) 14:06, 2 January 2010 (UTC)

I agree the distinction between conditional and unconditional is not absolute. Martin Hogbin (talk) 14:55, 2 January 2010 (UTC)

The conclusion is that we are all right. We cannot speak about the 'need' for conditional calculation; it is a matter of probability itself. In most cases we quietly agree on the use of it, but here we found an example in the twilight zone of effectiveness and efficiency. Even Nijdam is right (;), taking absolutely no risk at all. Using this strategy he could e.g. simply refuse to solve any problems for which the solution takes more energy than it produces. At least from his own perspective. Heptalogos (talk) 14:24, 2 January 2010 (UTC)

That is why I would like to start with a simple and convincing solution that does not mention conditions then, after that has been properly explained, we can move on the discuss the issue of conditional probability. Martin Hogbin (talk) 14:55, 2 January 2010 (UTC)

Where is it mandated that 'probability' is the only discipline that may be used to solve this problem? The article says it's a 'probability problem'. Some of it, yes: the original 1/3 distribution and the total = 100%. Beyond that, it's logic, as per the simple solutions. That's why, imho, it does not require conditionality. A mental check that I am indifferent to which door the car is behind does not make the conditional formulas a requirement. Nobody, after all this time, has answered my Huckleberry question. You all just go on with all this OR, and classroom stuff. ITS A GREAT PARADOX BECAUSE LAY PEOPLE CAN START OUT CONFUSED AND THEN GET INFORMED IN A SIMPLE MANNER. AND THAT HASN'T CHANGED SINCE 1975. Glkanter (talk) 15:44, 2 January 2010 (UTC)

You may be both right if you state that this very old paradox has been taken hostage by fighting groups because of a silly (only in an absolute way) question being asked in Parade. The old unconditional dilemma is wounded by someone accidentally conditioning a door number. On the other hand, the old dilemma is honoured to be the center of a new paradox, which is about the use and sense of conditional probability methodology. The first and original paradox however is now totally snowed under by the new paradox.

What I like to suggest is to divide this article into two parts. The first one describing the old paradox, unconditional. It should indeed not mention conditional probability and logically explain why chances are not 50-50. And the second part to actually present another paradox, which started almost immediately after the initial commotion that vos Savant started. It is about the exact (un)conditional questions and their solutions, mainly the unconditional (popular) and conditional approach.

Finally something could be said about the risk of asking less usefull questions. Because the conditional solution is correctly taking no risk in judging the relevance of any condition, but the unconditional approach is wisely using the common sense of understanding what not to ask. Heptalogos (talk) 21:49, 2 January 2010 (UTC)

There's no 'new paradox'. Whatever Morgan may have come up with is not even 'a' paradox, and certainly did not supplant Selvin's as 'the MHP paradox'. And Rick has said Selvin came up with 'it' first, way back in 1975. Check this out: The Meta Paradox. Glkanter (talk) 22:37, 2 January 2010 (UTC)

It's not at all something that Morgan comes up with; it is treating any given condition implicitely as a 'relevant' condition, just because it's presented as a condition, when the question is taken literally. What else should science do? Whatever the answer is, it will be in the outskirts of science, or abroad. I think your problem is in fact that you overestimate science. Try to keep it within your framework, instead of being the twist in it's framework. You are in another dimension. Heptalogos (talk) 23:22, 2 January 2010 (UTC)

You may be right that paradox is not the right term. It's about judging the methods of judging, about finding the answers to ask the right questions, about the usage of logic, which may or may not be determined logically. I used the term iteration to give the idea of not starting with a bird or an egg specifically, but to start with 'shady shapes' that gradually become eggs and birds. It means that even 'pure logic' doesn't start from scratch, but always uses assumptions and common sense at a certain level. We only have to be clear about that and not pretend that our solutions are perfect. The shortcoming of science is it's choice of conditional questions to answer, which may not be science at all. Heptalogos (talk) 23:48, 2 January 2010 (UTC)

There is one and only one paradox, and it's paradoxical precisely because it's on the edge between conditional and unconditional. Because of the way the problem is phrased, lay people create a mental model of a (conditional) situation where there are two closed doors and one open door (the player has picked door 1 and the host has opened door 3). In this case, there are two unknowns and people naturally jump to the assumption the probability must be equal. These people are willing to argue that the simple solution is wrong not because they don't understand it but because it does not match their mental model of the problem. In particular, it requires them to ignore the specific case they're thinking about - or, more sophisticated, see that the specific case and the general case must be the same. Most people have a very hard time switching mental models after one has been created, and an even harder time thinking about two models at the same time and seeing that they're equivalent, and therefore find the "simple" solution unconvincing.

Rather than force them to switch their mental model or bludgeon them into agreeing that the question is actually talking about the overall probability rather than the probability in the specific case they're thinking about, another approach is to show them exactly how even in a specific concrete case (such as the one where the player has picked door 1 and the host has opened door 3) the probability of the two closed doors is not the same. This is what the conditional probability approach does. -- Rick Block (talk) 00:28, 3 January 2010 (UTC)

Thank you for that outstanding piece of OR, Rick. The only facts are: Selvin came out with this great paradox in 1975. Morgan and others wrote papers critical of it, perhaps unaware of the meaning of 'Suppose you're on a game show...' Other papers are subsequently written using Selvin's original solutions. It remains a great paradox. Glkanter (talk) 08:59, 3 January 2010 (UTC)

I agree with Glkanter here. Rick's statement, '...it's paradoxical precisely because it's on the edge between conditional and unconditional' is not supported by any source that I am aware of. Although Rick is fully entitled to express his personal opinion here it should not be allowed to control the article. Martin Hogbin (talk) 11:24, 3 January 2010 (UTC)

The one and only paradox, the old one, has little to do with the specific question asked in Parade, which mentions door numbers. Starting the article almost immediately with this specific question, it already introduces a complexity which kept us all busy for years, while the simple paradox has not even been explained yet.

The complexity in solving any specific question asked, divides people mainly into two groups: those who insist on the need of conditional methodology and those who prefer the unconditional approach when this seems effective and efficient. It was only yesterday that I found that the discrepancy is not really contradictory because people argue at different levels.

The controversy over the conditional approach should actually not be based upon taking the question literally or not, but upon taking the question seriously, i.e. regarding it as usefull. It is the usually unquestioned scientific strategy that may be failing. Science is always searching for better questions to ask, using it's autonomy to improve knowledge. Sometimes it is asked by society to answer specific questions. And sometimes it is playing around, choosing a specific question to solve, for let's say, demonstrating it's superiority. It is such random strategy that to my opinion is felt by people as arrogant and useless. And it is interesting to see that when science loses it's legitimate drive, it risks the blame of being corrupted. Heptalogos (talk) 12:42, 3 January 2010 (UTC)

To rephrase: the paradox of the three door problem has in principle nothing to do with conditional probability. There also doesn't have to be a shady edge between conditional and unconditional. The shadow is actually on the benefits of conditions in questions, which is intrinsically the same as the choice of answering such a question.

Glkantor, btw, Morgan did not write about the great paradox, but about the specific question asked in Parade, which is significantly different. Heptalogos (talk) 13:07, 3 January 2010 (UTC)

You may be right. I've read it numerous times. I still don't know what their point is. But the Wikipedia article is about The Monty Hall problem, and the Paradox Selvin describes in 1975. Really, these 4 great scholars pick apart a general interest Sunday newspaper supplement brain teaser, and find it insufficiently rigorous for a Probability text in its problem statement? I'm shocked. Shocked! And then they feel this analysis is worthy of being published in a peer-reviewed professional journal? No wonder Seymann wrote his companion piece. And still, Morgan's paper becomes the focal point of this whole article? Unimaginable, really. Glkanter (talk) 13:44, 3 January 2010 (UTC)

I agree. They kidnapped the old dilemma, introducing a new problem to it that they created themselves, shining a light on their own superior solution to that. That's maybe why they chose to answer a question that is quite silly taking it literally. It is silly because there is no meaning to the door number, unless in providing the scientists an opportunity to present their ultimate elegant solution that needs no assumptions. In which they failed for that matter, because they assumed static door numbering. Anyway, I am very thankful for the event to happen, because I got inspired by it, and I know you all experience the same. Heptalogos (talk) 14:09, 3 January 2010 (UTC)

There Are No Previous Game Plays to Study

"Suppose you're on a game show..." does not imply that the show existed before, or will exist after you play it. There's no history to possibly study. Just this one playing to decide upon. Glkanter (talk) 16:59, 3 January 2010 (UTC)

That's why assumptions are made about the host's strategy. It's all in the strategy of that moment, even without the past or future. Heptalogos (talk) 18:23, 3 January 2010 (UTC)

So, can Huckleberry do better than he already did using a simple solution, if someone tells him about the equal goat door constraint equaling 1/2? And then explains what that means? I don't see how. Glkanter (talk) 18:49, 3 January 2010 (UTC)

No, his common sense creates the same result. But Morgan state that the door number is presented as a condition, so it should be used as a condition, even if there is no use or sense in it otherwise. It's a formality that either gives them a chance to validate themselves, or is the result of some imaginary rules of science. Heptalogos (talk) 20:10, 3 January 2010 (UTC)

Quite. Martin Hogbin (talk) 09:54, 4 January 2010 (UTC)

Explaining the paradox

Let us return to the original paradox. I believe that many intelligent people get fooled because of the specific question asked. You see, not all necessary information is given right away. This is accepted by most people, because information can be completed by making reasonable assumptions. However, these assumptions are just not reasonable enough to be made right away. It's a strategy of concealing information. What I am saying is that a significant part of the paradox is actually an illusionist's show.

In the graphics below can be seen that all necessary information forms a rectangle (1). Taking out some information randomly still shows the rectangle (2). We fill in the missing outlines intuitively. In our mind, we probably don't even notice it missing. However, taking out certain information creates an imaginable square for many people (3). It is only after realizing that we are missing information, that we think all given information through and are able to make the reasonable assumption, thus showing the big picture (4).

In the Monty Hall problem, the significantly missing information is obviously the fact that the host always opens another door than the one chosen by the contestant. He is not just opening another door, another door is always opened. It's only a reasonable assumption if you get the notion to make an assumption anyway. For some people it will still not make any difference to have an explicit idea of the exact host strategy, but I'm very curious what a difference it does make. Let's try to describe the dilemma without the necessity of any additional assumption:

---

Imagine a game show presenting three doors to a contestant, who may open one door, receiving whatever is behind it. One car and two goats are placed randomly, each behind one door. These are the game rules: first the contestant chooses a door randomly, which keeps closed. Then the show host always opens another door with a goat behind it, choosing randomly if possible. Finally the contestant is offered to open one of the remaining closed doors. To win the car most of the time, should contestants stick to their choice or open the other door?

---

Heptalogos (talk) 10:51, 8 January 2010 (UTC)

I agree that this is the Monty Hall problem as it should be stated. This means that the problem and solution are uncomplicated by the host's door preference, making it a simple mathematical puzzle, which most people still get wrong. The problem for us as editors of Wikipedia is that what we say should be based on reliable sources. Are there any sources which state the problem the way you do? It would be very good if you can find one. Martin Hogbin (talk) 09:53, 4 January 2010 (UTC)

I just added ", choosing randomly if possible", which I forgot. Anyway, I think it's fine that the question (at first) is quoted as it is asked most of the times, historically. It would only be nice to know what happens if we ask the question above, in order to understand the paradox better. I haven't found any example. Maybe we could arrange an experiment, in which different questions are asked to people. If the outcome would be (very) significant, media would like to write about it. Voila. Heptalogos (talk) 20:21, 4 January 2010 (UTC)

What experimental conditions are required for the 2/3 answer

I have a question for all you anti-conditionalists out there. Assuming the door numbers are persistent, e.g. Door #1 is always the leftmost door on the stage, Door #2 is always the middle door, and Door #3 is always the rightmost door (if you'd like to argue about whether this is a reasonable assumption can we please do it in a different thread?), if I want to run an experiment where the observed frequencies of winning by switching or staying are the same (in the limit, as the number of trials approaches ∞) for each of the 6 possible combinations of initial player pick and door the host opens, then what experimental conditions must I ensure are true? There are 12 frequencies I'm talking about, i.e.:

1-2-stay: (number of players who pick Door 1 and see the host open Door 2 and stay and win) divided by the total number of (players who pick Door 1 and see the host open Door 2 and stay)

1-2-switch: (number of players who pick Door 1 and see the host open Door 2 and switch and win) divided by the total number of (players who pick Door 1 and see the host open Door 2 and switch)

1-3-stay: (number of players who pick Door 1 and see the host open Door 3 and stay and win) divided by the total number of (players who pick Door 1 and see the host open Door 3 and stay)

1-3-switch: (number of players who pick Door 1 and see the host open Door 3 and switch and win) divided by the total number of (players who pick Door 1 and see the host open Door 3 and switch)

etc.

I'll start the list.

a) the car is initially uniformly distributed

b) the host, who knows what's behind the doors, always opens a door showing a goat

c) the host always makes the offer to switch

My desired experimental outcome (as n approaches is ∞) is

1-2-stay = 1-3-stay = 2-1-stay = 2-3-stay = 3-1-stay = 3-2-stay (= 1/3)

1-2-switch = 1-3-switch = 2-1-switch = 2-3-switch = 3-1-switch = 3-2-switch (= 2/3)

Aside from the list I started, is there anything else I need to ensure is true to get my desired experimental outcome? I think we all know that the answer is yes (the host must also choose between two goats randomly), but I just want to check and make sure there's no disagreement about this. -- Rick Block (talk) 20:30, 6 January 2010 (UTC)

Is the contestant aware of the situation (bias) if the host doesn't choose between the goats randomly? Is this consistent with 'Suppose you're on a game show'? Hosts and producers don't tell contestants where the car is located. The puzzle says nothing of any prior plays, from which to learn a history or pattern. So, I disagree. But I'm not a professor. And this is OR, anyways. Glkanter (talk) 20:53, 6 January 2010 (UTC) Glkanter (talk) 20:55, 6 January 2010 (UTC)

I, the experimenter, am aware of what door the contestant picked and what door the host opened. I'm measuring the success of switching or staying in the situations I've enumerated above. The question is what must be true for this to be 1/3 win by staying and 2/3 win by switching in each case I've enumerated. Yes, it is OR (of a sort), but I posted this on the Arguments page. -- Rick Block (talk) 01:13, 7 January 2010 (UTC)

Yes, if specific doors are specified in the problem statement (as you have done above) then I think that you are right, the host must choose a legal door randomly, in addition to the conditions a) to c) you quote, for the 1/3 vs 2/3 result to be true.

If the problem statement were to be that the player chooses any door and the host opens any one of the other doors to reveal a goat then the host is not required to choose randomly and your condition a) can also be dropped, the car could always be behind door 1 for example and the host could always open the leftmost legal door, provided that the distribution of the players original choice of door is uniform. The result in this case is still swap 2/3 win, stick 1/3 win. Martin Hogbin (talk) 22:39, 6 January 2010 (UTC)

Sounds like "yes" to me. I'm using specific doors in my experiment. I haven't connected this is a statement of the MHP yet, just talking about an experiment that could be run in the real world (well, n won't get particularly close to ∞ - but close enough to show a meaningful trend). -- Rick Block (talk) 01:13, 7 January 2010 (UTC)

Why do you want to run an experiment? Martin Hogbin (talk) 16:59, 7 January 2010 (UTC)

Any other takers? -- Rick Block (talk) 01:13, 7 January 2010 (UTC)

What is your realistic experiment? Is it a computer simulation, or did you collect humans randomly, or what? If it's a simulation, what does 'knows what's behind the doors' mean? The switching choice should be random. I miss a description of all actions, so I cannot tell if all requirements are met, if that's what you mean by 'condition'.

If it's a real show experiment, requirements would be about the knowledge and participation of the contestants, for instance. I miss a full description of the experiment.

Why should the host choose randomly, or should the car be uniformly distributed? Anyway, I am willing to accept the last requirement if that's simply the case. Heptalogos (talk) 22:12, 7 January 2010 (UTC)

The question is if I were to run such an experiment, what experimental setup would be required. It's the same point as the "Huckleberry works for the FCC" scenario above, in the forward direction rather than reverse. Each n-m-stay frequency measured by the experiment is, of course, the frequency probability of winning by staying given you've initially picked door n and have seen the host open door m while each n-m-switch frequency is the frequency probability of winning by switching in the same cases. It is a way to objectively measure probability, independent of "state of knowledge". Based on this experiment I could say, for example, what the (objectively measurable) probability is of winning by switching given the player picks door 1 and the host opens door 3.

Part of the problem here may be that you all are arguing that "state of knowledge" is the ONLY way to determine probability and you are specifically excluding knowledge of which particular door the player picked and which particular door the host opened from the player's state of knowledge. This makes "the probability" 1/3:2/3 (stay:switch) regardless of the host's preference (I'm perfectly willing to agree with this). However, if the doors have persistent numbers the actual probability (actual in the sense that we can measure it) in the 6 individual cases must be 1/3:2/3 (stay:switch) if and only if the host chooses between two goats randomly (in addition to the other conditions) regardless of the player's "state of knowledge". I'm not exactly sure how to reconcile these two statements, both of which seem to be true. One approach is to say the frequency probability is from a different state of knowledge. But I think another might be to say that if the doors are persistently numbered and the contestant knows which door she initially selected and which door the host opened, then the host's bias is within the contestant's state of knowledge. This information definitely makes the host's bias knowable to the contestant.

As the article on probability makes clear there are two ways of defining probability, Bayesian (state of knowledge) and frequentism. I tried to discuss this issue long ago.

In the Bayesian approach we need to consider a particular well-defined state of knowledge. This could be the SoK of a character within the problem scenario or the SoK of us, the problem solvers. The important point is that we are consistent and stick with the same clearly defined state of knowledge. If we agree that we are going to consider the problem from the state of knowledge of the player (which many here think is the only valid option, bearing in mind Whitaker's question starts, 'Suppose you're on a game show') then exactly what the player knows or does not know is of critical importance. As I have said before we have an option here, we can either consider a real-world situation, which is very complicated, or take the problem to be mathematical puzzle (again many here think this is the essence of the problem), where certain reasonable assumptions are made, such as that the cars are initially randomly placed and thus we have no knowledge about where they might be, and that we have no knowledge of the host's door opening policy and we must therefore take this to be random.

An alternative, and more formal, approach is to use only information given in the problem statement. This is the approach used by Morgan et al. However, once we adopt this approach we must continue with it consistently. What we think the player might know is not important, thus using this approach, it does not matter whether the player knows the host preference or not, but it also does not matter whether the player sees specific doors opened or not. All that matters is what the problem statement tells us.

A frequentist approach does not avoid any of the issues discussed above. The answer still depends on how you set up your experiment. We need to decide what changes and what stays the same between trials. Specifically you need to consider which of, the initial car placement, the initial player door choice, the host door choice policy, changes between trials. Obviously if they are all fixed the result is always the same. The point is, as Seymann says, we must decide on exactly what the question is before we can provide an answer. Once the question is properly settled there is no need for simulations (except perhaps to convince sceptical members of the general public) the mathematics is well understood. Martin Hogbin (talk) 13:03, 8 January 2010 (UTC)

"we must decide on exactly what the question is before we can provide an answer." This is the main source of confusion. How do we strip (communicated) phenomenon from irrelevant complexities, resulting in elementary code? Is Seymann a reliable source? Did he state that the decision mentioned is a main source of discussion? Heptalogos (talk) 20:50, 8 January 2010 (UTC)

The point I have been making all along is that, whichever of the approaches to probability you use, provided that you are consistent, there is no justification for taking the host to choose a legal goat door other than randomly. Martin Hogbin (talk) 13:03, 8 January 2010 (UTC)

In any event, the main point is and continues to be that the conditional probability in any of the 6 specific cases is mathematically different from the unconditional probability. These can have the same numeric value, or they can be different. An unconditional solution tells you one but not the other. Since many sources present both solutions, the article should as well. -- Rick Block (talk) 01:34, 8 January 2010 (UTC)

I think the essence of your message is that not all information is used in the unconditional approach. A consequence is that conditional solutions "can have the same numeric value, or they can be different." This is generally true. But certainly not in the Monty Hall case, because the door number has no statistical dependency on the requested probability. In plain logic: there is no relevance in the number. As there is no relevance in the knowledge of the host, for which reason Morgan et al. chose to ignore this information also.

The dispute is not about maths, but about it's practice. Three basic elements are involved: phenomenon, theory and communication. Maths can be communicated (written) fully objective; phenomenon cannot. Language is interpretated to strip the phenomenon from all irrelevant information (game, show, host, door, number) into objective maths. The math calculation is fully valid, but the risk is in the interpretation. Real experiment (show, host, game, contestant) has less risk of interpretation, but has the risk of bias.

Knowledge is the only way to determine probability. By knowledge you mean knowledge that influences the phenomenon or knowledge to determine the probability with. Actually the difference in 'door knowledge' is not essentially in the number of the door; it might also be e.g. the knowledge of the door closest to the chosen door. Another issue: regardless of the contestant's knowledge, it is not enough to require random host behaviour, because in that case the contestant might simply know where the car is. Or she may know when the host is behaving randomly, concluding that both unchosen doors have goats.

Is it true that bias means systematic distinct from random? Knowledge of the host's choice in a single event does not give information about the existence of such bias, does it? Heptalogos (talk) 11:16, 8 January 2010 (UTC)

Making the specific general, explicitly justified.

The probability of winning by switching is not at all effectively influenced by the knowledge of the particular door that is opened; it doesn't matter a bit whether door 2 or door 3 is opened. If this (extensive) logic is included in the correct unconditional solving method, it is fully proof. One could even state that this kind of solving is conditional eventually, because it does use the specific information, which is then identified as part of a bigger, more general, pool of events with uniform effect on the requested probability. Uniform meaning that each specific event in the pool is as likely to happen; all types of specific events are equally distributed. By accurately replacing the specific event with the more general pool of events, all possible consequences are fully accounted within the entire scope of the problem at hand.

Do you agree? Heptalogos (talk) 14:44, 10 January 2010 (UTC)

Have a look at this page, which I am writing with Nijdam (although he seems not to be around at the present). This does what I think you are talking about, it starts with the general case and considers various specific cases. The most important thing is to know which of: the producer's initial car placement, the player's initial door choice, and the host's legal door choice are to be considered random. Your comments, on the associated talk page, are welcome.Martin Hogbin (talk) 15:23, 10 January 2010 (UTC)

OK, I'll check it later anyway, although it doesn't seem familiar at this point. Heptalogos (talk) 20:19, 10 January 2010 (UTC)

I think Heptalogos's point is that if

$P({\text{car behind door i}}|{\text{player picks door i and host opens door j}})\,$

is the same for all i,j in {1,2,3} (with i != j), and these are non-overlapping events that exhaust all possibilities, then these all must be the same as

$P({\text{car behind door i}})\,$

And, if an "unconditional" solution is accompanied by a logical argument showing the relevant statements hold (all conditional probabilities must be equal and the conditions exhaust all possibilities) then this solution is "complete". This is, of course, true. The issue some sources have with the "unconditional solutions" is that rather than make such an argument they implicitly assume BOTH of these statements, sometimes using wording implying they are always true. Falk (in particular) goes into much more detail about this. For example, many solutions say something like "since you already know one of the two unchosen doors must be a goat, by opening one of them the host has not given you any additional information about the probability that your chosen door hides the car" - as if it is a mathematical axiom that P(A|B) = P(A) if B "only reveals information you already know". There is no such axiom. The reasoning is WRONG. The host picking the leftmost door if possible variant is the common counterexample showing how this is false. What is true is that P(A|B) = P(A) if A and B are independent.

"P(A|B) = P(A) if A and B are independent" is fine and I do not disagree. The problem is that it is somewhat tautologous. To say that event A is independent of event B is to say that the occurrence of event B does not affect the probability of event A (and vice versa). So this gets us nowhere. We have to use out judgment to decide whether events are independent and thus we have to use our judgment to decide whether a given event must be treated as a condition of the problem. There is no simple rule that answers this question. The only basis on which we can objectively address this question is a Bayesian one. That is to say does the occurrence of event B give us any more information about the probability of event A? There is no other formal basis. Martin Hogbin (talk) 16:51, 10 January 2010 (UTC)

This is exactly the point! You can't just "use your judgment" because people's intuitions about whether events are independent or not are often completely wrong. If the problem gives you a condition, you shouldn't ignore it but rather you should verify whether it affects the solution or not. -- Rick Block (talk) 21:07, 10 January 2010 (UTC)

The only formal basis: yes. Intuition not adequate: yes. Plain logic not objective: wrong. Every formal mathematical axiom has naked logic as it's mother. The essence here is: seemingly solid intuition needs explicit arguments when asked for it. Heptalogos (talk) 21:48, 10 January 2010 (UTC)

Rick, if you are saying that it is a good precaution to treat any event which might affect the probability of interest as a condition just in case it does then I agree, as my statement at the end shows. 86.132.191.65 (talk) 22:46, 10 January 2010 (UTC)

If the host always picks the leftmost door if possible, his choice actually does reveal new information. So I don't see how this could prove the 'axiom' false. Furthermore, some true axiom's may still be missing in maths. Accurate logic should be enough. Heptalogos (talk) 20:29, 10 January 2010 (UTC)

Yes. I had not noticed Rick's claim to the contrary.

The common argument is that because you already know one of the doors hides a goat, the host showing you that a specific door hides a goat does not give you any additional information. This is the statement for which the leftmost host variant is the counterexample. If you precisely define "no new information" to be the same as independent, then you're OK - but the intuition and the mathematical precision often don't align. And, if you're using a "new" axiom, you really need to prove it is true first (this might be what you mean by using accurate logic). -- Rick Block (talk) 21:07, 10 January 2010 (UTC)

The leftmost variant is not valid with the assumption of random door choice. Indeed, 'no new information' means 'no new relevant information' means 'no information of statistical dependence'. That's the difference between logic and maths; logic is much more tricky in it's many interpretations. So, I do hope that you consider non-maths logic as a valid solving method anyway? Heptalogos (talk) 21:26, 10 January 2010 (UTC)

Another example is many solutions say something like "your initial chance of picking the car is 1/3, if you never switch then after the host has opened a door your probability must also be 1/3", as if it is an axiom that P(A) = P(A|B) regardless of B! The "host forgets" variant is a common counterexample showing how this is false.

Except in that case the host action does give you information. Martin Hogbin (talk) 16:51, 10 January 2010 (UTC)

The assumption must be that the host always opens another door with a goat (randomly), so it doesn't really matter what the host forgets. Heptalogos (talk) 20:39, 10 January 2010 (UTC)

Heptalogos, the "host forgets" variant is a standard term here for the case where the host may reveal a car. Martin Hogbin (talk) 21:00, 10 January 2010 (UTC)

Yes, which is not possible with the reasonable assumptions made. Heptalogos (talk) 21:37, 10 January 2010 (UTC)

Perhaps your intuition is better than most people's - but I don't think it's at all obvious how the host's action gives you information in this case but not in the normal one. You know it MUST be true since the probability is different, but that gets back to your tautology issue. -- Rick Block (talk) 21:07, 10 January 2010 (UTC)

If, instead of this sort of wrong reasoning, an "unconditional" solution were accompanied by an argument that all the conditional cases are equally probable and they exhaust all the possibilities then the solution is indeed addressing the conditional case. -- Rick Block (talk) 16:17, 10 January 2010 (UTC)

I would accept this statement. If there is any doubt as to whether a given event B might affect the probability of event A occurring then it might be wise to treat B as a condition and deal with the problem on that basis. If it turns out that B makes no difference that is the answer. Martin Hogbin (talk) 16:51, 10 January 2010 (UTC)

Nice to agree on that one! I think this opens possibilities. Heptalogos (talk) 20:41, 10 January 2010 (UTC)

I think a better way to say this is if the problem asks for P(A|B) and you're going say the answer is P(A) you HAVE to show B makes no difference. -- Rick Block (talk) 21:07, 10 January 2010 (UTC)

Exactly, why didn't I say so. Heptalogos (talk) 21:39, 10 January 2010 (UTC)

Rick, in the case of a random legal door choice by the host you can do just this. If the host chooses a door, when there is a legal choice, randomly the door he opens can convey no information. This is a fundamental fact of information theory. If no information is conveyed then the probability cannot be affected. As I agree above, it might be a wise precaution to treat the problem conditionally but you cannot insist that the problem must be treated this way if it can be shown that the condition makes no difference. Martin Hogbin (talk) 22:50, 10 January 2010 (UTC)

Of course Rick can say, and has said for years, it must be conditional. So did a bunch of those Mathematics guys last spring, and Nijdam and Kmhkmh to this day. Because Morgan calls the simple solutions 'false' in a published source. Glkanter (talk) 23:26, 10 January 2010 (UTC)

Do you want us to agree anyway? Showing the condition not to be dependent can be regarded as a first phase conditional solution, because in that phase it changes the problem from conditional (how it is indentified) to unconditional. Heptalogos (talk) 12:00, 11 January 2010 (UTC)

Your argument isn't with me. Rick and I agree on nearly nothing at all. I'm trying to play within the rules of Wikipedia, and unfortunately, Morgan is reliably published. I'm trying to demonstrate that Morgan adds nothing, and should, therefore be minimized. I've advocated for simple solutions since October, 2008. Morgan calls them 'false'. We have a consensus. Let's get through Mediation, Arbitration, or whatever. Glkanter (talk) 15:43, 11 January 2010 (UTC)

Rick is very close to what I mean. Using his method, which is very nice, I come up with this:

$P({\text{car behind door 1}}|{\text{host opens door 2 or 3}})\,$ is the same as $P({\text{car behind door 1}})\,$

$P({\text{car behind door 1}}|{\text{host opens door 2}})\,$ is the same as $P({\text{car behind door 1}}|{\text{host opens door 3}})\,$

so $P({\text{car behind door 1}}|{\text{host opens door 3}})\,$ is the same as $P({\text{car behind door 1}})\,$ Heptalogos (talk) 20:16, 10 January 2010 (UTC)

That seems a good argument for not worrying about the 'condition' if the host chooses a legal door randomly. Martin Hogbin (talk) 21:04, 10 January 2010 (UTC)

Yes, but the question is why are

$P({\text{car behind door 1}}|{\text{host opens door 2}})\,$

and

$P({\text{car behind door 1}}|{\text{host opens door 3}})\,$

the same? I think you can't conclude this without mentioning something about the host's preference in the case where there are two goats (unless you're going the "I can't tell the difference between 2 and 3" route). -- Rick Block (talk) 21:07, 10 January 2010 (UTC)

Indeed, we're still using the most reasonable assumptions here. So, then you agree with this method? Heptalogos (talk) 21:30, 10 January 2010 (UTC)

Agree in what sense? It is logically correct, however (as far as I know) there's no source that actually treats the problem this way. The popular sources (no doubt following vos Savant's lead) gloss right over this. The more academic sources solve the conditional case directly (using Bayes' theorem or a probability tree or some other conditional case analysis mechanism). Without a source it becomes OR to put it in the article (even though it's true). From the standpoint of the article, I think we should do what Kmhkmh and several others have suggested which is "stick to summarizing all reputable literature in a representative and readable fashion". -- Rick Block (talk) 21:55, 10 January 2010 (UTC)

True, but let's not bother about the article for a moment; we're still on the arguments page. Now I'll try to write down the maths that we agree about in normal language, given the same reasonable assumptions:

The chance that the car is behind door 1 is still the same, given the goat revealed behind door 3, since the host would anyway open door 2 or 3 revealing a goat, for which the chances are both the same.

Would such an answer be correct?Heptalogos (talk) 22:12, 10 January 2010 (UTC)

"...for which the chances are both the same". Marilyn didn't say so. She used the example of the sea shells.

• MvS: "Then I simply lift up an empty shell from the remaining other two. As I can (and will) do this regardless of what you've chosen, we've learned nothing to allow us to revise the odds on the shell under your finger."
• Morgan: "You may do this regardless of my choice, but you can still give a hint on the winning shell."
• MvS: "Good heavens, we made so many reasonable assumptions, isn't my random behavior simply another reasonable assumption?"

(Actually, the table below the sea shells example, with all 6 possibilities, shows the random behavior.)

Formally, Morgan are right in their criticism: one can never be sure. But then they make the same mistake of assuming things within their own solution. The lesson is that it's all in the interpretation, for every phenomenon described. Otherwise it would be written in mathematical code. Heptalogos (talk) 23:36, 10 January 2010 (UTC)

The problem is vos Savant's quote, "As I can (and will) do this regardless of what you've chosen, we've learned nothing to allow us to revise the odds on the shell under your finger.", is exactly one of the kinds of statements that assumes P(A|B) = P(A) if B is "known". She makes this claim without even hinting that how the host chooses between two goats might matter (it's not in her problem description or any of her followup columns). It's not enough to know that the host can and will show a goat. The host also has to choose between two goats randomly. This is not a minor point. What she should have said is something like "As I can (and will) do this regardless of what you've chosen and the specific shell I flip tells you nothing, we've learned nothing to allow us to revise the odds on the shell under your finger. Without some "and" here at least implicitly saying the host will choose randomly if he has a choice, this statement is demonstrably false (per the leftmost preference variant). -- Rick Block (talk) 17:00, 11 January 2010 (UTC)

So, out of all other assumptions, which seem reasonable to you, you find this one not reasonable: the host choosing randomly between goats. Why not? What else would be more reasonable? Heptalogos (talk) 19:48, 11 January 2010 (UTC)

Bear in mind that Whitaker's question was from a member of the public to a column in a popular general interest magazine. It can hardly be expected that he frame his question with mathematical precision. Vos Savant made an excellent job of interpreting it. Morgan, on the other hand wrote a paper for a peer-reviewed statistics journal. Martin Hogbin (talk) 23:47, 10 January 2010 (UTC)

Quite. Vos Savant is effectively trying to solve a realistic problem; Morgan are trying to explore, analyze and educate the domain of scientific statistics, using the opportunity of public attention. Heptalogos (talk) 12:13, 11 January 2010 (UTC)

Firstly, I would say that vos Savant was trying to solve a mathematical puzzle. It terms of mathematics and probability theory there is nothing at all new in the Morgan paper. All it does is conjure up an additional layer of pointless obfuscation for a simple mathematical puzzle. Martin Hogbin (talk) 14:03, 11 January 2010 (UTC)

It's not a maths problem basically. Maths may be used though. Which offers the opportunity to make it a maths problem. MvS would have liked to avoid maths, as she did initially. I'm not even sure if she used it anyway, I guess not. Heptalogos (talk) 21:04, 11 January 2010 (UTC)

Editing break

I completely disagree with this. What the Morgan et al. paper does is argue that the problem is conditional. Per Krauss and Wang most people (nearly all) initially think about the problem in terms of the player having picked door 1, door 2 remaining closed, and the host having opened door 3 - i.e. they think about the specific conditional case given as an example in the problem description. The unconditional solution is a completely different approach requiring a different mental model (per Krauss and Wang) that people have a hard time accessing after they've created their initial mental model (also per Krauss and Wang) and (IMO) this is why people argue that the unconditional solutions must be wrong. I also think that by ignoring the conditional case the popular solutions have contributed to probabilistic innumeracy reinforcing WRONG intuitions like "if the outcome of B is known, P(A|B) = P(A)". The fact that I know B is going to happen does not mean it is statistically independent of all other events. Many of the popular solutions confuse prior and posterior probabilities so much that it's hard to argue anything other than that they are simply wrong. -- Rick Block (talk) 17:00, 11 January 2010 (UTC)

Morgan show that in a specific case the legal door opened by the host could be important, in which case the problem must be solved conditionally. I agree, in the case they initially consider, that the host may choose a legal door non-randomly, and the doors are identified, the problem must be treated conditionally.

In the more realistic, and consistent, case that the host opens a legal door randomly, the Morgan paper shows us that we might be wise to treat this case conditionally too, just in case the door opened matters. I agree with this too.

But, Morgan or you are claiming that this is the only possible way to treat the symmetrical case then I disagree. We know from symmetry of the situation that the the result for either door opened must be the same. We also know that if a door is opened randomly, knowledge of the actual door opened cannot give us any information and thus cannot alter the probability the car is behind the initially chosen door (Rick, do you agree that this is true?). We therefore have a sound basis on which to say that the probability the car is behind the initially chosen door is independent of the opening of a door by the host and thus it need not be included as a condition in our calculation. Martin Hogbin (talk) 18:05, 11 January 2010 (UTC)

Martin has explained this very nicely. I would like to add towards Rick the following. The difference in mental models is of course completely true, but it makes no sense to state that only one party fails to understand one of them. Also, I see nobody here claiming the unconditional solution to be wrong. It's just not very efficient because it uses conditions which are not dependent. I'll start a section on the dependency. Heptalogos (talk) 20:33, 11 January 2010 (UTC)

Martin, in your last section you just formulate that the conditional probability to be calculated, because of the symmetry, equals the unconditional. But it is the conditional one the decision is based on, as we mentioned over and over!Nijdam (talk) 18:13, 11 January 2010 (UTC)

Even if door numbers are specified, I am not obliged to set up my sample space using them, as they are not significant. Suppose I set up a sample space with just two events: player chooses a goat(2/3), player chooses the car(1/3). So, I start with of the options one of the events occurs, the host opens a door, which I have shown is independent of the original choice, I now have to calculate the probability of winning by swapping but I know for sure that I will get the opposite of my original choice, which I still know is 1/3 chance of getting the car. Martin Hogbin (talk) 19:06, 11 January 2010 (UTC)

The important point to add to this, which was raised by Morgan, is that if the host might choose non-randomly the above method fails, and conditional probability using a sample set based on door numbers must be used. Martin Hogbin (talk) 10:25, 12 January 2010 (UTC)

If door numbers are specified, you are obliged to formulate a sample space - relating to the discussed problem - taking account of the door numbers. Nijdam (talk) 13:15, 12 January 2010 (UTC)

I disagree, it might be wise to use door numbers, it may have great instructional value to treat the problem conditionally using door numbers, but there is never an obligation to treat any mathematical problem is a specific way. For all I know there may be a geometric solution to the problem, but I cannot declare it invalid just because it is not my chosen solution. There is also no obligation to use redundant information in any probability problem.

Please tell me the flaw in my solution.

I set up a sample set containing two events and assign probabilities as shown. The player initially chooses a goat(2/3), the player initially chooses a car(1/3).

>>Okay: events G(oat) and its complement C(ar). Now describe to me in your probability space: "the host opens door 3 and reveals a goat" or "the host opens a door and reveals a goat".Nijdam (talk) 14:46, 12 January 2010 (UTC)

These events are not in my probability space any more that the event that the host sneezes is. Non of these events affects the probability that the car is behind the door originally chosen. They can be ignored in the probability calculation. They cannot be ignored in deducing that the player always get the opposite of her original choice if she swaps, but that has nothing to do with the probability calculation. Martin Hogbin (talk) 19:03, 12 January 2010 (UTC)

This set meets the requirements for a sample set, namely: '...the probability function f(x) lies between zero and one for every value of x in the sample space Ω, and the sum of f(x) over all values x in the sample space Ω is equal to 1', taken directly from the probability theory article. There are also no possibilities omitted from the set.

Now the host opens a door to reveal a goat, choosing randomly when he has a legal choice. At this point we are asked to calculate the probability of winning the car by switching doors. We note that the events in our sample set are independent of the event that the host has opened a door to reveal a goat on the following basis:

1) This event happens after the player's initial choice of door it cannot possibly affect the player's initial choice of door.

2) It is always possible for the host to reveal a goat, so the revealing of a goat cannot reveal any information that might change the probability that the player has originally chosen the car.

3) As the door number opened has been chosen randomly, the number of the door opened cannot reveal any information that might change the probability that the player has originally chosen the car. (Not that this condition fails if the host might choose non-randomly)

We now simply have to observe that if the player switches she gets the opposite of her original choice with probability 1.

What is the problem? Martin Hogbin (talk) 14:23, 12 January 2010 (UTC)

If you must have door numbers then change, 'The player initially chooses a goat(2/3)' to 'Door 1 has a goat(2/3)', and 'the host opens a door to reveal a goat' to 'the host opens door 3 to reveal a goat'. Martin Hogbin (talk) 14:36, 12 January 2010 (UTC)

Nijdam, you are still arguing that even if the host chooses a legal door evenly, the simple solution is wrong. I have just given a valid solution above. You might not like it and maybe you would not choose to do things that way but that is not reason enough to say that it is actually wrong. Martin Hogbin (talk) 17:09, 14 January 2010 (UTC)

Door numbering matters

Consider the following game. Roll a (fair) dice, call the outcome X. Put as many white balls as the outcome in an urn and complete with black balls till 6 balls all together. Then before you draw a ball from the urn, predict its colour. Then draw a ball; if your prediction was right you win a car, otherwise a goat. What will be your prediction? Nijdam (talk) 17:04, 11 January 2010 (UTC) [Copied from the main talk page]

If I am not aware of the dice throw then white. Otherwise, I would, of course, predict the colour which had the most balls in the urn. Martin Hogbin (talk) 17:47, 11 January 2010 (UTC)

How can you compose the urn without being aware of the dice throw? Would you say the outcome X matters?Nijdam (talk) 18:06, 11 January 2010 (UTC)

Yes, it matters if I know what it is. Martin Hogbin (talk) 18:42, 11 January 2010 (UTC)

As Martin says, the unconditional probability says the choose white. Why? Because by your problem, the possible ball combinations (each with 1/6 probability of occuring) is: (w/b) 1/5, 2/4, 3/3, 4/2, 5/1, 6/0. Since the probabilities are equal, there is 1+2+3+4+5+6 (21) probability of picking a while ball vs. 5+4+3+2+1+0 (15) probaility of black (21/15) = 1.4:1 odds of while. I'm not sure if this was your intent, Nijdam; if your question was "add balls until there are SEVEN (not six), there would be equal odds of black or while being picked.

The bottom line is that this has NOTHING to do with the Monty Hall Problem. Your problem is analoguous to: ALWAYS being six doors, but the number of cars vs. goats are unknown. What I'm not sure of is what you mean when you say "door numbers matter".

Are you suggesting that the number of total doors affects the probability of winning by switching (it does - 2/3 with 3 doors, 3/4 with 4 doors, 99/100 with 100 doors)? Or are you suggesting that it matters which door number is picked by the contenstant or host? (eg: matters if I pick door 2 and host opens door 1? As stated in the FAQ re: conditional vs. unconditional, this matters, but the answer only changes in non-standard MHP questions. Or are you actually saying that the number of goats vs. cars matters, which is what your analogy matched? (it certainly does affect the odds of winning by switching). TheHYPO (talk) 21:28, 11 January 2010 (UTC)

I think Nijdam has not finished yet. Heptalogos (talk) 21:47, 11 January 2010 (UTC)

Title reads: Door numbers matter, not: Door number matters. Nijdam (talk) 13:07, 12 January 2010 (UTC)

Nijdam, I have given my answer to your question. Is there more? Martin Hogbin (talk) 13:50, 12 January 2010 (UTC)

@Martin: No, my comment is meant for TheHYPO. As for you: do you see that the player on stage knows which door she has chosen and which door has been opened? Nijdam (talk) 14:39, 12 January 2010 (UTC)

I cannot tell what the player on stage knows for sure but I expect that they will probably know the door opened by the host, they are very unlikely to know the host's door opening policy. The problem statement tells us the door number opened by the host, but not the host's policy. So, yes, on strictly the information given in the problem statement, the door opened by the host is known. However, this information is not necessary in order to calculate the probability of winning by switching if you make consistent assumptions about unknown probability distributions. Martin Hogbin (talk) 16:01, 12 January 2010 (UTC)

You almost become a fan of Morgan! The player has chosen door 1 and the host has opened door 3. Should she switch? As you mention, she may not know the host's policy. Hence she models it using a parameter q. The rest you know. Nijdam (talk) 16:14, 12 January 2010 (UTC)

This so-called analogy is meaningless. First off, knowing what door numbers are chosen at any point does not, per se, affect the answer. It is only knowing how door numbers affect the random choices that can influence the results. Nijdam, you apparently wanted to include that information in the probability you asked for. But to do that, you had to include how the information was used in your problem statement. No such information is included in the MHP brainteaser, which is why this analogy is meaningless. Now, had you asked "A random number of white balls, 0<=W<=N, are placed in an urn by a game-show host, based on the roll of a die (which you see). He then places enough black balls into the urn to make N total. What is the probability of drawing a white ball from this urn?", the answer is 1/2. Note, please, that I did not say you know (1) What N is, (2) What W is, or (2) How the host might use the die to pick W. This is the form of your question that is analoguous to the MHP. The probability is 1/2 because you don't know these things, even if you see the die roll.

Anyway: if you were the one who rolled the die and composed the urn, would you know the outcome of the die roll? And in the other form, with 7 balls in total (which I deliberately didn't choose) what would you predict if the die showed 1? Nijdam (talk) 23:34, 12 January 2010 (UTC)

It crossed my mind we may actually play the game (I mean the one with the 7 balls, and hence a probability 1/2 on a white ball). I pay you 10 pound, and roll a die. After the composition of the urn I predict the outcome of the draw af a ball. When right, you pay me 20 pounds. Seems fair from your point of view. Are you willing to play this game for let's say a 100 times?Nijdam (talk) 12:18, 14 January 2010 (UTC)

Since the MHP (which, by the way, describes a game that has never appeared on an American TV game show, so it is purley hypothetical) does not include any information about bias, no such information can be assumed when answering the question. A real host might have a bias, but a different host with the opposite bias is just as likely. Since it is hypothetical, we don't have a real host, and we must represent all possible hypothetical hosts. As a result, we must assume that all such choices are made uniformly, which represents the average of all such biases. The car is placed with 1/3 probability behind each door; and when the host has N possible doors he could open, he opens each with probability 1/N. And no published source addressing the MHP has suggested that these probabilities are any different. Some have allowed them to be defferent by deriving a parametric solution, but any actual application of that solution to the MHP (as opposed to specific hypothetical setting with stated biases) has always set those parameters to the uniform values. And to repeat: This doesn't mean we are assuming that we know what the probabilities are, or that they actually are 1/2. It means we are assuming we don't - in fact, can't - know, and use 1/2 to represent that lack of knowledge.

Anyway: if you were the player on stage, would you know which door you'd chosen and which one is opened? And what would your calculations be? Nijdam (talk) 23:34, 12 January 2010 (UTC)

(1) Door numbering can not matter, unless we are explicitly told how the numbers affect probabilities. (2) If you assume you can use unknown probabilities, then the MHP is unanswerable because the car might not be placed uniformly. The fact that Morgan ignored this invalidates their analysis - their answer is wrong by the assumptions they claim to have (and explicitly claim to not have) made. (3) But we know (see Seymann, and MvS' response to Morgan) that without a statement of bias, the host must be treated as an agent of chance. That means we assume he acts without bias. (4) We know that MvS did not intend the door numbers to be part of the problem. We know this, because it is the only interpretation difference between MvS and Morgan, and she said they did misinterpret. And finally, (5) We also know that (see K&W), the door numbers mentioned are not sematically part of the problem statement. They are examples, that can be used without loss of generality if and only if no bias is allowed. Which is good, because no bias is allowed. JeffJor (talk) 17:32, 12 January 2010 (UTC)

If Morgan take it that the host legal door choice might not be random they must also take it that the initial car placement and the players initial door choice might not be random. This makes a solution impossible. The only logical and consistent assumptions are to do as Jeff says and take all the undefined distributions to be uniform. Martin Hogbin (talk) 18:49, 12 January 2010 (UTC)

He's just stringing you guys along. It is my understanding that Boris left this discussion only after he had demonstrated to Nijdam's satisfaction that the simple solutions are not inferior to the conditional solutions in any way. Do I have that right? Glkanter (talk) 17:41, 12 January 2010 (UTC)

Independence

In probability theory, to say that two events are independent intuitively means that the occurrence of one event makes it neither more nor less probable that the other occurs.

Event A = contestant wins
Event B = host opens door number 3

other data:

• three doors with fixed numbers (1,2,3)
• car and goats placed randomly, each behind one door
• contestant picks door number 1
• host always opens another door
• host always reveals a goat
• host always chooses randomly between goats
• host always offers to switch
• contestant switches after the host opens a door

Statement HL1: event B makes it neither more nor less probable that event A occurs. Heptalogos (talk) 20:33, 11 January 2010 (UTC)

In fact, as the door opened by the host is random, the number of the door opened is automatically independent of all other events. Martin Hogbin (talk) 20:44, 11 January 2010 (UTC)

Small chance that he opened the door randomly. Heptalogos (talk) 20:52, 11 January 2010 (UTC)

I do not understand what you mean. You say above, 'host always chooses randomly between goats'. Would you prefer me to say that the goat chosen by the host is random and thus independent of any other event. Martin Hogbin (talk) 22:33, 11 January 2010 (UTC)

I mean that the host most of the times has no other choice than to reveal the goat, which is no random choice. So only between goats he chooses randomly. But I now see that you may mean that overall, through repeated experiment, doors 2 or 3 are equally distributed in opening? Which you call random? Heptalogos (talk) 20:03, 12 January 2010 (UTC)

Yes, I usually say the host opens a legal goat door randomly with legal being a short way of saying that the host always opens an unchosen door to reveal a goat. Thus of the doors available to the host to choose from, according to the rules, he chooses one randomly. Sometimes he only has one to choose from. I should have said ' the host always chooses randomly between goats, when he has a choice'.

Also, for completeness, I might have mentioned the case where the host has no choice of door to open. In this case the door that he must open is decided by the random initial placement of the car, but nobody ever asserts that the door he opens could be significant in this case, as far as I know.Martin Hogbin (talk) 14:58, 13 January 2010 (UTC)

That's right. Whoever is responsible for the initial placement of the car would be equally likely to have a contrived bias as the host. Glkanter (talk) 16:16, 13 January 2010 (UTC)

Perhaps you could also confirm that the revealing of random information (note that is not the same as revealing of information randomly) cannot affect the probability of an earlier event. Martin Hogbin (talk) 22:44, 11 January 2010 (UTC)

Is "random information" a term generally used? If not, it's tricky without definition. I think I know what you mean by it, but why not call it independence? It's also not a valid general statement about "random information", since the host may randomly open a door with a car, which for sure affects the probability of the earlier event. Heptalogos (talk) 20:11, 12 January 2010 (UTC)

Yes, I am sure that there is better terminology that I should have used. The point that I am getting at is that, if the host opens a door to reveal a goat randomly, when he has the choice, revealing of the door number opened by the host cannot affect the probability that the player has initially chosen the car. Martin Hogbin (talk) 20:08, 13 January 2010 (UTC)

What do you mean precisely with: ...revealing of the door number opened by the host cannot affect the probability that the player has initially chosen the car? Because here lies the only point you seem not to understand. Nijdam (talk) 12:24, 14 January 2010 (UTC)

Independence doesn't play a big role in the MHP, except the independence of the placement of the car and the first choice of the player of course. Nijdam (talk) 13:10, 12 January 2010 (UTC)

And as explained above. Martin Hogbin (talk) 14:49, 12 January 2010 (UTC)

If A and B are independent, P(A|B) = P(A). I know your opinion is that, even when P(A|B) = P(A), the problem still needs to be solved as P(A|B). Maybe because they may coincidentally be the same in this case, but not consequently, or proven. But what if we can prove that P(A|B) = P(A) in the given case, even without calculating P(A|B)? I think I also know your answer to that, because you wrote that "one is obliged to take account of whatever is specified", which is probably a holy dogma of statistics. On that matter your opinion differs from Rick's, who maybe has/uses another discipline. The problem here may be that different persons use different methods, some of them implicitly judging that their discipline is the only true one. Heptalogos (talk) 20:28, 12 January 2010 (UTC)

Why would you prove that P(A|B)=P(A) without calculating P(A|B)? Indeed do different persons use different methods, even false methods.Nijdam (talk) 23:39, 12 January 2010 (UTC)

Science

Science at it's best is proven formula, nothing to argue about. Sense or logic is of no use, other than changing the formula or rules. Which makes sense. So within the wide scope of science, there may be consensus about the discipline used to solve a specific problem. In our case, this may be statistics. In that case, statisticians totally rule. Even if others argument reasonably that some rules of statistics may be changed to become more efficient, this is another discussion than solving the actual problem scientifically.

In that case I don't understand why Nijdam participates in the discussion at the point where we're stuck in the scientific dogma. Because there's nothing to explain really. It's simply: "one is obliged to take account of whatever is specified", because that's the rule. Which is a good rule in science! It's scientific, yes!

So, what if Einstein explained his theories, and most of us would really understand it. We would fully, logically understand, and be able to share our thoughts, write them down, over and over, backwards, diagonal, and it would all fit. Then the practicing scientists of that time would not agree, because it's no proven law (yet). End of discussion. Heptalogos (talk) 21:00, 12 January 2010 (UTC)

What is so very nice about the history of this problem is that science and logic are really in debate: best proven system (science, Morgan) against best proven process (intelligence, Vos Savant). So which one is leading? Intelligence is constantly improving (and therefore disproving) science, but science is constantly disproving (claims of) intelligence. Which one is mostly leading in which field? Then there is no science without intelligence. Science as nothing but proven formula is not real. Even mathematicicans use their intuition to decide what formula to use on what problem, and no rules for that even exist in many cases.

Instead of "popular solution" and "probabilistic solution" I think it's best to use "logical solution" and "scientific solution". Heptalogos (talk) 21:57, 14 January 2010 (UTC)

This is such a great story, because Vos Savant may not be the most intelligent person existing, but scientifically she has the highest proven intelligence! Then we have some scientists who actually agree with her logic. They may be very intelligent to share this logic, but they confused logic and science. Then we have some scientists who disagree on the method. They may be very intelligent to see that the logical arguments are not scientific, but they confused science with truth. They should not have criticized the logic of Vos Savant, but only the scientific proof. Which is about what Seymann was trying to explain. Heptalogos (talk) 09:19, 15 January 2010 (UTC)

Thoughts from user:Raccoon.zip

After spending hours of sorting out where my confusion was, I'd like to add the following which is not obvious on the wiki page. Granted, after the host revealed a goat door, at that moment, the probability whether the player's initial pick is a winning choice becomes 1/2.

No, that is not true. After a goat has been revealed the players chance of holding the car remains at 1/3. It would change to 1/2 if the host had opened a door at random and had just happened to reveal a goat by chance, but, under the usual rules, the host knows where the car is and always picks a goat, knowing where one is.

Racoon.zip (talk) 22:27, 14 January 2010 (UTC) I think this is a little misguided answer. Whether it makes a difference that the host had the inside knowledge, would depend on the question we want to answer. His inside knowlege would have had no impact on the outcoming, if we first agree that the question at hand is what the palyer's chance is *NOW* in this *particualr*, *single* run of game (vs. in an arbitrary game taken from the "statistically sufficiently large set" of game outcoming data space). Then it is 1/2 after the host reveals a goat gate in *this* particular run of the game. On the other hand, if we are talking about the player's chance in a statistical sense over the "statistically sufficiently large set" of game data space, then his *overall* (or statistic) chance is 1/3 if the host happened to have opened a goat gate with no inside knowledge. Note that in this latter case, the host could have picked a car door as well, in which case it obviously makes no sense to ask the player whether he wants to make a switch, and the player obviously ends up with a goat in 2/3 of the cases statistically.

No, the chance that the player has the car is 1/3 plain and simple. I am not sure of the distinction that you are trying to make here by talking about this run if the game. I presume you agree that the players chance of holding the car before the host opens a door is 1/3 for this run of the game?

After the host opens a door (I will assume that he opens it evenly when he has a choice of two goats, even though you may think this does not matter) the probability for this game is still 1/3. Why do you say it is 1/2? Martin Hogbin (talk) 23:35, 14 January 2010 (UTC)

There are two different cases I was referring to, and I did not explain well. The case where we differ is as follows. In the original game, let's assume the host has no inside knowledge, and we start the game for just one run. At step 2 the host happens to opena goat gate. Now, the question I'd like to pose is: what is the chance of the player having the prized car in such instances of games that satisfies this sequence of events? Note that I have changed the game outcoming space by eliminating 1/3 of space where the host happened to have picked up the car door because he had no prior knowledge. Surely, the player's winning chance was 1/3. But wait, what's the chance of the unchosen door having a car behind? It's also 1/3. So the player's chance vs. the unchosen door is 1/3:1/3, which means 50:50 chance for getting the car if the player chooses to make a switch. Or look at it another way, since in my newly defined game spaces I have removed 1/3 of the original full outcoming space of the game where the host happened to have picked up the car door because he had no prior kowledge, the player's chance to win is 1/2 as long as he sees the host open a goat door. Racoon.zip (talk) 01:34, 15 January 2010 (UTC)

And, whether the host opens a door with "inside knowledge" or not affects the probability. This point is very hard for many people to wrap their head around. If the host opens a door randomly and fortuitously reveals a goat, the chances are 50/50. If the host opens a door knowing what's behind the doors (picking randomly if he has a choice of two goats) the chances are 1/3:2/3. There's an argument that from the player's perspective if she doesn't know the host "forgot" where the car was and the host fortuitously (but randomly) revealed a goat the player's chances are 1/3:2/3, or conversely if the player thinks the host is opening a door randomly but wasn't, then from the player's perspective the chances are 50/50 - but this only comes up if we're trying to evaluate the situation based solely on what the player knows as opposed to what we know as the problem solvers. -- Rick Block (talk) 00:03, 15 January 2010 (UTC)

Your last case is somewhat similar to my variation of the game above, at least in how the 50/50 number would come about. Racoon.zip (talk) 01:34, 15 January 2010 (UTC)

Raccoon, the answer to the question in your first sentence above is 1/2. Te make clear what we are talking about let me restate the facts. The car is placed at random, the player picks a door at random, the host now opens either of the two remaining doors at random, which happens to reveal a goat. Given the above information, what is the probability that the the car is behind the door originally chosen by the player? Answer 1/2.

Now let me change the facts to represent the Monty Hall problem. The car is placed at random, the player picks a door at random, the host now opens either of the two remaining doors that he knows conceals a goat, if there are two such doors he chooses at random between them. Given the above information, what is the probability that the the car is behind the door originally chosen by the player? Answer 1/3.

As Rick says, whether or not the host is known to have inside knowledge of where the goat is does affect the probability that the car is behind the door that the player has initially chosen. Martin Hogbin (talk) 10:01, 15 January 2010 (UTC)

Maritin, I agree with both your answers. However, even your two cases can be read in different ways (and raise questions about your answer). As discussed in this thread, the answer can be 1/3 for your first case (if you add clearly that the host has no inside knowledge, which I believe is implied as it's the only difference between the two csaes you stated).

I think all the troubles with MHP is that there are multiple ways to misread it, some are more incorrect than others. I think it would be very helpful to list them in the main wiki page about common confusions and misinterpretations. For example, we may misread the MHP as to derive

(1) the player's winning chance in the sub-space of the game where the host has no inside knowledge but only happened to have picked the goat door (thus 1/3 of the space where the original pick was a goat gate needs to be excluded):- it's 1/2*0 + 1/2*1 = 1/2 (after removing 1/3 of game space, each of the original 1/3 becomes 1/2).

(2) the player's winning chance by assuming that the game is equivalent to that we re-started it after the host's pick by first asking the player to place his initial pick back into the unchosen door pool (thus 1/2:1/2 = 50/50)

(3) the ratio of winning probability of player's original pick vs. the unchosen door (1/3:1/3=50/50 if the host has no inside knowledge, or 1/3:2/3=1/2=50% if the host has inside knowledge)

(4) the player winning chance by double counting goat-1 and goat-2 end results as of the same weight as car door end result if the player makes a switch (thus 2-loses:2-wins = 50/50 chance) Racoon.zip (talk) 06:24, 17 January 2010 (UTC)

However, what his winning possibility becomes now is not the problem we are trying to solve. What we want to solve is what is the winning chance if he makes a switch. If his initial pick was a goat (remember, he had a chance of 2/3 to have picked the goat), he'd win by switching as we are certain the unchosen door has a car behind. This gives us the answer of 2/3 because if his initial pick was a car, he would lose 100% of the time by switching.

That is correct.

The answer is numerically correct, but not the argument. Before the opening of the door by the host the chance of hiding a car is 1/3 for each door. Afterwards however this has changed. It's obvious for the opened door: chance 0. It has to be, and may be, proven that for the originally picked door also afterwards the chance is 1/3 to hide the car. From this one may conclude the 2/3 chance for the remaining door. (Not this has also changed.)Nijdam (talk) 16:43, 14 January 2010 (UTC)

You have yet to show that if the host chooses evenly, the simple argument is wrong. I mention the non-symmetrical case below and refer to them the appropriate section of the article. Martin Hogbin (talk) 17:04, 14 January 2010 (UTC)

You have to understand the difference between before opening the door and after. I explained it a thousand times, and even somewhwere above I left you with the question how you would explain the difference. You didn't react. But anyway, here we go again: before opening: probability to hide the car: 1/3, 1/3, 1/3 for the 3 doors. And now for something completely different: after opening: in order of chosen, remaining and opened door: x,1-x, 0. That's all we know for the moment. x will turn out to be 1/3, showing the chances: 1/3, 2/3, 0. They must be different probabilities, as you will understand, because they differ from the original uniform 1/3. And they belong together!! They belong to the situation after opening. Symmetry may be helpfull in calculating x, but it doesn't help in avoiding the notion of x, which is in technical terms: a conditional probability. I find it extremely important that students who read the article understand this. And happily, or better, naturally, there are sources to rely on. A final remark about the simple solution. It does not address the above problem, but a much simpler version: the player knows what is going to happen, and before she even made her first choice she is forced to make a decision about switching later on. I'm actually convinced this is not the Parade version, and certainly not K&W and others. Unfortunately MvS gave the simple solution, and as I understood it, when attacked about it, tried to find a way out and invented the "unconditional" version.Nijdam (talk) 00:19, 15 January 2010 (UTC)

This is not the place for this discussion. I have referen the OP to Morgan. Martin Hogbin (talk) 11:56, 15 January 2010 (UTC)

Racoon.zip (talk) 22:27, 14 January 2010 (UTC) I agree that whether the host picks the two goat doors evenly affects the formula, but the end result is the same. It suffices for the original problem to assume the host does his pick evenly.

I happen to agree with your last statement above but there are plenty here who do not and there has been a long running argument about this. I suggest that you leave this subject until we have finished the bit about the chances of the player holding the car after the host has opened a door. Martin Hogbin (talk) 23:35, 14 January 2010 (UTC)

One may argue (as I had), that since we have two possible outcomings of goat 1, and goat 2, depending on what the host picked, for the case where the initial pick was car and he switched, we should count the two different outcomings as "two losing games" when compared to the two winning outcomings for the case where the initial pick was a goat and the player switched. The fallacy here is that the goat 1 and goat 2 cases together should only be counted in terms of the weight of the "car pick" chance (1/3). While they are indeed distinct outcomings, they are confined to the event space of "initial pick is a car", which occupies 1/3 of the outcoming space.

Yes, this is shown in the second large diagram.

One may argue (as I had), that for the case where the initial pick was car and a switch is made afterwords, since we have two possible outcomings of ending up with goat 1, or goat 2, depending on which goat the host had picked, we should count the two different outcomings as "two losing games" when doing tally, as done with the two winning outcomings for the case where the initial pick was a goat and the player switched. This would then yield a tally of (2-losses + 2 wins) which means 50% of winning if the player switched. The fallacy here is that the tally of cases of ending with goat 1 and goat 2 should only be done in terms of the weight of the "car pick" chance (1/3). While they are indeed distinct outcomings, they are confined to the event space of "initial pick is a car", which occupies 1/3 of the outcoming space, with the other 2/3 being the case of initial pick of a goat. In probability terms, the losing probability with initial pick of car and making a switch, is (1/3 for initial pick of car) x ((1/2 for goat 1 picked by host) x (100% for switching car with the remaining goat 2) + (1/2 for goat 2 picked by host) x (100% for switching car with goat 1 that is left)) = 1/3 x (1/2 x 100% + 1/2 x 100%) = 1/3 x (1/2 + 1/2) = 1/3. Since the player is guaranteed to win in the case of initial pick being a goat and making a switch after, i.e., his losing chance is 0 there. Combining the two, the player's total losing probability with a switching is 1/3 + 0 = 1/3, meaning a winning probability 2/3.

Yes. Note that if the host does not choose evenly between the goats the above calculation becomes more complicated. This is what is discussed later on in the 'Probabilistic solution' section. Martin Hogbin (talk) 16:17, 14 January 2010 (UTC)

[END of Racoon.zip (talk) 00:05, 14 January 2010 (UTC)]

Summary

I'm aware that much of what I summarize here is a repetition of what is said earlier. But it may be of help in organizing our thoughts in the mediation process. I also have to confess that from a private discussion with Gill I learned that even some experts consider a rather simplified version as the MHP. And much of the confusion and discussion stems from the diffent views of what is considered the MHP. It mainly comes down to:

MHP

We are the audience and on stage we see three doors. Also on stage is the player and the host. The rules are explained and

This is already not the MHP, which begins, 'Suppose you're on a game show...' 'We' are not the audience. 'We' are the contestant. Glkanter (talk) 12:58, 15 January 2010 (UTC)

There is no "the MHP". Why would it be the Parade statement and nothing else? Heptalogos (talk) 20:20, 15 January 2010 (UTC)

A: before anything happens on the stage we (with no additional information) are asked to answer the question: "should she switch (in the end)".

B: the host asks the player whether she will switch in the end.

B': the player makes her initial choice and is then asked the same question.

C: the host opens the door with the goat and then offers the player to switch.

Remarks

Some consider A to be the MHP (to my surprise). I know of course it is a situation one may like to analyze, but I never would call it the MHP. The version A is the (fully) unconditional situation.

This is certainly a valid view. Seymann's comment at the end of the Morgan paper reflects a standard view of statisticians. In order to answer any statistics question we must consider the intent of the questioner. It was quite likely that Whitaker actually wanted to know the answer to this question. This is addressed in detail on my Morgan criticism page. Martin Hogbin (talk) 11:12, 15 January 2010 (UTC)

Of course it is a valid view, whatever that is supposed to mean. My point is, most people have a different view of what the MHP is. Nijdam (talk) 17:14, 15 January 2010 (UTC)

Some may consider B the MHP. It is equivalent to B', as the player will know which door she will choose. It is what Boris (I'm too lazy to look back) called "semi-conditional", as it formally needs conditioning on the choice. As the choice and the placement of the car are to be considered independent, this may in a certain way also be treated unconditional.

My opinion is that C is to be considered the MHP. We see on stage the player pointig to a door and an opened door with a goat. There is now the choice between two still closed doors. This picture causes the confusion, and generates the 50/50 idea. C needs to be solved with conditional probabilities. It doesn't allow the simple solution, nor the combined doors solution, nor the many doors solution. I consider the Parade version as well as K&W as from the form C.

You continue to say, without proof, that C needs to be solved with conditional probabilities, but I have shown a proof on the arguments page starting with a sample set of two events, noting that both these events are independent of the opening of a legal goat door randomly by the host, and proceeding to a valid solution.

I just know. Nijdam (talk) 17:14, 15 January 2010 (UTC)

Part of the controversy stems from the different opinions on what to consider the MHP. Another confusion comes from people who think of C as the MHP, but solve it as being A.

Maybe some people are confused about this but I am not. View A is certainly a valid one. Martin Hogbin (talk) 11:25, 15 January 2010 (UTC)

I didn't suggest you are confused, hope you aren't. What is the content of your remark?Nijdam (talk) 17:14, 15 January 2010 (UTC)

As the only person in these discussions who read vos Savant's columns in real time, as they were published, I can assure everyone that the only question the readers had was, 'Why isn't it 50/50?'. Nobody gave two hoots about which door. Per my recollections, the door #s never came up in her columns, other than as examples. That's not what 10,000 people wrote letters about, or why 1,000 PhDs told her she was wrong. Then changed their minds. Glkanter (talk) 13:14, 15 January 2010 (UTC)

I notice Nijdam's examples steer clear of any 'host bias'. So, using symmetry, wouldn't A - C all be equivalent? I thought Boris proved this to Nijdam's satisfaction. I demonstrated with 'Huckleberry' that Monty may open either door in each playing of the game, and Huckleberry will win or lose based solely on his original choice, regardless of 'when' he makes his decision.

Nijdam keeps saying that the original 1/3 is not the same as the ending 1/3. But it is. It never stopped being 1/3, then became some other value, then returned to 1/3. If so, when does it change, what does it change to, and when does it change back? And why?

And without a contrived 'host bias', of what value is Morgan's paper? Selvin already provided a conditional solution 16 years earlier. Glkanter (talk) 13:35, 15 January 2010 (UTC)

The symmetrical problem can be solved without conditional probability. I have done it and no one, including Nijdam can state what is wrong with my solution. Martin Hogbin (talk) 15:06, 15 January 2010 (UTC)

The point here is again: do you mean A, B, C or even something else?Nijdam (talk) 17:14, 15 January 2010 (UTC)

Enlightening example

Now why do I think C is the MHP? That's why I gave above the following example: You are to roll a die. Put as many white balls as the outcome in an urn and complete with black balls until a total of six balls. You are to draw a ball from the urn, but before you do this you may predict the outcome. Of course you win a car if your prediction is right, otherwise you get a goat. How will you solve this problem? Do you say: the overall probability of a white ball being drawn is 21/36, hence I predict always white? Or do you say: if the die shows 1 or 2, I predict black. If it shows 3 it is indifferent. Does the die show 4, 5 or 6 then I predict white. I would know what to do!

I understand your troubles about the page to post this issue on. My reaction would first of all simply be: is it OK for us, editors, to present the MHP in a way that we find consensus on, or should we use resources? Problem with the consensus is the small timespace we're in, practically. Formally this may be OR. Heptalogos (talk) 11:00, 15 January 2010 (UTC)

Nijdam, your example may correctly represent what you understand the MHP to be but this does not explain why you believe that it must be option C. I agree that it may be option C, and many sources, including Morgan, clearly take this to be the case, but that may not be what the actual questioner wanted to know. You keep ignoring the fact that the question was not in a statistics exam it was a letter to a popular magazine. It is highly unlikely that the writer had considered events in such detail. There are other sources that, indisputably, give a correct answer to question A. We might assume that this is what they take the question to be. We have no right to assume that all those who give the correct answer to question A actually intended to answer question C but got it wrong.

All you are doing here is essentially restating your opinion. Martin Hogbin (talk) 11:38, 15 January 2010 (UTC)

This example is of course not enlightening at all, because your winning chance here is in fact dependent on the information in the number, which is not the case in the number of the door in MHP. Heptalogos (talk) 11:24, 15 January 2010 (UTC)

This example is completely invalid. In the example we are presumed to know how many ball are in each urn, in all versions of the MHP we are not told the hosts door policy, thus we cannot use that information. Martin Hogbin (talk) 11:41, 15 January 2010 (UTC)

I only ask you both: what would be your solution?Nijdam (talk) 17:18, 15 January 2010 (UTC)

(rearranged)Nijdam (talk) 14:46, 15 January 2010 (UTC)

If I knew the throw of the die, I would do as you do and choose according to whether there were more white or black balls in the urn. How about you answer my questions below. Martin Hogbin (talk) 17:24, 15 January 2010 (UTC)

Why not do the same in the MHP??Nijdam (talk) 17:39, 15 January 2010 (UTC)

Because, in the MHP, we do not know the hosts door preference. Martin Hogbin (talk) 19:34, 15 January 2010 (UTC)

Because, in the MHP, we assume random host behavior. Heptalogos (talk) 20:28, 15 January 2010 (UTC)

Quite contradictory. isn't it? Of course I take C to be the MHP. That is comparable to my example. Just like what you propose in my example, the player will use all the info she can get. It doesn't help her to know that on the average there is a 2/3 chance to win the car when switching. She wants to know what the chance is in her situation. With the random host behavior also in her situation (conditionally) the chance is 2/3. And now your reaction, Martin. When you don't know the hosts preference, you have to model it. Read Morgan. Nijdam (talk) 23:06, 15 January 2010 (UTC)

I have read Morgan thoroughly as you well know, I was the first to find an actual error in it, which no one has noticed in nearly 20 years. How is that going by the way?. I do not see why you do not understand my argument about unspecified distributions. There are three unspecified distributions in the MHP. The original car placement, the player choice (which is not that important here) and the host door choice. There are only two consistent ways to treat these unspecified distributions.

1) Model them all, as you call it. So we start with a parameter C1 representing the probability that the car is behind door 1, and then C2 plus parameters representing the host choice. In this case the problem is insoluble.

2)Take them all as uniform.

There are no other consistent ways to deal with the problem. There is no logical basis on which to assume one unspecified distribution is uniform but the other might not be and should therefore be parametrically modeled. Martin Hogbin (talk) 23:24, 15 January 2010 (UTC)

"With the random host behavior also in her situation (conditionally) the chance is 2/3." So you agree on that, but you don't agree on the assumption of random host behavior, is that it? Heptalogos (talk) 11:25, 16 January 2010 (UTC)

Whom is this comment addressed to, Heptalogos? I do not understand it. Martin Hogbin (talk) 11:38, 16 January 2010 (UTC)

To Nijdam, as can be seen by the 'tab' (an extra :), the same as you did. Heptalogos (talk) 11:40, 16 January 2010 (UTC)

Thanks for making that clear, Heptalogos, not everybody is as consistent with their indenting as you. You seem to be the only on that understands one of the principal deficiencies of Morgan's paper.

If we treat Whitaker's question (somewhat perversely in my opinion) as a formal probability question then it cannot possibly be answered without deciding how to treat the information missing from the question. Most people, take it that the host always offers the swap and the he never reveals the car (the statement does give is a strong clue here, that the host knows where the goats are). Even when these standard rules are agreed we still have missing information.

1. What is the distribution for the producer's initial car and goat placement?
2. What is the distribution for the player's initial choice of door?
3. What is the distribution for the host's choice of goat door when the player has originally chosen the car?

The problem statement tells us nothing about any of these these distributions. Strictly speaking, we should say that we cannot answer the problem, after all, for all we know they may all be non-uniform, the car might always be behind door 2, the player may always choose door 1, and the host may always open door 3.

The only way to get an answer is to apply the principle of indifference to all the above distributions on the basis that we have no information to prefer any one possibility over any other. This means we must take the host legal door choice as uniform.

Does anyone disagree with this? Martin Hogbin (talk) 12:41, 16 January 2010 (UTC)

I see these valid options:

1. Not make any assumption and solve it conditionally the way Morgan tried. (but they failed)

I have just shown above that this is completely impossible. If you make no assumptions and only use the information given in the question then the answer (probability of winning by switching) is from 0 to 1. Not a very useful result. Is that what you mean by (but they failed)?

They failed because they made assumptions. I meant the options as general options, but even in the MHP I don't think you proved that it's impossible to solve without making assumptions. I agree that it seems unlikely, but this may be another sort of discussion as about proving independence: it may seem so logical, but where's the solid proof? Consider that Morgan came up with a very creative way to get around some problems that you probably also couldn't imagine. For practical reasons however we may skip this option since nobody succeeded in this. Heptalogos (talk) 14:40, 16 January 2010 (UTC)

It is obviously impossible to solve the problem with no additional assumptions. The two cases below are both are consistent the problem statement:

The car is always be behind door 2, the player always chooses door 1, and the host always opens door 3. Chance of winning by switching 1.

The car is always be behind door 1, the player always chooses door 1, and the host always opens door 3. Chance of winning by switching 0.

Martin Hogbin (talk) 14:50, 16 January 2010 (UTC)

2. Make assumption, by these rules, in order:

a. Make reasonable assumptions. (reality approach)

b. Treat anything undefined as random. (mathematical approach)

c. Do a, but do b if there's argue about what's reasonable.

If there's no consensus on (within) 2, it cannot be solved. Heptalogos (talk) 13:37, 16 January 2010 (UTC)

Fine, we could base our answer on what we might expect in a realistic situation for a hypothetical game show. My assumptions would be:

1 The distribution of the car's initial placement car is approximately uniform. A stage hand is probably told to to place the car randomly but I doubt that they would use any true randomising process, like throwing dice.

2 The distribution of player initial door choice is probably approximately uniform. There is no reason for the player to prefer any particular door but people do not actually choose uniformly is such circumstances

3 The distribution of host's legal door choice is approximately uniform. The host has no reason to prefer any particular door when he has a choice and probably would just choose on the spur of the moment. The choice would be unlikely to be completely uniform.

So in the realistic case we cannot actually solve the problem at all. An approximate solution is 2/3.

We have only one choice. Treat everything undefined as random. Nothing else produces a solution. Martin Hogbin (talk) 14:21, 16 January 2010 (UTC)

OK, now you're being very, very realistic. One could question anyway how realistic a calculated probability is. For a single event, it doesn't make sense actually. We can only map the phenomenon to a logical system that we find consensus on as being the most reasonable system to use. This is a system which includes repeated experiment, within which we can state that event X is happening with a frequency of y as part of a total frequency of Y. Which we then call the probabiliy of X to happen, but actually it's not the same.

This logical system may assume random behavior, which is then perfect random behavior, as the most reasonable logical mapping of what the realistic behavior is supposed to be. So we don't need to mention 'approximately random' or alike. Heptalogos (talk) 14:56, 16 January 2010 (UTC)

An alternative example

How would you answer this question? There are 6 white balls and 6 black balls. Six of these are randomly chosen and placed in an urn. You then pick a ball from the urn. What is the probability that it is white?

A better example

This is a much more realistic example, already given above. You did not answer it. There are two identical urns. One contains 10 white balls and 20 black balls, the other contains 20 white balls and 10 black balls. A person picks a ball from one of the urns. Using only the information given, what is the probability that it is a black ball? Martin Hogbin (talk) 14:36, 15 January 2010 (UTC)

In essence, the Morgan argument is that the host, Monty Hall, may be an imperfect human being.

Monty might have a bias. Hard to argue with that. Of course, the contestant (that's us), may be imperfect, too. But, there are producers and legal guys and network guys all around to make sure things are on the up-and-up. No matter, as they are all likely to be imperfect human beings, too.

Late edit by Glkanter: Hey, what about the guy who originally places the car behind a door? He might be imperfect, too. Morgan writes a paper about how the host might be an imperfect human being with a bias, and completely ignores the identical likelihood for the guy who originally places the car? How can they do that?

And so might the guy who places the pea under a shell be an imperfect human being.

Or the factory worker who makes the black and white balls for all these urns. Maybe the imperfect human being removing the balls from the urns can somehow feel the difference between the black and white balls made by this imperfect factory worker.

Do I even need to mention the guy who deals the cards from the deck might be an imperfect human being, or have a bias?

Maybe one of you qualified people could publish this idea somewhere. Then every puzzle could be fouled by this ridiculous line of thinking, as has the MHP. Glkanter (talk) 13:26, 16 January 2010 (UTC)

You have a point, saying that assumptions always need to be made. Otherwise we in fact assume crazyness, chaos, which is also an assumption. (Is this a paradox?) There is no elegant solution that needs no assumptions. There may be elegant solutions making only assumptions which are undisputed, and at the same time treat any undefined information in a consequent manner. These can at least exist in theory. Heptalogos (talk) 13:57, 16 January 2010 (UTC)

Comparative Likelihoods Of Bias: The Host v The Car Placer

Anybody want to tell me the difference?

And then defend, to a Wikipedia reader, how Morgan treats them differently?

That ain't the paradox. Glkanter (talk) 14:20, 16 January 2010 (UTC)

Please try to be economical with new sections. This one could be included in the section above. Also you might ask less and state more yourself. Your implicit statement was already that Morgan are not consequent in assumptions they make. This statement is not new; it is actually already presented as an error in the Morgan paper. Heptalogos (talk) 14:27, 16 January 2010 (UTC)

This is not the actual error in the paper that I mentioned. It is a serious and problematic inconsistency. Martin Hogbin (talk) 14:53, 16 January 2010 (UTC)

The errors can be found on this page in section "Issues about the Morgan paper". Heptalogos (talk) 15:04, 16 January 2010 (UTC)

An interesting result

My analysis page now confirms a fact that have always suspected which is this: Within the standard rules (which we all know), for the answer (chances of winning by switching) to depend on the host's door opening strategy and be anything other than exactly 2/3, all the following conditions must be true. Specific doors must be mentioned in the problem statement, the host choice of door must be non-uniform (in other words he must be assumed to choose non-randomly) but the initial placement of the car must be uniform (random).

It is this last condition that I bring to your attention. It has sometimes been claimed that, provided the player chooses randomly, the answer may still not be exactly 2/3. My (with Nijdam) complete analysis of the problem shows that this is not the case. As is often the case, this is obvious when thought about in the right way.

Suppose the initial car placement is non-uniform, let us take an extreme example, let us say the producer always puts the car behind door 1. If we ask what the answer is if a player chooses a door and the host opens another door to reveal a goat, the answer is exactly 2/3. If, on the other hand, we restrict ourselves to the case that specific doors are given in the question, say the player chooses door 1 and the host opens door 3, the answer clearly must depend on the probability that the car was initially placed behind door the door chosen by the player, in this example 1. In this case the player is certain to lose if she switches. If you are not convinced by this simple argument have a look here. [[1]]

The point I am making is this. We cannot use a random player choice to 'randomise' the initial car placement. For the answer to depend on the host door choice strategy the producer must initially place the car randomly but the host must choose (within the standard rules) non-randomly. This assumption cannot be justified on any logical grounds. Martin Hogbin (talk) 10:26, 3 January 2010 (UTC)

"...the host must choose (within the standard rules) non-randomly." AND convey this is some way to the contestant. But he can't. Because it's a problem about a game show. And game show hosts don't tell contestants where the car is. Glkanter (talk) 16:55, 3 January 2010 (UTC)

They don't? But keep in mind: In telling her the rule they tell her where a GOAT is. They tell her that "1/3+1/3" never is conclusive. Never. As written above: She cannot choose BOTH goats, and there is only one car. She made her choice, and she definitely separated the group of three doors in two groups: Into one single door and into a PAIR of two doors. So the refused PAIR of two doors never can contain two cars. This pair of two doors always is to contain ONE GOAT, at least: Just ONE unavoidable goat!

Exactly ONE unavoidable goat. No more and no less. No matter what door she has chosen. And no matter "which one" of the two refused doors contains a goat. Remember: Because "one goat" will be shown there anyway, within this PAIR of two doors, still before the questeion "which one?" even can be put! In any case: Exactly ONE unavoidable goat with a chance of zero. Only one! Since the time she made her first choice. Doesn't matter if (in 1/3) BOTH goats will be there. Remember: Exactly ONE unavoidable goat. And when "one goat" with a chance of zero is shown there: A second goat never is "mandatory", for only ONE is "given" there, with a chance of zero. She knows that, and she pays attention to this fact. Hope you will finally pay attention to this fact, too.

One mandatory goat with a chance of definitely zero. Exactly ONE mandatory goat. Maybe there will be even two goats. Maybe. But remember: There is only ONE mandatory goat. Exactly one. She knows this, the rule says so. If one goat is shown there, the other door can impossibly be a "mandatory" door, also. Because there's ONLY ONE mandatory goat. No matter which one of the two doors is concerned. No matter, for one goat will always be shown there. Regardless behind which door! Only one mandatory goat. The "other door" however always is free to contain whatever it contains. In 1/3 the second door will also contain a goat, but never a mandatory one. Please pay attention to this fact: Never a mandatory one. And guess what in 2/3 the second door will contain? The combined chance of the PAIR of two doors is 2/3. And only ONE of them is a mandatory goat with a chance of zero. Read the rule. So it is always given that the 2/3-chance of this PAIR is always and in any case solely consolidated by "the other" door, even if it contains (in 1/3 of events) the second goat. And she knows which one is the other door, "the privileged door", because that "one goat" has already been shown there.

The chance of the "other" door offered, of that "privileged door", always is 2/3 (not 1, but 2/3, FROM THE BEGINNING, when she made her choice). Hence: Those two doors never are equal "1/3+1/3", but the rule says: In each and every case, those two doors have to be "0+2/3", from the beginning, since she made her decision. She, at least, pays attention to the fact that the refused PAIR of two doors is to contain exactly ONE mandatory goat. The position of the "mandatory goat" never is relevant, it will be known when "one" goat is shown there. Because there's only ONE mandatory goat. If you ignore this fact (0+2/3) you unnecessarily will be operating with "conditional" chances, regrettably. If you still ignore the fact that she has definitely been told where a GOAT is. She knows it since she made her decision. You don't? If you ignore the fact of "one unavoidable goat" within this pair of two doors you unnecessarily will continue to claim "conditional chances" a necessity. And you are wrong if you argue that, from the time of her decision on - not "0+2/3" but "conclusively 1/3+1/3" - is all you know. Obsolete. Sorry for that.   Regards,   Gerhardvalentin (talk) 20:27, 20 January 2010 (UTC)

An unconditional solution to the symmetrical problem.

We start with two events: the player initially chooses a goat (G), the player initially chooses the car (C). These two events form the whole of our sample space space. The probabilities associated with these two events are 2/3 and 1/3 respectively and they are independent of all other events.

So the sequence of events is.

1) One of the two events in our probability space occurs with probabilities as shown.

2) The host opens a door to reveal a goat, when he has a choice he chooses randomly. We may be told that, in fact, he opens a door with a number 3 on it. The probabilities of the events in our sample space are independent of this event, thus we do not need to take any account of it or include it in our sample space, just as we do not include any other events of which the probabilities of C and G are independent.

3)The probability that the player still has a goat is 2/3.

4)The player now makes the decision to swap or not.

5)A player who has a goat will with certainty get the car if she swaps, and vice versa.

6)The probability that the player gets the car if she swaps is 2/3.

Martin Hogbin (talk) 15:28, 15 January 2010 (UTC)

I'm sorry Martin, I guess you are an expert on the speed of light, but not in probability theory. This sounds to me like the following will sound to you. Suppose you are on a train and you notice a source of light next to you. The train is increasing speed every time the driver blows his whistle. Connected to the wheels is a mirror, rotating with double frequency. I look into the mirror and notice that the speed of light reduces every time the mirror has made one turn. So it is obvious nothing can go faster than the speed of light. Nijdam (talk) 17:09, 15 January 2010 (UTC)

I make no claim to be an expert in probability theory but your reply might be more convincing of you told me what is wrong with my solution. Martin Hogbin (talk) 17:21, 15 January 2010 (UTC)

Okay, remove it from here and put it on User:Martin Hogbin/Monty Hall analysis. Nijdam (talk) 17:37, 15 January 2010 (UTC) ‎

You still have not told me what is wrong with my solution. Martin Hogbin (talk) 19:35, 15 January 2010 (UTC)

I more liked to discuss this on your analysis page, but okay here we go. You said: we start with two events G and C (=complement of G). What do you mean by "start with", are these the only events to consider, or will you introduce others later on? Nijdam (talk) 17:50, 18 January 2010 (UTC)

Martin, better not to talk about sample space and then exclude events that do happen from it because they're independent. That indeed doesn't make sense when using these mathematical terms that have a different meaning. However, you and others have been very clear in showing logically that there is not at all any relevance in the door number, not any influence on the probability of winning by switching. Nobody ever disproved that logic, and in 99% of the time nobody even tries. All they do say is that the number should be used as it is given as a condition. Furthermore they use examples in which it does indeed matter to use the given conditions, as if they don't see the difference. So, how should they see the difference using math rules? They can't, because those rules don't exist. You are trying to show something to a blind man, who is btw perfectly professionally blind because he shouldn't see all kinds of things which are 'apparently so logical' or something like that. Heptalogos (talk) 20:51, 15 January 2010 (UTC)

There are always events that we exclude from our sample space because they are independent. Trivial events like the host sneezes, for example.

I agree that it is good practice to include events that you are not sure about just in case it turns out that they are important, but I am not claiming that this is the best way to solve the problem, just that it is a valid way. Martin Hogbin (talk) 23:28, 15 January 2010 (UTC)

I don't think that any event mentioned and happening in the experiment can be ever excluded from the sample space. The host sneezing is not such. Heptalogos (talk) 11:04, 16 January 2010 (UTC)

Of course it can. We always have to use our judgment to decide which events to include in our sample space, based or our assessment of whether they might affect the probability of interest. For example in the MHP problem statement we have the event that the host says the word 'door'. It seems fairly obvious that just saying the word 'door' will not affect the probability of interest (this is not an entirely trivial example, words spoken by the host could easily affect this probability) so we leave this event out of our sample space. The sample space does not create itself, we create it using our judgment. Martin Hogbin (talk) 11:46, 16 January 2010 (UTC)

Let's rephrase: any effect mentioned, with a probability of happening, that influences the sample space should be specifically measured. In reliable sources, you won't find any exception. But this is indeed no explicit rule, apparently. It's very typical that I have to mention this, while Nijdam is replying with funny metaphors. I think he is not explicitly aware of this; it's rather something that grew in his system over time which is so familair and obvious to him that he cannot even externalize it. Heptalogos (talk) 12:32, 16 January 2010 (UTC)

I think we are in basic agreement? I would not necessarily recommend the above solution because it relies too much on intuition. One of the strengths of mathematics is that by systemising things we have to think less and therefore are less prone to errors. Some may think that a particular approach, which may be a good one, is the only approach. Martin Hogbin (talk) 12:57, 16 January 2010 (UTC)

If A and B are independent, P(A|B) = P(A). We already agreed on that. There is also no disagreement on specific independence in the MPH. Rick is even willing to accept unconditional solution when independence is proven or assumed. However, Rick and Nijdam are not willing to apply logic to prove the independence. They need some law, whether implicit or explicit. That's all I can make of it. Which is not wrong, but it would have helped if they told so earlier. Heptalogos (talk) 13:22, 16 January 2010 (UTC)

There is law. Hosts don't tell contestants where the car is. It's illegal in the US. Which is where Selvin's and vos Savant's and Morgan's papers were all published. Everybody in the US knows this, as we watch game shows, beginning at the age of 2, instead of reading books or playing outside. That the contestant gains no knowledge from which door Monty opens is a premise of the puzzle, as it starts, 'Suppose you're on a game show...' Glkanter (talk) 13:33, 16 January 2010 (UTC)

That's cool. Btw, the MHP is not about Monty Hall. Heptalogos (talk) 13:43, 16 January 2010 (UTC)

If A and B are independent, P(A|B) = P(A). We already agreed on that. There is also no disagreement on specific independence in the MPH. Rick is even willing to accept unconditional solution when independence is proven or assumed. However, Rick and Nijdam are not willing to apply logic to prove the independence. They need some law, whether implicit or explicit. Rick, is this correct? Is this the essence of argue? Heptalogos (talk) 23:18, 17 January 2010 (UTC)

Are we talking about the article here, or personal beliefs? For the article I need sources. I think what we have are 1) sources that explicitly say the unconditional solutions don't address the (conditional) problem and, 2) no sources using an unconditional solution explicitly saying it addresses the conditional problem. I've said this before, but personally I think the crux of the Monty Hall problem is that (no matter how it's exactly phrased) people are led into thinking about the conditional situation and that unconditional solutions require a different mental model that most people find difficult to switch to. Frankly, I don't think ANY unconditional solution (no matter how convincingly presented) will satisfy most people if they are initially presented with anything remotely like the "standard" version of the problem. As a matter of logic, if you're going to address the conditional situation with an "unconditional" solution you have to make some argument for why it applies which must implicitly or explicitly use the "equal goat" assumption. Again personally, I find a direct conditional solution a far simpler approach. -- Rick Block (talk) 03:28, 18 January 2010 (UTC)

The 'equal goat' assumption is a premise of the MHP. It begins, "Suppose you are on a game show..." This precludes you, the contestant, from knowing about any potential host bias. The only assumption you can make is that the host choice is uniform. That means symmetry. And with symmetry, we know that the simple solutions are equivalent to the conditional solutions. Glkanter (talk) 04:16, 18 January 2010 (UTC)

Rick, not talking about the article here, but trying to find the essence of argue. Your answer: "if you're going to address the conditional situation with an "unconditional" solution you have to make some argument for why it applies which must implicitly or explicitly use the "equal goat" assumption". I agree. We might simply explicitly assume the equal goat assumption. Then we are indeed stating that P(A|B) = P(A), when A and B are independent, is that true? I believe that we already agree, implying that the argument for the independence is possible, is that also true? Heptalogos (talk) 09:13, 18 January 2010 (UTC)

The equal goat assumption is exactly what makes A (car is behind door 1) and B (host opens door 3) independent, and A and B being independent implies the equal goat assumption, so if one of these is assumed then there are plenty of ways to show the other. Is that what you're asking? Maybe I'll start another section on this, but I think the mathematical crux of the MHP is that P(A) is obviously 1/3 (which requires initial random car placement) and we're asked to think about whether P(A|B) is the same or different. It seems to me if the problem is stated in such a fashion that we can mathematically model the answer as P(A) completely ignoring P(A|B) we've entirely missed the boat. I mean, the whole question is does P(A) = P(A|B), and if so why and if not why not? What I and I think Nijdam are objecting to about the "unconditional problem" is it turns the question into an obscurely worded way of asking whether P(A) = P(A). Well, duh. This makes it hardly paradoxical at all, but more akin to childish word problems like "a plane crashes exactly at the North Pole, where are the survivors buried?"

The player picks a door (say Door 1) and clearly has a 1/3 chance of having picked the car - P(A) is obviously 1/3.

The host now opens a door, say Door 3. Does this have any effect on the player's initial chance - is P(A|B) the same or different from P(A)?

versus

The player picks a door (say Door 1) and clearly has a 1/3 chance of having picked the car - P(A) is obviously 1/3.

The host now opens another door. Considering all possibilities (i.e. keeping the initial sample space the same as the final sample space) what is the chance the player has picked the car - what is P(A)?

-- Rick Block (talk) 15:58, 18 January 2010 (UTC)

Rick, you seem now to be saying that you think the answer to the unconditional problem is obvious. Well, I suppose it is. The question just boils down to, what is the probability that you initially chose a goat. But, with a little diversion, in the form of opening a door which makes no difference at all, most people get it wrong. That is the whole point of the MHP, it is really easy when looked at correctly, but for some reason most people do not do, and sometimes cannot do, this.

As to why P(A|B) = P(A) in the symmetrical case, I have given two reasons, based on sound established mathematical principles, why this must be so. Firstly symmetry, secondly no information is conveyed by a random event. Martin Hogbin (talk) 16:45, 18 January 2010 (UTC)

What I'm saying is that the "unconditional problem" boils down to a completely uninteresting mathematical tautology. You're saying people don't understand that the MHP is simply an obscurely worded way of asking whether P(A) is the same as P(A) and if you look at it "correctly" this is what you see, i.e. they have trouble arriving at the mathematical question that is asked. I'm saying people understand the problem (whether it's technically worded this way or not) as asking whether P(A) is the same as P(A|B) and the issue is not an incorrect mathematical representation of the problem statement but an intuitive failure in evaluating the conditional probability. Your response seems to be saying that you think the one and only "correct" view is to "understand" that the problem is actually asking whether P(A) is the same as P(A). I'm saying there's ANOTHER approach, which is to show them how P(A) is indeed the same as P(A|B). This is Morgan's and Nijdam's point when they say the problem is inherently conditional. In their view, the problem asks whether P(A) is the same as P(A|B), NOT whether P(A) is the same as P(A). I think we keep talking past each other because you keep ignoring or not understanding this point. -- Rick Block (talk) 20:56, 18 January 2010 (UTC)

P(A|B) = P(A) because indeed the car behind door 1 and the host opening door 3 are independent. Thanks for answering. However, that's not what was asked in the problem statement. The question was more about the probability of door 2, which is in fact dependent! Now the trick used in the unconditional method is that the chances of door 2 are easily derived from those of door 1, using the combined door theory, which is possible because of symmetry when assuming equal goats. So it's still a conditional problem solved by unconditional method, or whatever you want to call it. Indeed this all has nothing to do with the paradox. And indeed, you are not saying that the popular solution is wrong. Nijdam is however, but that's a formal issue as far as I'm concerned. I'm glad that we finally or hopefully almost agree on the theory behind. Heptalogos (talk) 21:10, 18 January 2010 (UTC)

The title of this section is: An unconditional solution to the symmetrical problem., but the issue here is: A symmetrical solution to the conditional problem.. I'm glad you admit it is a conditional problem, meaning it has to be solved by calculating a conditional probability. In what way this is done is of no interest to me. You come up with the term "unconditional method", a term you invented yourself. You may use any (correct) method you like, unconditional, unconventional, unlogically, whatever. I never objected this. In fact, if you have followed the ongoing discussion, somewhere I showed how to calculate the desired conditional probability, by the use of the symmetry in the problem. But the popular solution is wrong, as is the combined door solution, the many doors solution, most of the simulations, etc. I.e. as a solution to the "conditional problem", the one I (and not just me) consider the MHP. Nijdam (talk) 23:35, 18 January 2010 (UTC)

When you say the popular solution is wrong what do you mean? It gives the right answer. Sure, it would not apply to a more complex version of the problem (the non-symmetrical case) but this is often the case in mathematics. For any solution to any problem in mathematics it is always possible to find a more general case to which the solution is not valid. (I am not really sure that it is always true but I conjecture that it might be so, it is certainly often true). The simple solution does not give the correct answer just by chance, it gives it because of a symmetry that most people recognise. Martin Hogbin (talk) 00:48, 19 January 2010 (UTC)

@Rick The first point that most people do not spot is that P(getting a car on swapping) = 1 - P(having the car), as you know this is always true. The misdirection of opening a door makes this simple fact not obvious. The combining doors solution makes this fact obvious again.

Next, some people think that P(having the car after the host has opened a door) <> P(originally choosing the car|the host has opened door 3) even when the host is specified to choose randomly when the player has originally chosen the car. Many people think that P(having the car after the host has opened a door) = 1/2 because there are two doors and one car. Again, the opening of a door seems to confuse them. Nobody ever thought that the door opened by the host was important.

Morgan's point is whether P(A|B) = P(A|C), where C is the event that the host opens door 2. If P(B)=P(C)=1/2 I would think that it is obvious to most people that P(A|B) = P(A|C), but most people do not even think about this possible complication. Martin Hogbin (talk) 22:05, 18 January 2010 (UTC)

No, you're completely missing Morgan's main point which is that the question is about the difference between P(A) and P(A|B). As it turns out, because vos Savant treated the question unconditionally she completely overlooked specifying that the host must choose randomly between goats. Since she was answering "does P(A) = P(A)" this makes no difference to her answer. It DOES make a difference if you're addressing whether P(A) = P(A|B), and provides a handy way to show that these are indeed different questions. I will agree that most people don't consider the case where P(B) != P(C), but this is not at all the main point. -- Rick Block (talk) 23:00, 18 January 2010 (UTC)

What I cannot work out is whether you intend to distinguish between P(A|B) and P(A|B or C). Would you describe (PA|B or C) as a conditional probability? Martin Hogbin (talk) 23:29, 18 January 2010 (UTC)

IMO, P(A|B or C) defines the same sample space as P(A) since you're given one of events B or C must happen - this means P(A|B or C) is mathematically the same as P(A). Do you distinguish between these two? If they're not the same (mathematically) what is the difference? P(A|B) on the other hand is NOT the same sample space. If you think the MHP is asking about P(A|B or C) you're saying the question is asking whether P(A) = P(A). Of course it does. This is not paradoxical. The paradox is that P(A) = P(A|B). In words, the paradox is that the prior probability before the host opens a door is the same as the posterior probability after the host opens a door. The paradox is the same no matter which door the host opens, but that doesn't mean the posterior probability is P(A|B or C). The posterior probability is P(A|x) where x is one of B or C. This is entirely different than P(A|B or C), because P(A|B or C) is identical to P(A). -- Rick Block (talk) 01:02, 19 January 2010 (UTC)

Combined doors

On the german Wikipedia one of the discussiants gave the following funny version.

When the host offers the player to make her choice, she says: "Let's be practical: open doors 2 and 3, and give me what is behind." The host says he can not do this, but the player tells him she will choose door 1, and then he has to open one of the other doors, where after she will switch and open the remaining door, so in the end door 2 and 3 will be open. So why not immediately open door 2 and 3.

Logic?Nijdam (talk) 11:07, 17 January 2010 (UTC)

This was just meant to amuse you. Did it?Nijdam (talk) 17:54, 18 January 2010 (UTC)

But ... do you see the flaw? Nijdam (talk) 12:03, 20 January 2010 (UTC)

Your example itself lacks logic, as with 2 doors revealed, the placement of the car is always known. The contestant would alter her decision based on what is seen. Besides, the 'Combining Doors' solution shows one open door revealing a goat, then the contestant decides.

Here's a better one: You get one play of the game, with the doors opening exactly as per Whitaker. Whoever gets the car, keeps his life. Regardless of when you are asked, are you going to switch? Glkanter (talk) 12:22, 17 January 2010 (UTC)

Here's the best one: as an honoured scientist, you get two plays of the game, which are broadcasted on live television. Firstly, the host opens door 3 and offers you a choice to switch, after which you demonstrate your sophisticated calculation method, resulting in acceptance of the offer. In the second play, the host opens door 2. Do you recalculate or not? Heptalogos (talk) 12:39, 17 January 2010 (UTC)

No, for symmetry reasons (in the case of random hast behavior, of course) I know the (numerical) answer is the same. But both are conditional probabilities. Nijdam (talk) 16:54, 17 January 2010 (UTC)

Now you confirmed that you actually don't use the information in the number of the door shown. Heptalogos (talk) 18:16, 17 January 2010 (UTC)

How do you know symmetry applies in this case? Is a 'host bias' somehow ruled out? Glkanter (talk) 17:39, 17 January 2010 (UTC) (comment moved to improve clarity of dialog between Nijdam and Heptalogos)

Maybe Rick Block would also answer Heptalogos' question? Glkanter (talk) 18:21, 17 January 2010 (UTC)

Now that you've corrected the anecdote, Nijdam, the answer already exists. You've just described the rationale behind another of the simple solutions. The host is effectively offering the better (or equal) of the 2 doors, as he always reveals a goat. All the business about revealing a goat is just carnival-like misdirection, intended to make the reader think its 50/50. This is old news, Nijdam. Glkanter (talk) 12:49, 17 January 2010 (UTC)

Nijdam, what is your answer to Heptalogos' excellent question above? Martin Hogbin (talk) 12:51, 17 January 2010 (UTC)

He answered: "for reasons of effectiveness and efficiency, I do of course not recalculate, but to keep my honour in the world of science, I would have told the viewers that formally I would have recalculated". In other words, there's no use in recalculating, but these are the rules of science at this moment. Which is not true by the way, because there are no rules, but he simply clampses to his bigger brothers' intuition and experience for using the right method. Heptalogos (talk) 14:03, 18 January 2010 (UTC)

You say, 'he answered'. Where was that, and why not here? Martin Hogbin (talk) 16:49, 18 January 2010 (UTC)

I'm sorry for my sarcasm. I was translating the answer above. Please don't worry about not being civil; nothing bad will happen. Heptalogos (talk) 19:08, 18 January 2010 (UTC)

(outindented)New idea! Combine the chosen door and the door to be opened. The probability that the car is behind this pair of doors is 2/3. Now after the door is opened, the probability is 0 it hides the car, hence the chosen door has chance 2/3 to hide the car. So stick to the initial choice!???Nijdam (talk) 13:10, 21 January 2010 (UTC)

If door "A" is chosen and (from the moderators scope: "B=0+C") he opens "B", then winning chance for "C"=2/3. Your suggestion to "combine chance A+B=0" is based on another game, where door "C" must have been chosen and door "B=0" was opened also. But you forgot the moderator's scope: In your game it would not have been "B=0+C" like in the game before, it must have been "A+B=0" in "your" game. And it is true, in that latter game the chance for "A" would be 2/3, indeed. But it was another game, so I rate your new idea a hoax. -- Gerhardvalentin (talk) 06:42, 25 January 2010 (UTC)

The same mistake again. Maybe you're unable to understand. But, once more, in what seems your terminoligy: chosen door A, before opening B: Ab+Bb+Cb=1; Ab=Bb=Cb=1/3 {Ab means A before etc). Now comes the tricky part: after opening B: Aa+Ba+Ca=1; Ba=0(<>Bb), hence Aa+Ca=1 (Aa means A after opening of B, etc.). But we may not just assume Aa=Ab!! Nor may we reason: Bb+Cb=Ba+Ca. So we really have to calculate Aa, which turns out to be 1/3, and hence Ca=1-Aa=1-1/3=2/3. Got it? Nijdam (talk) 02:04, 26 January 2010 (UTC)

Sorry you do not face the facts and the mistake in your plea. Salu -- Gerhardvalentin (talk) 04:04, 26 January 2010 (UTC)

Gloomy view, Nijdam? Wrong combination. Put on your pair of glasses. The door opened never has been opened within a pair of "one door selected" and "one door refused", neither within a group of three doors. Read the rules. Remember: It always has been opened within the pair of those two refused doors. Put on your glasses, bye. --Gerhardvalentin (talk) 14:19, 21 January 2010 (UTC)

Can't follow you. Every door has chance 1/3 on the car, hence every pair a chance 2/3. Do you agree? Nijdam (talk) 14:56, 21 January 2010 (UTC)

Nijdam what? You can't follow? Please don't wimp out. You are strongly and seriously prompted to follow. Please read the rules. Every pair has a chance of (only) 2/3 as an average. And just ONE door of any pair of doors unevitably has to be a goat. You forgot? Just ONE. No matter which one. But never BOTH doors! And moreover I accuse you to neglect the rule concerning the host's target. The host is always given a "certain" pair of doors. He cannot escape: A "certain" pair of doors is given to the host. And in each and every "pair of doors" there necessarily has to be just ONE unavoidable goat. Agree? So your approach must include, and it positively has to include the following constellation:

The host "is given" (by whom?) a certain pair of doors, one of those two doors priorly having been selected already by the player, and that door selected by the player is a goat indeed. Agree? And the second door "given!" to the host might be the CAR. You forgot? ONLY ONE inevitable goat in any pair of doors. So this possibility that the second door is a car has to be considered. Then what? He cannot show that car. So your unneeded proposal to "combine" is a dead-end street, Nijdam. No more blind alleys, please. Regards, -- Gerhardvalentin (talk) 17:19, 21 January 2010 (UTC)

Nijdam is attempting to show the logical fallacy of the usual "combining doors" solution (he is a professor of mathematics at a university). Perhaps another attempt at this.

1. Each door has a 1/3 chance of hiding the car.
2. Therefore, any pair of doors, not just the two doors the host must choose between, has a combined chance of 2/3.
3. Therefore, door 1 and door 3 have a combined chance of 2/3.
4. Therefore, if the player picks door 1 and the host opens door 3, since the chance of door 3 is now clearly 0 the chance of the player's door must be 2/3.

Like Nijdam, I'm not saying this is a valid argument, only that it is just as valid as the usual "combining doors" argument. The reason this argument is NOT valid is the same reason the usual combining doors argument is not valid. -- Rick Block (talk) 19:52, 21 January 2010 (UTC)

The combining doors argument works perfectly well in cases where you can show that the host action does not change the probability of the car being behind door 1. As I have explained on several occasions, there are two methods of doing this if the host chooses a legal door randomly. They are symmetry and information theory. Martin Hogbin (talk) 21:50, 21 January 2010 (UTC)

No Martin, what you are referring to is not the "combined doors argument", as this argument doesn't involve the chosen door. Think. Nijdam (talk) 22:44, 21 January 2010 (UTC)

Back to reality: The group "selected door+open door" has a chance of exactly 1/3, whereas the group "refused door+open door" has a chance of exactly 2/3, and the group "selected door+alternatively offered second closed door" has a chance of exactly 3/3 as a fact. You know about this fact and about the causal chain? Seriously: The reason therefor is clear and evident, curious who can explain it − sorry abt objectionable style, Nijdam, just tried to copycat   :- )   Kind regards -- Gerhardvalentin (talk) 23:39, 21 January 2010 (UTC)

Well, why has the group "refused door+open door" a chance of exactly 2/3? Nijdam (talk) 00:17, 22 January 2010 (UTC)

You can tell it in a hundred ways, e.g. because - after the host showed a goat behind the third door - there are only two remaining doors now, the one originally chosen by the player with its unchanged chance of 1/3 and the still closed "priviledged door" in this game.

Here lies your mistake. Two doors are left to hide the car in this new situation. For neither of them it is clear what the new probability is. You say: the one originally chosen by the player with its unchanged chance of 1/3, but that's what we do not know. We have to calculate this new probability. Although it will turn out to have the same value as the old one, it is a different probability. Study hard, and you may understand it. Nijdam (talk) 17:51, 24 January 2010 (UTC)

"Here lies your mistake?" Sorry Nijdam, you aren't addressing the issue, you go past the issue. You need to see and evaluate the situation and draw the appropriate conclusion, and you have to look straight forward, never backwards and never changing your view. Remember: Those three doors have already definitely been devided in two groups, with a chance 1/3 for the door originally selected and with a chance 2/3 for the denied rest of the doors (it was a pair of two doors). Remember? This rest contains one inavitable goat, at least. It has a chance of 2/3, though. No matter if there are two goats, and no matter if one goat is behind the one door or behind the other door or if there are two goats hidden behind both doors. No point of interest. At least one inevitable goat. This pair has a chance of 2/3 though. If you tell no you're wrong. Everyone knows who can count to two: A pair of two doors, only one car in the game.(Have a look). Whether or not one goat behind this remaining pair of doors is still hidden or will be (randomly) shown later, does'n even matter! Opening one door showing one goat without giving any additional information is NO NEWS in this stage. Proven millionfold. Chances remain 1/3 to 2/3. Repeat: Whether one door showing a goat is still closed or already has been opened: No difference regarding the chances 1/3 to 2/3. Because showing one goat there is absolutely "NO NEWS". Of course, you can try the math, it will bring nothing new, it will only confirm the reality. No "different" chances. "Different probability" is bothering you alone, is not bothering the reality. What counts is the correct result, not the arduous detour which you take to reach the same result. Understanding the paradox is what counts. Math and to read it's history in this MHP is important to know, but not necessary to understand the paradox. You assert "mistake!" without a proof. Respectfully, -- Gerhardvalentin (talk) 22:03, 24 January 2010 (UTC)

Both joined together have a chance of hiding the car of 3/3. Of course you can ignore the situation that has developped and say "two doors, a car and a goat", 50:50. Or you can misinterpret the chance of the door originally selected, forgetting about the correct question and apply a false view, saying "any two doors have a joint chance of 2/3, also the door opened (0/3) and the door originally selected (thus 2/3), so the chance of the door offered as an alternative is 1/3. You can use different false viewing directions, a false line of sight and pretend to ignore your deviation that is so easy to reveal. You can misinterpret anything to an alarming extent, but you will be convicted. -- Gerhardvalentin (talk) 23:16, 23 January 2010 (UTC)

Gerhard - What, exactly, about the reasoning I provide above is in error? I assume you're saying the group "refused door+open door" has a chance of 2/3 because (if the player picks door 1) the probability of door 2 and door 3 sums to 2/3. I'm saying any two doors have a combined chance of 2/3, for the same reason. Actually, this is in fact true (i.e. this is NOT the error in the logic above). There is most certainly a problem because this reasoning can be used to show both (in the the case the player picks door 1) that door 1 has a 2/3 probability and door 2 has a 2/3 probability. To show how similar this is, here's a "combined doors" argument using the same logical steps.

1. Each door has a 1/3 chance of hiding the car.
2. Therefore, the two doors the host must choose between have a combined chance of 2/3.
3. Therefore, if the player picks door 1 and the host opens door 3, since the chance of door 3 is now clearly 0 the chance of door 2 must be 2/3.

I'm guessing that you might think the above is valid reasoning. But if it is, my reasoning about door 1 is valid as well. I don't know if you're saying you don't know what the problem is and you would like someone else to explain it. If this is the case, just ask and I'll be happy to explain what I think the problem is. If you think you know, please explain it. -- Rick Block (talk) 00:34, 22 January 2010 (UTC)

Outintended
Sorry, but I would rather like not playing stupid, believe me. Sorry to find obsolete and untruthful arguments, impossible to agree to those offside embellishments. Listen: Each and every "virgin door" in the standard MHP has a chance of 1/3, two doors have a chance of 2/3, guess what a group of 3 doors will be like. So: If you take a "virgin door", e.g. the door originally selected by the player, it has a chance of 1/3 to hide the car. If you doubt I will give you evidence. And I hear uncaringly saying: "Listen, I am going to combine your virgin door with a demonstrable goat, and believe me, I promise your chance will rise to 2/3. Trust me, I double your chance by adding a goat to your virgin door. No error in the logic." Unbelievable.

You can give numbers to the doors (not necessary) and combine them:
Car behind door 1, player choses door 1, switching hurts
Car behind door 1, player choses door 2, switching wins
Car behind door 1, player choses door 3, switching wins
Car behind door 2, player choses door 1, switching wins
Car behind door 2, player choses door 2, switching hurts
Car behind door 2, player choses door 3, switching wins
Car behind door 3, player choses door 1, switching wins
Car behind door 3, player choses door 2, switching wins
Car behind door 3, player choses door 3, switching hurts

and so on. But you can spare this sortilege by saying: There are three doors: door 1, door 2 and toor 3, and they contain:

Car Goat Goat - Player selects door 1, host opens door 2 or door 3, switching hurts, chance=zero. If you stick: chance=1. Even combined chance with open door 2 or open door 3=1
Car Goat Goat - Player selects door 2, host opens door 3, switching wins, chance=1. If you stick: Chance=zero. Even combined chance with open door 3 showing a goat=zero
Car Goat Goat - Player selects door 3, host opens door 2, switching wins, chance=1. If you stick: Chance=zero. Even combined chance with open door 2 showing a goat=zero

Swithing: 2 of 3 wins the car. Sticking: only 1 of 3 wins the car, even "combined with a goat".
Please understand: I never will follow your suggestion to combine with a "demonstrable goat"! Believe me. - No error in logic!
Why do you try to sell demonstrable goats as "virgin doors"? Suppose soon you will say:
"Any two doors have a chance of 2/3. And even if the host shows two goats, then their chance will be 2/3, also. Believe me. No error in logic."
Looking blank. Regards, -- Gerhardvalentin (talk) 02:02, 22 January 2010 (UTC)

No one is asking anyone to play stupid. The point is to identify the precise problem in the "combining doors" solution. You've switched from combining doors to a different (better!) analysis that correctly shows the probability of winning by switching is 2/3 and the probability of winning by sticking is 1/3. Perhaps paradoxically, by itself this does not mean that if you pick door 1 and see the host open door 3, that door 1's probability is now 1/3 and door 2's probability is now 2/3. If we walk through the false solution that shows the door 1 probability is 2/3, we'll see the same issue in the combining doors solution.

1. Each door has a 1/3 chance of hiding the car. Absolutely true, although to be precise we should say each door has a prior probability of 1/3 of hiding the car, where prior probability means the probability before the host has opened a door.

? If the door will be opened according to the rule, without giving any unauthorised addidional information, the chance of the door originally selected will not change. Whether the door is still closed or if it is open later: No difference, bedause showing a goat there where a goat has to be is absolutely no news in this respect.

2. Therefore, any pair of doors, not just the two doors the host must choose between, has a combined chance of 2/3. Also absolutely true, although these are also prior probabilities.

3. Therefore, door 1 and door 3 have a combined chance of 2/3. Also absolutely true, also referring to prior probabilities.

4. Therefore, if the player picks door 1 and the host opens door 3, since the chance of door 3 is now clearly 0 the chance of the player's door must be 2/3. This is where the problem occurs. The probabilities we're talking about here are posterior probabilities, i.e. probabilities after the host has opened a door. It is the sum of the prior probabilities that is 2/3 (by step 3), not the sum of the posterior probabilities. We haven't said anything about the posterior probabilities yet. We know the posterior probability of the car being behind door 3 is clearly 0, but to say the sum of posterior probabilities must be 2/3 because the sum of the prior probabilities is 2/3 does not follow - and is indeed false in this case.

Not correct. 1) the guest has devided the 3 doors in two diffeent groups. You have to look straight forward, not backwards. You never can combine door 3 with door 1 anymore after the doors have been devided in two different groups. Of course the car can be behind the door selected, doors 2 and 3 hiding two goats. But this will happen only in 1/3 of all cases.

Similarly, the problem in the combining doors solution is the 3rd step:

3. Therefore, if the player picks door 1 and the host opens door 3, since the chance of door 3 is now clearly 0 the chance of door 2 must be 2/3.

This statement is also talking about posterior probabilities where the previous ones were talking about prior probabilities. This one is saying the posterior probabilities of the group "refused door+open door" must be 2/3 because the sum of the prior probabilities is 2/3. The "door 1 must be 2/3" false solution shows that this is invalid reasoning. We know the posterior probability of door 3 is 0, but that doesn't say anything about the posterior probabilities of door 1 and door 2. This is the topic of the 2nd question of the FAQ at the top of this page. The bottom line is that unless you assume or know how the host picks between two goats (if given the chance) all you know is that the posterior probability of door 1 is between 0 and 1/2 (with an average of 1/3), and the posterior probability of door 2 is between 1/2 and 1 (with an average of 2/3). To make these definitely 1/3 and definitely 2/3 you have to say or assume the host picks between two goats randomly (with equal probability), and then use this fact in your reasoning. Combining doors doesn't do it. Even enumerating the cases (as you have) doesn't do it! What this shows is the average probability of winning by switching or staying, but not the posterior probabilities of the chosen door and the refused door. -- Rick Block (talk) 03:20, 22 January 2010 (UTC)

But: (Have a look).

Tank you, Rick, for your efforts. Will be back tomorrow. Until then. My talks in german WP with Nijdam, I really like him very much, esp. for his efforts, had exactly this issue. I showed him millions of "tests", all with the same result. The more tests, the more precisely the results. The chance of the door originally selected remains what it originally was before: Exactly 1/3. And the chance when switching (door open or even before, as soon as once she has selected her original door) will always be 2/3. From the time the guest makes her choice and devides those 3 door into two groups. It has been proven millionfold. Will be back tomorrow. Bye, -- Gerhardvalentin (talk) 03:56, 22 January 2010 (UTC) P.S.: Chances always from the guest's view.

I wrote a simulation a while ago that shows results as well - specifically that the overall probability of winning by switching is 2/3 but this can be true at the same time that the probability of the unchosen door and refused door are not 1/3 and 2/3. See /Archive 1#Excel simulation of difference between "random goat" and "leftmost goat" variants. I'd be happy to discuss this simulation with you if you'd like. -- Rick Block (talk) 04:25, 22 January 2010 (UTC)

Thank you, would really like it! Wanted to go to bed (now 5:30 in the morning), but no, couldn't make a break. Your words "Therefore, if the player picks door 1 and the host opens door 3, since the chance of door 3 is now clearly 0 the chance of the player's door must be 2/3. This is where the problem occurs." made me amend my "routine" and I tried: If the door originally selected immediately receives any second "virgin door" as a compagnon (from the group of the two doors originally refused), then the chance will rise from 1/3 to 2/3. But after the host has opened one door and has shown a goat within the pair of the two doors not selected, then the chance will stick on 1/3 in case the goat is given as a compagnon, and the chance will rise to 3/3 in case the still closed door is added as a compagnon. So: Any two doors have the double chance of one door, provided they are "virgin doors", i.e. the compagnon is not a demonstrable goat (shown) nor the other closed door offered as an alternative. So we have to destinguish "virgin doors" and "prestressed doors" to add as a compagnon. Any "virgin door" (1/3) added will double the chance from 1/3 to 2/3. Any "prestressed door" will have other effects: Adding zero in case a prestressed demonstrable goat (0/3) is joined, and adding 2/3 (total now 3/3!) if a prestressed "still closed alternative door" is added. So you are right: Adding a second door will double the chance. But it has to be a "virgin" door also, and it may never be "prestressed". Results will be different ones, then, as shown above. My "routine" clearly showed it. Regards, -- Gerhardvalentin (talk) 04:43, 22 January 2010 (UTC) BTW: Did you see my contrib. above in An interesting result "The rule doesn' say to the guest where the car is, but the rule says to the guest where a GOAT is"? Valid! Bye -- Gerhardvalentin (talk) 05:00, 22 January 2010 (UTC) P.S.: Chances always "from the guest's point of view"

Rick, let me tell you about virgins. I read your words written miles above:

"I think we're on the same page except for the bit about not identifying doors. My view is that the problem is either conditional, i.e. applies to a specific player standing in front of two closed doors and a goat (in which case the doors are inherently identified) or unconditional (applies only in some "average" sort of way). You seem to be thinking there's a third option - "conditional" but without specifying which door." To avoid struggle it could be helpful to correctly name the issues, the items, the circumstances of the case we are mediating.

I hear "unconditional" and "conditional". In MHP only the host does know everything, guess he can peek behind closed doors. If you are the host and the guest selects one door she will do that randomly, and two doors remain unselected/refused. The door she selected has a chance of 1/3. Because it had been selected randomly between three dors. Imagine the host offers her the opportunity to add a second door to her first door, and she would pick out a second door. She also would do this "randomly". Only the host knows what's behind the two doors she originally refused: At least one goat behind one of the doors, and behind the second door, the "priviledged door" in 1/3 a second goat in 2/3 the car. Let's suppose for this example, it was the CAR. The guest did NOT know what's behind the two unselected doors and would select RANDOMLY. In 2/3 of all cases she would pick a goat, in 1/3 of all cases she would pick the car. That's what I mean with "randomly". If she choses the CAR it would help, if she chooses the goat: no effect on her chance. Back to the rules: As the total chance of both unselected doors (randomly) always is 2/3 the chance of each unselected door is 1/3. She could double her chance by adding a second door "randomly": in 50% of the cases the left one, (generally doubling her chance), in 50% of all cases the right one (in general doubling her chance). The host knows what is behind the closed doors. If the host opens one of the two unselected doors now, say the LEFT DOOR showing a goat, the chance of this door is zero. But zero only in this one game. Next game: Other distribution. In the long run she doubles her chances by picking a second door. If it is chosen randomly. This is true "in general". May not apply to one specific case. After the host has opened a door showing a goat the chance of this door is 0 for this specific case. Other times other distribution, chance of this door in general and on the long run: 1/3. We have to be a little more precise to express exactly what we mean. And have to avoid statements like "if the player picks door 1 and the host opens door 3, since the chance of door 3 is now clearly 0 the chance of the player's door must be 2/3" and similar statements. Because it has no effect on the probability of the initially chosen door, and so on. (Chances always from the guests view)

I hear "unconditional" and "conditional". Fine. But if you find precise words to express what you mean you never need maths with "conditional probabilities". Am Austrian, my language is german, not English. Necessary to say exactly what we mean. I am sure we will find excellent names for the phenomenons we want to express, even without mathematics.   Regards   -- Gerhardvalentin (talk) 11:56, 22 January 2010 (UTC)

Rick, if a host bias may exist, we indeed do not know the posterior winning chance of door 1. But why should it be between 0 and 1/2? However, when 'everything' happens randomly and therefore symmetrical, we know for sure that the exact chance of door 1 remains 1/3. Doors 2 and 3 together then have a chance of 2/3. One can of course only compare the prior combinations of doors together (like 1+3). Posterior comparisons like 1+3 don't make sense, no matter if Nijdam makes them.

He is not a professor anyway. If he is the same Nijdam as on Wikibooks, he was a maths lecturer (PhD) at the University of Twente until 2004. Heptalogos (talk) 16:43, 22 January 2010 (UTC)

His contribution to the discussion isn't much more than a repeated "it is a conditional problem, meaning it has to be solved by calculating a conditional probability", which is of course true and done by the combined door solution. The key here is the exact definition of the condition: is it another door with a goat or is it number 3? With the proper assumptions there's no problem with most solutions. And no reliable source says so, but Nijdam does. Without arguments. Heptalogos (talk) 17:43, 22 January 2010 (UTC)

Posterior winning chance of the door originally selected by the guest

Intro: The humor of the host does not matter. He unnecessarily never will lift his left hand nor his right hand high in the air. And he never lifts his right leg nor his left leg noticeably and unnecessaryly high. And - having the choice between two goats behind both doors - he never will prefer neither the rigt nor the left door. He is not grinning broadly and he does not look grumpy, alternately. Useless to say: He never may reveal any unauthorized and irregular information. Take that for granted.

My view:
For the guest, the winning chance of his door originally selected is 1/3 (the host knows better anyway, no point of interest here). But for the guest the winning chance is 1/3.
And the chance of at least ONE goat hiding behind the pair of two doors refused is 3/3 "at least  :-)" The host may see two goats there in 1/3 of all cases: No point of interest here. The guest knows about the only car, so the guest knows: At least ONE goat is hidden behind the two doors not selected.

So: Regardless of WHAT door the Host will open showing ONE goat behind this pair of doors not selected, this will never add any relevant new information for the guest regarding the chance of his originally selected door. The host knows more than the guest, no question. Of course he does. But the chances for the door originally selected by the guest has not changed in any way - from the guest's view. Any assumption that - for the guest's view - this chance could have changed by opening one door is ridiculous.

Am I right or am I wrong? Regards, -- Gerhardvalentin (talk) 17:54, 22 January 2010 (UTC)

If the host is given or is assumed to choose between two goats randomly, then (and only then) it turns out you're right. No one has ever said anything attempting to contradict this. The sources that talk about this are talking about a version of the problem where it is NOT given that the host chooses between two goats randomly. It is by thinking about this sort of version that the weaknesses in many of the arguments used to show the result of the standard version can be exposed. -- Rick Block (talk) 04:54, 23 January 2010 (UTC)

You are right, he chooses randomly if he should have two goats. Therefore I said "Take that for granted". -- Gerhardvalentin (talk) 23:25, 23 January 2010 (UTC)

Is the unconditional problem too easy?

Unless I am misunderstanding his reply, Niijdam at least considers the if the problem is clearly described the solution is obvious.

Can I ask if there is anyone who thinks that the solution to this, I hope clearly unconditional, problem is obvious. It is based on the K&W problem statement.

You are going to be on a game show with the following rules. You’re given the choice of three doors. Behind one door is a car; behind the others, goats. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door.

What would you expect to be the chances of winning the car for players whose policy is to always switch? Martin Hogbin (talk) 20:48, 17 January 2010 (UTC)

Before I am starting to study this, is it the exact K&W problem statement? If not, we cannot use it as our problem statement anyway. Heptalogos (talk) 22:49, 17 January 2010 (UTC)

I am not proposing to use it. I am just asking whether editors think the answer is obvious when the question is put like this. Martin Hogbin (talk) 23:05, 17 January 2010 (UTC)

Sorry to maybe change the subject, but I understand your strategy in this section is to find out why we keep disagreeing. IMO it might be helpful to arise from the doors and draw the bigger picture. In a section above I asked Rick about the essence of argue. Apart from the article, he is willing to use unconditional method for solving conditional problems, as long as we make the right assumptions. This is IMO also what Morgan implies. The willingness to agree may not be too big because of possible consequenses for the article (my interpretation), but I agree with them that the sources are leading. I think we actually agree on most probability issues now. Heptalogos (talk) 13:53, 18 January 2010 (UTC)

A simple Bayesian analysis

Situation: door 1 is picked

• Winning by switching.

P(A): car behind door 2 = 1/3.
P(B): door 3 is opened = 1/2.
P(A|B) = P(A and B) / P(B) = (1/3*1) / 1/2 = 2/3.

In this line you seem to use two different values for P(B). Martin Hogbin (talk) 23:04, 17 January 2010 (UTC)

That's correct. You can read the reason in the explanation directly below. Heptalogos (talk) 23:07, 17 January 2010 (UTC)

Explanation for P(A and B): given the car behind door 2, door 3 must be opened, so P=1.

But that is P(B|A). P(A and B) = P(A)P(B) they are not independent. Martin Hogbin (talk) 23:10, 17 January 2010 (UTC)

Looks a bit dodgy to me. See what Nijdam thinks. Martin Hogbin (talk) 23:13, 17 January 2010 (UTC)

• Losing by switching.

P(A): goat behind door 2 = 2/3.
P(B): door 3 is opened = 1/2.
P(A|B) = P(A and B) / P(B) = (1/3*1/2) / 1/2 = 1/3.
Explanation for P(A and B): two goats behind doors 2 and 3 (P=1/3) and door 3 opened (P=1/2).

Is this correct? Heptalogos (talk) 22:54, 17 January 2010 (UTC)

Nijdam's corrections, but now underneath:

Situation: door 1 is picked

A: car behind door 2; P(A)= 1/3
C: car behind door 1; P(C)= 1/3
B: door 3 is opened;

P(B|A)= 1. P(B|C)= 1/2.

P(B)=P(B|A)P(A)+P(B|C)P(C)=1*1/3+1/2*1/3=1/2

P(Winning by switching|B)=P(A|B) = P(A and B) / P(B) =

=P(B|A)P(A)/P(B)=(1*1/3) / 1/2 = 2/3.

No need to calculate "Losing by switching", but if you like:

P(Losing by switching|B)=1-P(Winning by switching|B)=1-2/3=1/3

(That's what Nijdam think) Nijdam (talk) 17:37, 18 January 2010 (UTC) Heptalogos (talk) 18:55, 18 January 2010 (UTC)

Nijdam, what's wrong with my solution? Heptalogos (talk) 18:57, 18 January 2010 (UTC)

I love it.

Situation: door 1 is picked

P(A): car behind door 2 = 1/3.

P(B): door 3 is opened = 1/2.

P(A|B) = P(A and B) / P(B) = 1/3 / 1/2 = 2/3.

A and B are dependent. Together, their chance is 1/3. The overall probability of B = 1/2. It's so logically true. Why would anyone want to watch the theatre below? Heptalogos (talk) 21:37, 18 January 2010 (UTC)

Nothing wrong for the good willing reader. It is clumsy written. That's why I helped you with correct terminology. Instead of "together", which could be interpreted as "united", you better use "simultaneous". And I calculated the (overall) probability of B for you. Glad you begin to show some interest in probability theory. Nijdam (talk) 22:45, 18 January 2010 (UTC)

This is the essence of the conditional solution. Nijdam's solution essentially provides more detailed reasoning. Welcome to the dark side (or maybe welcome out of the darkness). Being picky there are two issues. How do you know P(B) is 1/2, and how do you know P(A and B) is 1/3? Hint: look at the tree diagram or the expanded figure in the Probabilistic solution section. Nijdam's solution derives these from more elemental values, specifically from the assumptions P(A)=1/3, P(C)=1/3, the given that the host opens Door 3, and the "host strategy" rules that P(A|B)=1 (the host must open Door 3 if the player picks Door 1 and the car is behind Door 2), and P(B|C)=1/2 (the host picks randomly between Door 2 and Door 3 if the car is behind Door 1). If these are directly given in the problem statement Nijdam's solution is a fully rigorous proof that P(A|B) = 2/3. -- Rick Block (talk) 22:31, 18 January 2010 (UTC)

Just another straight question for you Rick, with no hidden agenda. By 'conditional' are you just referring to the fact that a door is opened by the host, in which case argument over, or do you mean the condition as to which door has been opened by the host is significant. Martin Hogbin (talk) 22:36, 18 January 2010 (UTC)

I mean a door, but it is inherently a specific door (because the host, and the player, and everyone in the audience can see which door it is). This means if it does (or even might) matter which door, we now know which one it is. We can examine the situation for any specific door we want, and if it doesn't matter which door, then the analysis will presumably have a result that is independent of door. If it does matter, then the result will show the dependency on the individual door. -- Rick Block (talk) 03:45, 19 January 2010 (UTC)

It looks like you are using the term conditional to mean that it matters, or at least that it might matter, which door the host opens. This is of course true, in some cases it does matter.

So, this is essentially what we are arguing about. In the symmetrical case it turns out that the door opened by the host does not matter (change the result). Are we obliged to treat this as a special case of the more general case in which the host may open a legal door non-randomly?

I know of no mathematical principle which states that we must do this, that we must treat any specific problem as a special case of a more general one. For example, you might argue that many the proofs of Pythagoras' theorem are invalid because they do not apply to triangles that are not right-angled. The only true proofs are those that generate relationships between the sides of all triangles of which the right-angled triangle is a special case. Proofs which use the specific properties of right-angled triangles are false.

We can solve the symmetrical MHP by using specific and well-known properties of the uniform distribution of legal host door choice by virtue of either symmetry of the fact that random information is no information. Although learned papers and text books can recommend more general methods as being advantageous, they do not have the power to prevent proofs specific to the symmetrical case from being valid. Martin Hogbin (talk) 17:00, 19 January 2010 (UTC)

Is the sample space in which you're evaluating the prior probabilities (that are in effect before the host "opens a door") the same as the sample space in which you're evaluating the posterior probabilities (those in effect after the host opens a door)? I think this is what is meant by saying the player's initial probability "does not change". If this is what you mean, then I don't think there's any argument. The probability of the player's door is 1/3 and the probability of the other two doors is (combined) 2/3. However, if you want to describe the situation after the host has opened a door then I think you have to be talking about a different sample space. It is only in this different sample space that it makes any sense at all to say the probability of the door the host has opened is now 0 and the probability of the "other" door is 2/3. This sample space cannot be the same as the original sample space since the probability of the door the host opens has changed.

A specific example might help. If the original sample space is 3000 samples of the game (where the player has picked door 1), then in this sample space we'd expect about 1000 of the players to have initially selected the car P(A)=1/3. P(car behind door 2) and P(car behind door 3) are also 1/3 (about 1000 each). After the host opens a door if we're still talking about all 3000 samples, none of these probabilities have changed. However, if we want to talk about one of the cases where the host has opened a door, say Door 3, we're not talking about all 3000 cases any more but (in the symmetric case) only half of them. Now we're talking about 1500 samples, not 3000 samples. The probability the car is behind door 1 in this (new) sample space may numerically equal the probability that the car is behind door 1 in the original sample space but these are different probabilities because they are in different sample spaces. The "unconditional solution" only addresses probabilities in the original sample space - that's what it means to be unconditional. What it is saying is of the 3000 initial samples, whatever the host does we'll still have about 1000 players who've picked the car. The probability of the "other door" is always the probability of BOTH other doors (1/3 + 1/3), not the probability of either individual "other door". The probability of, say door 2, is (can be) different only in some reduced sample space - like the one of 1500 samples where the host opened door 3. -- Rick Block (talk) 19:30, 19 January 2010 (UTC)

A bit more on the above. IMO, the reason the MHP is paradoxical is because the sample space changes asymmetrically with respect to the 3 doors. Half the samples where the player's selected door hides the car end up in each sub-sample space (assuming the host picks equally between goats) but all of the samples where, say door 2, hides the car end up in one and all of the samples where the other door hides the car end up in the other. People are simply not used to this. -- Rick Block (talk) 21:01, 19 January 2010 (UTC)

Rick, you do not need to explain the method of solving the problem by setting up a particular sample space, conditioning it according to the door opened by the host and then using the resulting conditioned sample space to calculate the probability of interest; I already fully understand this method, and have done so for some time. This a perfectly good method of solving the problem, perhaps it is even the best method, but it is not the only method.

For any given mathematical problem there are usually many diverse methods of dealing with it. Sometimes, when the attempted solution of a difficult problem grinds to a halt, a way forward is found using a completely different approach. This new approach may involve a different mathematical discipline and sometimes it may depend on spotting an unrecognised symmetry or connection that enables progress to be made. This is true of the MHP, it can be solved by means other than the the method that you (and Morgan and others) prefer.

You seem fixated of a specific method of solution that involves conditional probability. I agree, it is a good method and it is one that is applicable to the more general case where the host has a known legal door opening policy. I agree also that the method is instructive in showing how conditional probability works, but, it is not the only method of solving the MHP. I do not believe that a mathematician would ever claim that there is only one possible method of solving a given problem. Martin Hogbin (talk) 10:51, 20 January 2010 (UTC)

Conditions again

I strongly have the feeling, you do not understand what it means when we say: the MHP is necessarily conditional. This has nothing to do with the method used, but with the question asked. The question asked is after a conditional probability. The method used to calculate this conditional probability is unimportant, as I already told so many times. On the other hand: an unconditional problem, where one is asked to calculate an unconditional probability, may well be solved through conditional probabilities. What important is, is to understand that the simple solution, as well as the combined doors and similar reasoning, are not sufficient. Nijdam (talk) 12:00, 20 January 2010 (UTC)

I accept what you say in the trivial sense that the question is asked after the host has opened a door. If this is all you mean this fine, let us call the problem conditional, clearly one thing happens than another happens then we calculate the probability. If you want to insist that the term 'conditional probability' is used that is fine with me, so long as it does not determine what methods of solution are valid.

I still have not got an answer to what exactly makes the problem one of conditional probability. Is it because:

Something happens after the player chooses a door?

A door is opened after the player chooses a door?

A specified door is opened after the player chooses a door?

Perhaps you could enlighten me. Martin Hogbin (talk) 16:58, 20 January 2010 (UTC)

Happy to be of service. It seems you don't read thoroughly, as I already have indicated, me nor Rich will insist of the use of the word "conditional probability" in the introductory section. It is however a conditional problem and hence we object the simple solution, etc. without mentioning something about the conditional nature in some way. And we also never insisted on a specific method to be used, as long it is valid! We have discussed conditioning also over and over, so your first possibility is way out of line. Both the other two form a condition. Then it depends on the question asked if this condition is to be taken into effect. Nijdam (talk) 12:41, 21 January 2010 (UTC)

Let me tell you my thinking:

I say that a condition is any event mentioned in the problem statement which might affect the probability of interest. In the MHP we have these events:

1) Monty says the word 'door'. It is not inconceivable that Monty might use the word 'door' only when the player has initially chosen the car, perhaps he says 'this one' otherwise. Such things, however, are not the normal assumptions of mathematical puzzles and we might therefore take it that this event does not affect the probability of interest, even though it is mentioned in the problem statement. It is therefore not a condition.

2) Monty opens door 3. Initially it might be suspected that this event would affect the probability of interest, it therefore might me considered wise to take this event as a condition. If, on the other hand, it can be shown that, when the host chooses a legal door randomly, this event cannot possibly affect the probability of interest then this event need no longer be considered a condition. The problem is no longer, therefore, conditional. Martin Hogbin (talk) 14:22, 21 January 2010 (UTC)

Of course it is not up to you to decide what has to be considered a condition. But I try to follow your line of thinking. The opening of door 3 doesn't influence the probability of the car behind door 1, means, the probability after the opening is (in value) the same as before the opening. We also discussed this several times. The probability after is (formally) the conditional probability. It need not be said with so many words, but it is necessary to mention that it needs proof (may be logical proof) that there is no influence for door 1 (on the value!!). For the other doors there is influence! But for all doors there is influence on the nature of the considered probability. Before: unconditional, afterwards: conditional. Nijdam (talk) 14:51, 21 January 2010 (UTC)

It can only be up to the person answering the question to decide whether the problem is conditional or not, unless the problem statement mentions the word 'conditional' (which the MHP does not). You have never given me a sound basis on which this decision is to be made. My suggestion is that the person answering the question has to consider every event mentioned in the problem statement and decide whether it could possibly affect the probability to be calculated. If an event which might affect the probability of interest exists, then the problem is conditional. If no such event exists then the problem is not conditional. Do you agree?

Regarding the proof that opening door 3 (when the host chooses a legal door randomly) does not affect the probability that the car is behind door 1, we note that, if the car is behind door 1, the host must choose randomly (uniformly) between doors 2 and 3. In this case the door actually chosen gives no information about anything (this is a well-know property of a random choice), in particular the location of the car. Thus we can safely consider this event not to be a condition in our problem. Martin Hogbin (talk) 21:43, 21 January 2010 (UTC)

As soon as you consider the opening of door 3 there is a before and an afterwards, with corresponding probabilities. Whether the opening of door 3 gives info or not is irrelevant for this. It may however be relevant for the calculation of the values. What is it that withholds you from understanding? Nijdam (talk) 22:52, 21 January 2010 (UTC)

What 'prevents me from understanding' is that you cannot tell me what in the problem statement tells us that, "the host says the word 'door'" is not a condition of the problem, with a before and afterwards with corresponding probabilities, but "the host opens door 3 is". What exactly in the problem statement tells me this?

Do you believe that the probability of an event can change if no information about it is revealed? Martin Hogbin (talk) 18:24, 22 January 2010 (UTC)

We had this discussion before. Instead of "the host saying 'door'", you then spoke of "Clapping my hands" or something alike. I explained you that a condition means a reduction of the sample space, do you remember? And just as handclapping is no event in the MHP, so is saying 'door' no event. We only consider events that matters to the problem, and they are about placing the car, choosing a door and opening a door by the host. I showed you the actual reduction, and as I'm recollecting right, Rick also did. Nijdam (talk) 22:53, 22 January 2010 (UTC)

Yes, I remember the discussion well. Your argument that a condition is anything that reduces the sample space does, at first sight, seems reasonable. However, there is nothing about a sample space in the problem statement, we have to set one up based on our understanding of the problem, based on events that matter, as you put it. Thus, before we set up our sample space, we have to decide which events matter, in other words which events might possibly affect the probability that we are trying to calculate. For the MHP, I could choose to set up a sample space including both the set of events in which the host says the word 'door' and the set of events in which the host does not say the word 'door'. According to your first definition, the saying of the word 'door' now becomes a condition of my problem, since considering only the case in which the host says 'door' reduces the sample space. In some circumstances this could be a perfectly reasonable thing to do.

Now, I do agree that in the MHP it is wise to set up a sample space to include the host door choice, as it would, on the face of it, appear that this might affect the probability of interest. In the symmetrical case, we can then do our calculation and discover that it, in fact, does not. That is what you suggest, I believe, but this is not the only way of addressing the problem. If we can positively show that the host choice of door cannot affect the probability of interest then we are free to construct a sample space in which this choice is not represented, and thus, by your definition, is not a condition. In view of the extremely counterintuitive nature of the problem this is probably not a good idea, but we could do it.

I would also add that there may be ways of addressing the problem that do not use a sample space at all but use some other mathematical construct. Maybe there is a geometric way? Who knows? I do not accept that any paper or text book can demand that there is only one possible way to address a problem. This is not normal in mathematics, sometimes solutions come from completely unexpected directions. Martin Hogbin (talk) 11:16, 23 January 2010 (UTC)

Of course you may complicate things by introducing events that are meaningless for the problem. Although the host says the word 'door' somewhere, nowhere in the problem is this considered to be important. So let's stay down to earth. You're constant talking about the symmetrical case, as if your hope is, this can do without conditional probability. Well, it cannot. Calculation, based on all the necessary events(!) may show that some unconditional probability and conditional probability have the same numerical value, but that doesn't imply we may do without some events. Then about your remark on other methods. There you're right, and Gill for instance showed a game theoretical approach. But most people, most sources, MvS herself, speak of probability. So we have to present the MHP firstly as a probability puzzle. Agree? Nijdam (talk) 16:09, 23 January 2010 (UTC)

You talk about events that are meaningless for the problem. The problem statement does not tell us which events are important and which are not. We have do decide which are the important events. To be a little perverse consider this scenario:

Monty Hall is replaced by a new host who tries to be a little more helpful to the player, maybe the show is loosing audience share because the prize is not being won often enough. This new host gives the player a bit of a clue what to do by saying "do you want to change to this door?", with this door being said in a warm and inviting tone, when the player will win by switching. When the player has originally chosen the car, the host says uninvitingly, "or would you like to change your choice?". Now the saying of the word 'door' is an essential condition. A contrived scenario, maybe, but not impossible. One still might chastise Morgan for not including this possibility in their analysis. But, to be a bit more serious, now that you are aware of this possible scenario, do you insist that this 'condition' must be included in every solution of the problem? So it is with the host door number. If we can show that it is unimportant then we are free to propose a solution that does not use it.

I agree that probability is the natural branch of mathematics to address this problem, particularly the extended version which asks for the probability of winning by switching, but, we do not have to include a condition which we can show is not important in our understanding of the problem. Martin Hogbin (talk) 19:58, 23 January 2010 (UTC)

Well Martin, the MHP is a probability problem. The story it is packed in, is just to make it attractive and imaginable. I'm not interested in all strange scenario's that has nothing to do with the problem, and that clarify nothing.

The scenario that I suggest above is not in any way ruled out by the Whitaker's MHP statement. Martin Hogbin (talk) 20:40, 23 January 2010 (UTC)

You continuously seem to be confused about the nature of the problem and the nature of the solution. The latter is unimportant. The problem is described in terms of (A) placing the car, (B) choosing a door and (C) opening a door by the host. This determines the necessary events. It's about time you come to understand that. Nijdam (talk) 20:21, 23 January 2010 (UTC)

I am not confused, you are just repeating your assertion, without supporting argument. Do you not understand my point at all? You have to make a decision at the start of a problem as to what might be important. Suppose you were teaching a student how to decide what to consider a condition upon reading a probability problem. What would you say? Martin Hogbin (talk) 20:40, 23 January 2010 (UTC)

To make my point clear let me ask you this question, supposing the host always uses the two phrases I have described above. Suppose he always uses the word 'door' when the host has not originally chosen the car and never uses it when the host has originally chosen the car. What is the probability of winning by switching, given the Whitaker statement (and that the host opens a legal door randomly)? Clearly it is 1. Now what is the probability given that the host chooses randomly which phrase to use? Do you need to calculate this? Martin Hogbin (talk) 09:08, 24 January 2010 (UTC)

No one is interested in describing the MHP in other terms than my A, B and C. They are necessary and sufficient. Where are you heading at? Nijdam (talk) 17:42, 24 January 2010 (UTC)

You know where I am heading, I am attempting to demonstrate to you that what is a condition and what is not a condition in a probability problem must be decided by the person answering the question. There is no other way in which this can be done. Thus you cannot say that, in any given question, a particular event must be a condition. It may turn out to be irrelevant, like the sneeze. You do not need a calculation to do this. Martin Hogbin (talk) 00:47, 25 January 2010 (UTC)

(outidented)Martin, I really don't know. If it makes you happy, you may formulate a kind of MHP, in which sneezing, handclapping, mentioning the word door, walking your dog, eating a hamburger, etc. playes a role. In my MHP only the mentioned aspect A, B and C do. Before I discuss anything further, please tell me how you describe the MHP. Nijdam (talk) 02:47, 26 January 2010 (UTC)

Nijdam, do you know of any other published conditional solutions to the MHP? I think Chun's table is looking wobbly. Glkanter (talk) 23:01, 21 January 2010 (UTC)

Martin - most of the above was explanatory to make sure we're on the same page with regard to the question I asked you that you didn't respond to. Is the sample space in which you're evaluating the prior probabilities (that are in effect before the host "opens a door") the same as the sample space in which you're evaluating the posterior probabilities (those in effect after the host opens a door)? -- Rick Block (talk) 13:21, 20 January 2010 (UTC)

Yes, in the symmetrical case, for the two reasons that I have given previously. My sample space contains the same two events, both with the same probability, before and after a door is opened. You may want to start with a different sample space. Martin Hogbin (talk) 16:58, 20 January 2010 (UTC)

I was a witness to what happened.

vos Savant's column generated letters that she answered for months. In 1990, there was no e-mail. Parade magazine probably had a printing schedule with 4 - 6 week lead times. Unlike our debates, this all took place in slow motion.

The only question/objection ever raised in Parade was 'It's 50/50'. I was there. That's why I am so adamant that nobody cared at that time about the door numbers. All the furor was about the 50/50 vs 1/3 2/3. Nothing else. When I first read the Wikipedia article, I was floored by the revisionist history I found. You're all entitled to your own opinions, but not your own facts. Glkanter (talk) 04:25, 18 January 2010 (UTC)

Please see WP:V, which is right up there with WP:NPOV (i.e. one of the 3 core content policies that Wikipedia is based on). Your own recollection of anything is completely beside the point as far as Wikipedia is concerned. If it's not in a reliable source it might as well not exist. -- Rick Block (talk) 20:25, 18 January 2010 (UTC)

I started a new section about what the MHP is, purely based on an overview of sources. I do have the impression that Parade is the centre of gravity. Do you have an overview of all Parade editions handling the MHP? Heptalogos (talk) 12:42, 18 January 2010 (UTC)

The four columns are reprinted at [2]. The sections of text in red are Marilyn's responses. I'm not sure exactly how much of the intervening text was actually published (I only have a copy of the first column). -- Rick Block (talk) 20:25, 18 January 2010 (UTC)

Thank you, Rick. That column of letters and vos Savant's responses supports my statement of what transpired 100%. It was all 50/50 and 1/3 v 2/3. There was absolutely nothing about which door or a host bias. She probably had 10 million readers, give or take. How many people read Morgan? Glkanter (talk) 22:02, 18 January 2010 (UTC)

There is one other reference that is important, in addition to those three MvS columns. The one where Marilyn vos Savant said that Morgan's interpretation of the problem was incorrect. I don't know how to link to it (I think it requires special access that I have, but won;t translate to others), but it was published in The American Statistician, Vol. 45, No. 4 (Nov., 1991), pp. 347-348. While Morgan, et al, and Rick Block seem to be of the opinion that MvS was talking about interpretation of her answer, that is clearly not the case. Because there was no such misinterpretation of her answer. The only misinterpretation was of the question.

And while I'm here, let me stress one point: There is no published 'scientific analysis' that concludes the problem is conditional, based on door numbers. Three sources state an opinion that it is, but offer no arguments for why that is so (and one of those even states that sematically, it isn't so). Far more articles that cite the Parade article treat it as unconditional, starting with the original author of the problem (and as MvS has stated she edited the question for content, she has to be considered the original author) and continuing to Leonard Mlodinow. While we can't discount those three dissenting opinions, NPOV guildelines say the two differing viewpoints need to be be described clearly and with balance. That means separating the two viewpoints, and not concluding that "the conditional interpretation" is correct. Since it really isn't a "scientific" issue, but one of interpreting the problem (and then applying "science" to the interpretations), there is nothing NPOV about treating the two problems as separate problems, even in the same article; while it violates NPOV to suggest only one is correct.

You wouldn't consider Selvin "the original author of the problem"? Why not?

There's a Formal Mediation request out there. You'll be the final signee. Glkanter (talk) 17:38, 20 January 2010 (UTC)

And one finale comment, more on editorial style. The article stresses words and phrases like 'scientific', 'probabilititic' and 'correct solution' far too much. The approaches wouldn't be included if those words and phrases didn't apply. As is, the only point of including them seems to be POV on the part of whoever put them in, as if they add weight to their favored interpretation. JeffJor (talk) 17:27, 20 January 2010 (UTC)

Here are some interesting quotes from WP:V.

The appropriateness of any source depends on the context. So which is the best source regarding a question to a popular magazine, the regular writer of the column to whom the question was addressed or a bunch of statistics professors writing years later in a second rate statistics journal?

The original writer who say the full question, and has admitted she condensed it down to the concise statement that reached print. JeffJor (talk) 17:27, 20 January 2010 (UTC)

The most reliable sources are usually peer-reviewed journals; books published by university presses; university-level textbooks; magazines, journals, and books published by respected publishing houses; and mainstream newspapers. Nothing here tells us that one has the power to override the other. Martin Hogbin (talk) 22:58, 18 January 2010 (UTC)

The immediate "peer review," Seymann, contradicts Morgan. The fact that the controversy over the problem - which is the only reason it is famous and included in Wikipedia - had absolutely nothing to do with any conditional/unconditional issues shows that that is a minor sidebar to the article. JeffJor (talk) 17:27, 20 January 2010 (UTC)

Does Seymann contradict Morgan? Please show me where. I only read in his comment that without a clear definition of the problem, it cannot be solved properly. Quite an eye-opener to me. The K&W version BTW is clearly defined.Nijdam (talk) 20:09, 23 January 2010 (UTC)

Don't overlook two sentences later: Academic and peer-reviewed publications are highly valued and usually the most reliable sources in areas where they are available, such as history, medicine, and science. -- Rick Block (talk) 03:25, 19 January 2010 (UTC)