Talk:Monty Hall problem/Arguments/Archive 9

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Faulty

The above tables and conclusions are faulty. You forgot about the frequentness:

Staying strategy:

[W][L][L]	[W][L][L]	[W][L][L]	:Result	Frequency
[X][ ][ ]	[X][\][ ]	[X][\][ ]	:W	1/6
[X][ ][ ]	[X][ ][\]	[X][ ][\]	:W	1/6
[ ][X][ ]	[ ][X][\]	[ ][X][\]	:L	1/3
[ ][ ][X]	[ ][\][X]	[ ][\][X]	:L	1/3

[L][W][L]	[L][W][L]	[L][W][L]	:Result	Frequency
[X][ ][ ]	[X][ ][\]	[X][ ][\]	:L	1/3
[ ][X][ ]	[\][X][ ]	[\][X][ ]	:W	1/6
[ ][X][ ]	[ ][X][\]	[ ][X][\]	:W	1/6
[ ][ ][X]	[\][ ][X]	[\][ ][X]	:L	1/3

[L][L][W]	[L][L][W]	[L][L][W]	:Result	Frequency
[X][ ][ ]	[X][\][ ]	[X][\][ ]	:L	1/3
[ ][X][ ]	[\][X][ ]	[\][X][ ]	:L	1/3
[ ][ ][X]	[ ][\][X]	[ ][\][X]	:W	1/6
[ ][ ][X]	[\][ ][X]	[\][ ][X]	:W	1/6

Results:

1/3 Wins versus 2/3 Losses = 33 % Success rate in scenarios using the staying strategy.

Switching strategy:

[W][L][L]	[W][L][L]	[W][L][L]	:Result	Frequency
[X][ ][ ]	[X][\][ ]	[ ][\][X]	:L	1/6
[X][ ][ ]	[X][ ][\]	[ ][X][\]	:L	1/6
[ ][X][ ]	[ ][X][\]	[X][ ][\]	:W	1/3
[ ][ ][X]	[ ][\][X]	[X][\][ ]	:W	1/3

[L][W][L]	[L][W][L]	[L][W][L]	:Result	Frequency
[X][ ][ ]	[X][ ][\]	[ ][X][\]	:W	1/3
[ ][X][ ]	[ ][X][\]	[X][ ][\]	:L	1/6
[ ][X][ ]	[\][X][ ]	[\][ ][X]	:L	1/6
[ ][ ][X]	[\][ ][X]	[\][X][ ]	:W	1/3

[L][L][W]	[L][L][W]	[L][L][W]	:Result	Frequency
[X][ ][ ]	[X][ ][ ]	[ ][\][X]	:W	1/3
[ ][X][ ]	[\][X][ ]	[\][ ][X]	:W	1/3
[ ][ ][X]	[\][ ][X]	[\][X][ ]	:L	1/6
[ ][ ][X]	[ ][\][X]	[X][\][ ]	:L	1/6

Results:

2/3 Wins versus 1/3 Losses = 67 % Success rate in scenarios using the switching strategy.

your original tables and your original conclusions are faulty, see: that table. Regards, Gerhardvalentin (talk) 12:09, 13 August 2012 (UTC)

Gerhard is completely correct. Just because there are 4 possible cases where the player picks door 1 ([x][ ][/], [x][/][ ], [ ][x][/], [ ][/][x]) does not mean these are equally likely. In particular, the car is equally likely to be behind each door, so ( P([x][ ][/]) + P([x][/][ ]) ) must be the same as either of the others.

An intuitive way to get to the answer is to reason that if you completely ignore the host, pick a door and keep it (don't switch) then you must win with probability 1/3 - whereas if you completely ignore the host, pick a door and switch then you win if you originally picked a goat (i.e. with probability 2/3). Always staying wins with probability 1/3 and always switching wins with probability 2/3. Assuming all of the 6 combinations of player pick and door the host opens (player picks door 1 and host opens door 2, player picks door 1 and host opens door 3, player picks door 2 and host opens door 1, etc.) are equivalent, the stay and switch probability in each case must be the same and also must be the same as the always stay and always switch probability - i.e. 1/3 if you stay and 2/3 if you switch.

Another way to get to the answer is to figure out the conditional probability in a particular example case, like player picks door 1 and host opens door 3. This is what is done in the Decision tree section of the article. -- Rick Block (talk) 15:05, 13 August 2012 (UTC)

One step further; 4 doors 

I'm having difficulty understanding the results of these tests using 4 doors. If someone can explain it would be appreciated.

Scenario: There are 4 doors. One door has a car. The other 3 have goats. The host knows where the car is. The player will always have a choice to pick another door after the host reveals a goat (steps 3 and 5).

Simulation:

1) Player randomly picks a door.

Chance of this door 1/4

2) Host randomly opens a door with a goat.

Host opens one of the 3 remaining doors, and if he got 3 goats, he chooses one door uniformly at random ???

  3a) Player randomly picks a door from the remaining 3. — Does that mean: Including the door of his first choice as per 3b) ??? (Yes, including the player's first choice)

The door first selected by the guest has a chance of 1/4, and the group (entity) of all 3 host's doors had a chance of 3/4 in total.
Now the group (entity) of the 2 still closed host's doors retains this chance of together 3/4, so each one of them has a chance of 3/8.

  3b) Player stays with first choice.   Chance to win the car 1/4

  3c) Player randomly picks a door excluding the player's first choice (swapping to one of the 2 remaining doors).   Chance to win the car will be 3/8

Now this is an unclear status regarding the actual guest's choice: is it actually his first door selected as per 3b), or is it actually one of the two still closed host's doors as per 3c) ?

4) Host randomly opens another door with a goat.

A)  As per 3b) the chance of the player's first door is still 1/4, and as the host can only open one of his two still closed doors, thereafter his other (only) still closed door retains the chance of 3/4. 

B)  But as per 3c) the guest had just swapped to one of the two still closed host's door with a chance of 3/8 - and that will not change.
So, before step 4) the two still closed doors had a chance of 1/4 (first selected by the guest, but he left it) and of 3/8 (the only one still closed host's door), together 5/8.
Irrespective of the host showing a goat now behind the door originally chosen by the guest, or behind his unselected own door, the actual door of the guest retains its chance of 3/8,
because the host kept secrecy regarding the door that hides the car, and the other still closed door has a chance of 5/8 now.

5) Player swaps.

A) resp. 3b) After swapping his chance will be 3/4 (75 % or "6/8"))

B) resp. 3c) After swapping his chance will be "5/8" (exactly 62,5 %)

Gerhardvalentin (talk) 22:17, 14 August 2012 (UTC)

Results:

3a is trivial. The winning chance is 2/3 because it is essentially the Monty Hall Problem.

3b is also trivial. The winning chance is 3/4. (similar to the 1000000 door explanation)

3c has a winning chance of ~62.6%, just under 2/3. The is the one I don't understand.

--- Resolved. 3c has a winning chance of 5/8. The player's first door has a 1/4 winning chance. After step 2, the two remaining doors have a combined 3/4, with 3/8 each. Swapping to one of the doors gives a 3/8 chance, and after step 4, the remaining door now has a 5/8 chance. Thanks. Foreign guy (talk) 02:23, 15 August 2012 (UTC)

Foreign guy (talk) 16:35, 14 August 2012 (UTC) 

Cont: Refusal to believe 

Continued from here

I am not saying "I don't believe", but, I have also felt, the possibilities mentioned in this puzzle a bit difficult to understand. The problem says about increasing winning chances, but, I think to follow this logic, there should be an exact and opposite possibility of decreasing winning chances.
Suppose there are 2 contestants ("you" and "I") and you and I choose door number 1 and 2 respectively. Show host opens door 3 finds a goat and asks both of us to switch our choices. We both agree and pick the other door.
Now.

• In such condition if your chance of winning increases, then my chance of winning should decrease (or vice versa). Since a) there is only one prize, so, b) we both increase our chance of winning sounds weird.

Alright, now, I eliminate you out of the game (don't ask why, please accept), so, I am playing alone or I am the only contestant of the game (which was actually proposed in the game i.e. 1 player). But, just above, it was hinted that after switching the choice my chances of winning decreased! So? --Tito Dutta (talk) 10:18, 23 January 2013 (UTC)

It's certainly no magic: Two players, the one who stays with his first selected door will win the car in 1/3, and his partner / friend who, in the end of the show, always chooses the host's second still closed door in the end of the show, will win the car in 2/3 (2/3 "only", never always).

Important is the scenario and the history of development of this tricky "story", its course / sequence / devolution that leaves behind a given obvious paradox, as a result. Marilyn vos Savant's intended scenario is the following:

It is a (only one) game show, and all we know about the game, its development and about the host’s strategy, is what MvS tells us:

* The host randomly hides the one and only price behind one of the three doors.

* The guest randomly selects one of the three doors.

* Then the host, following his declared intention to purposely open a losing door from his two unselected doors, is determined in any case to offer a switch to his second unchosen door.

If the prize is behind #2, the host opens #3, and if the prize is behind #3, the host opens #2. So if the prize is behind #2 or #3, the player will win by switching.

Progression: After the guest first has selected "his" door, the entity of three doors has irrevocably been divided into two opposing groups.

Three doors, and only one of them will hide the car. (Instead of "1/3 2/3" we should write "Player Host") and instead of "1 2 3" we should say (or add) "1/3 1/3 1/3"

We should be "aware" of this fact, because it is useful to examine those two groups. Each group can hide one car at most.

The guest's group can hide one goat at most, and the host's group can hide two goats at most, but the host's group "must" hide one goat at least.

We should help the reader to "decode that tricky picture".

The reader should be motivated to correctly distinguish, to single out, to decompose the elements and to put them together again in an insightful manner, to understand and to "decode" the paradox, imo he should be able to *vary* his first impression from different perspectives, yes.

He should be able to see that the chances for the player who stays versus the player who accepts the offered second host's closed door are 1:1 with a host who does not care what's behind the door that he opens, because in 1/3 he will show the car, destroying the perfect chance to win by switching in that 1/3.

But with a host who follows his declared intention to purposely open a losing door, those chances will be 1:2. That makes the difference. Regards, Gerhardvalentin (talk) 17:05, 23 January 2013 (UTC)

Another reason for confusion may be the assumptions hidden in the phrase "We both agree and pick the other door". What if only one of us chose to switch? Could we both choose Door #1 and split the prize (improving your odds of getting a smaller prize)? Mcwatson (talk) 17:24, 23 January 2013 (UTC)

• Good point. If you switch your choice and pick the door I chose (i.e. door 2), following the problem's argument your chance of winning increases! But, since the door 2 is my pick too, so, does my chance of winning also increase by "not switching" the choice? --Tito Dutta (talk) 17:33, 23 January 2013 (UTC)

Yes, you are right, Tito Dutta. Supposed you (who is in the audience), although not being asked, do secretly prefer door #2, but your friend, who is the player on stage, selects door #1 and in the course you see that the host opens door #3 in order to show a goat, then the chance of your preferred door #2 did rise from 1/3 to 2/3. So you can hope that your friend switches to that door #2, the one that you did secretly prefer just from the beginning.  Gerhardvalentin (talk) 10:34, 24 January 2013 (UTC)

The point is the development / history of the story. On stage, there are three doors, one host and one player in a fictive imaginary one-time show that, in exactly that way, never was nor is to happen in reality. Anyone in the audience is free two make his / her own decision if he / she wants, but without any effect to the development of the story that happens on stage.
First, the player irrevocably makes his first decision in irrevocably selecting one of those three doors, and this results in irrevocably separating his one and only selected door from the group of two unselected doors. This result / constellation cannot be changed by anyone in the audience. Anyone of the audience may be free to secretly select the same or another door of HIS / HER preference, but this is of no influence whatsoever on the said given constellation of the game that happens on stage:

The door first selected by the player has a chance of 1/3 (it will only in 1/3 hide the prize), while the group of the two unselected host's doors altogether collectively has a chance of jointly 2/3 in that fictive one-time show:

1. in 1/3 there will be two goats,
2. in 1/3 goat and CAR, and
3. in 1/3 CAR and goat.

From this group of two unchosen doors the host, in following his declared intention, purposely shows one goat and then offers a switch to his
ONLY ONE second still closed door that hides

1. in 1 out of 3 one goat
2. in 1 out of 3 one CAR
3. in 1 out of 3 one CAR

The host's offer to switch to this still closed second door is addressed to the player on stage only. The player on stage decides to stay with his first selected door, or to switch to the unselected and still closed second host's door. And anyone in the audience is free to make his / her own secret decision, but never will take home neither goat nor car. Only the player on stage, who can distinguish his first selected closed door from the unselected still closed second host's door that he is offered now to switch on, will win in 1/3 by staying and in 2/3 by accepting the offer to switch. Anyone who did not follow the show e.g., and who (consequently) cannot distinguish the closed door first selected by the guest from the host's closed door offered to switch on, cannot profit from that knowledge of actual difference in chance. By deciding randomly, his chance is (1/2 x 1/3) + (1/2 x 2/3), so exactly 1/2. Isn't it?   Regards,  Gerhardvalentin (talk) 19:10, 23 January 2013 (UTC)

Comment from Rmsgrey

I'd like to look at this a slightly different way: In the standard problem, you start out not knowing which way round doors 2 and 3 are. You find out which is door 1 by seeing which the contestant chooses, then find out which is door 3 by seeing which the host opens. If you know from the start which is door 3, then there are only two doors the car could be behind, and it doesn't matter whether you keep the first door or switch.

If there is a second contestant, who chooses a door after the first contestant has chosen which is door 1, but before the host announces which is door 3, then the second contestant doesn't know which is door 2 and which is door 3, so some of the time he will choose door 3 by mistake. Either that, or when he picks his door he must pick it as though he knows something the first contestant didn't - which door the host would have opened in the original version. If the second contestant can pick the door which we would have called door 3, then you have a different situation from the original version - the host is no longer guaranteed to always be able to open an unchosen door to reveal a goat. If the second contestant knows that a specific one of the two doors he has left to choose from definitely has a goat behind it, then that extra information means he's not in the same situation as the first contestant, so there's no reason to expect them both to gain the same way by switching doors.

However you interpret it, it doesn't give you a counterexample to the correct answer - that, in the standard problem, the contestant should always switch. Rmsgrey (talk) 00:17, 11 February 2013 (UTC)

Yes, it is important that the reader is motivated to correctly distinguish, to single out, to decompose the elements and to put them together again in an insightful manner, to "vary" his first impression from different perspectives, enabeling him to understand and to "decode" the paradox. Gerhardvalentin (talk) 12:24, 11 February 2013 (UTC)

Chronological order of added information

Some editors here repeatedly say that it is not the task of Wikipedia to support the reader in understanding of the context, the task was only to report what some sources say. Nevertheless for the MHP, to be in the know of the paradox, it is serviceable to pay close attention to added information in any step of the process.

The prior distribution of 1/3 "chance" for each one of all three doors, but 2/3 "risk" for each one of all three doors (watch out: "and nothing else !") is valid only before the player has made his first selection.

As soon as the player made his first decision, those three still closed doors have irrevocably been split into:

one still closed door (out of three) first selected by the guest, versus a group of any two (out of three) unselected still closed host's doors.
The chance of each single one of all three still closed doors is still 1/3, yes, but "nothing else" is dramatically and damnably inaccurate at this stage. At this stage of the process, any pair of two still closed host's doors (out of 3 closed doors) inevitably has a risk to contain goats of (2/3 + 2/3),
so in effect precisely [1 + 1/3] resp. [1/3 + 1], and no back down. At this stage, any such pair of two unselected still closed host's doors (out of 3 closed doors) definitely has a chance on the car of (1/3 + 1/3),
so in effect precisely [0 + 2/3] resp. [2/3 + 0] as long as no additional information has been revealed.

Actually this is the "prior picture" of three still closed doors that already have irrevocably been divided into two separate partitions. The respective picture depicts the stage after the guest irrevocably has selected his first door, and before any further action. It can serve as an "eye-opener". Why should that be denied? What does it help to conceal that given fact? Please help to avoid danger of collusion. Gerhardvalentin (talk) 23:47, 20 February 2013 (UTC)

From the discussion here we see the reason why the article for years has been staying so confusing and so misty. The discussion here makes evident that  ab initio  it is of utmost importance for the article to clearly distinguish the quite differing information given to us in contradicting scenarios:

• in the unambiguous correct standard scenario of the MHP, where the paradox clearly arises  (1 : 2 and not 1 : 1)
• versus other quite deviant scenarios where the paradox cannot arise in such palpable and clear way.

Other discussed scenarios – outside the paradox – determine quite deviant information:

A host that is not known to deliberately open a different goat hiding door on purpose, in order to necessarily offer a switch to his second still closed door (he also could open the door selected by the guest), or a host who – in case he got no car but two goats – is known to be not equally likely to open just any of them, but who is known to prefer to open one of them to a certain known degree (given we exactly know "which one" and we know exactly to what degree).

Each and any of such deviant scenarios should clearly be shown in later sections on deviant scenarios. Gerhardvalentin (talk) 15:14, 28 February 2013 (UTC)

Combining doors pictures

[Continued from the main talk page] I would argue that it is the the equality of the prior and posterior probabilities of the two unchosen doors combined that is the important step. This is demonstrated by considering the host who reveals a goat by chance. He gives us information about what was behind the two unchosen doors thus allowing us to revise our 2/3 figure. In the standard MHP no such information is given to us. Martin Hogbin (talk) 10:51, 19 February 2013 (UTC)

The equality of the prior and posterior probabilities of the two unchosen doors is derived from the equality of the prior and posterior probabilities of the chosen door. The figure 2/3 is just the copmplement of this unchanged 1/3, nothing more. Combining leads to the fasle way of thinking that the combined probability IS 2/3 and HENCE the unopened door must have this probability. Why do you defend this unnecessary and misleading picture. Wake up! Nijdam (talk) 09:49, 20 February 2013 (UTC)

Nijdam, you seem to have one fixed approach to this problem stuck in your head. You say, 'The equality of the prior and posterior probabilities of the two unchosen doors is derived from the equality of the prior and posterior probabilities of the chosen door.', but this need to be the case. There need not even be a third door, or there could be many other doors and we could still derive the posterior probability for a pair of doors given the prior probability given the events which occurred.

Imagine an urn with either two black balls (with initial probability 1/3) or a white and a black ball (with initial probability 2/3)

I take a ball (at random) from the urn which proves to be black. We can now calculate the posterior probability that a black and a white ball were originally placed in the urn.

Alternatively, I tell you that I will look in the urn and remove a black ball, which I do. You can now calculate the (unchanged) posterior probability that a black and a white ball were originally placed in the urn. Martin Hogbin (talk) 10:08, 20 February 2013 (UTC)

You say:...but this need to be the case. ???? And then: There need not even be a third door,... I'm lost there. Maybe there ain't any doors at all, even no participant. Concerning the urn, I'll save it up for my ashes, but do not compare it with the MHP. Nijdam (talk) 14:24, 23 February 2013 (UTC)

The equality of the prior and posterior probabilities of the two unchosen doors can be derived directly from the events that occur.

As you seem to have affected an irrational hatred for urns I will use doors, with cars and goats behind them. We have two doors and I roll a fair die. If I throw a 1 or a 2 a goat is placed behind each door. If I throw 3,4,5, or 6 a car is randomly placed behind one door and a goat behind the other.

I randomly open a (unspecified) door which proves to hide a goat. What is the probability that that there were originally two goats?

Alternatively, I look behind the doors and tell you that I will open a door with a goat behind it and then open a (unspecified) door to reveal a goat. What is the probability that that there were originally two goats?

I presume that you can answer these questions without invoking the need for a third door?

No doubt you would prefer to answer a different question is which the two doors are numbered 2 and 3 and the goat is revealed behind door 3. I am sure that you will not find that too hard, again without the need for a third door. Martin Hogbin (talk) 15:33, 23 February 2013 (UTC)

It may be interesting to discuss a thousand or more problems with 1, 2, 3 or more doors, with cars, goats, even differentiate between she- and he-goats, or urns and balls, or whatever you may come with, but it will lead us astray from the MHP. Let's concentrate on the MHP. And firstly only, I repeat only, on the role of the first picture in the so-called 'combining doors' explanation. The role of a picture here is to illustrate, or selfcontained to explain, but a picture in this role should not need a lot of caption, as to make clear what it stands for. Now what does the first picture communicate to the reader in order to assist them in understanding? My answer is: NOTHING, and worse it misleads the reader to a false type of argument. Now your answer to this question, Do not give your vision of the combined doors nonsens, but just answer the question. Nijdam (talk) 10:49, 25 February 2013 (UTC)

I have explained before, the first picture shows how the two unchosen doors combined have a probability of 2/3 of hiding the car. Do you agree with that fact?

This is useful because 2/3 is the answer that we are looking for. To answer you question as directly as I can the first picture communicates the numerical value of 2/3 to the reader in a way that the reader can understand and believe.

Provided we know (as described above) that this combined probability does not change when the host reveals a goat, this then leads us to the correct answer for the problem. Martin Hogbin (talk) 17:51, 25 February 2013 (UTC)

You may have expected my comment: nonsense! You mention a fact, but your explanation how it contributes to any understanding, i.e. that the answer is also 2/3, is absolute ridiculous. Why not show a picture with in big symbols just the fraction 2/3 on it, maybe that will help. As I guess this is the best you can do, it proves my point: it is only misleading!Nijdam (talk) 21:09, 25 February 2013 (UTC)

Step by step - where is the problem?

Nijdam, you have not explained your objection at all. Rather than just saying 'nonsense' it would make this discussion easier and shorter if you wer to tell me exactly what is nonsense. Let me go through the argument step by step. Please tell me which step is wrong.

1) The probability of hiding the car is 1/3 for each door.

2) After the player chooses, the probability for the two unchosen doors combined is 2/3.

3) Picture 1 shows the probability for the two unchosen doors combined being 2/3.

4) After the host opens a door the probability for the two unchosen doors combined is still 2/3.

5) We know this (the probability for the two unchosen doors combined is still 2/3) because the revealing of a goat tells us nothing about whether the car was originally placed behind on of the the two unchosen doors.

6) Picture 2 shows the probability for the two unchosen doors combined remaining 2/3.

7) We know it is not behind the opened door, therefore it is behind the unopened, unchosen one with probability 2/3.

8) Picture 2 also shows this.

Please tell me exactly where the problem lies. Martin Hogbin (talk) 22:26, 25 February 2013 (UTC)

As I repeatredly said before, it's not about being wrong, but about being misleiding. Your step-by-step explanation is correct, be it that some steps need proof. If you want to show this in pictures, you need, let's say, 8 pictures. I'm prettty sure the pictures are meant, originally, to show the following wrong way of reasoning (in short): the combined probability is 2/3, the openened door has probability 0, hence the remaining door must have probability 2/3. This is how Devlin, and I guess also others, originally argued, and it's what I find in many simple texts about MHP. I absolute don't accept this. What you're doing is trying to repair the meaning of the pictures. If the meaning of the first picture is no more then showing the equal probabilities for the doors, than show 1/3 for each door. No need to combine any two doors. The combining stems from the erroneous reasoning. Do you have eny objections to replace the first picture by one showing just 1/3 for each door? BTW: are the pictures well-sourced?? Nijdam (talk) 08:56, 26 February 2013 (UTC)

I do not think it is correct to call the way of reasoning that you propose 'misleading'; it does, in fact, lead the reader to the correct answer. 'Incomplete' would be a better term, but this is a comparative term. Every solution is incomplete in that further explanation or detail is always possible.

I do not understand your objection to the first picture. I can understand that there is an objection to the fact that no explanation is given as to how the second picture follows from the first. It seems obvious and is, in fact, correct but there is an argument missing.

The first pictures show that the probability (of hiding the car) of the two unchosen doors maybe combined into a single probability of 2/3. This leads people to believe that, before the host has opened a door, the combined probability is 2/3, which is indisputably (within reason) correct. Why we should want to lead people to believe this is another matter. I want to do it because it gives us a figure of 2/3, which you may see as a bad purpose but the picture itself is surely not misleading. If you disagree please tell me what the picture leads the reader to suppose that is, in fact, incorrect. Martin Hogbin (talk) 09:28, 26 February 2013 (UTC)

Do you or do you not agree or even understand that this way of reasoning: the combined probability is 2/3, the openened door has probability 0, hence the remaining door must have probability 2/3. is plain wrong? Nijdam (talk) 09:56, 27 February 2013 (UTC)

Yes, of course: ' hence the remaining door must have probability 2/3' is not correct. I have discussed two cases where this statement is plainly false. One is where the host opens an unchosen door randomly and just happens to reveal a goat, the other is where we take the question to be specifying door numbers and the host has a known preference for a particular door. I have suggested that we should not just tell our readers that the argument does not apply in these two cases but also explain to them exactly why it does not apply. This leads on to an explanation of why the argument does apply for the standard MHP.

None of this tells me what you find misleading about the first picture. This only shows that the combined probability of the two unchosen doors hiding the car is 2/3, before the host opens a door. What is wrong with that?Martin Hogbin (talk) 10:36, 27 February 2013 (UTC)

Patience: step-by_step. Do you notice the similarity of the pictures with the agreed wrong way of reasoning? Nijdam (talk) 22:55, 27 February 2013 (UTC)

So what is wrong with the first picture? Martin Hogbin (talk) 23:40, 27 February 2013 (UTC)

Please, answer the question. Nijdam (talk) 07:38, 28 February 2013 (UTC)

Yes of course I see a similarity between the pictures with the agreed wrong way of reasoning', that is why we have an explanatory caption for the second picture.

Now perhaps you can answer my question, without jumping ahead. So what is wrong with the first picture? Martin Hogbin (talk) 16:29, 28 February 2013 (UTC)

Then you'll understand and support my critisism of using this two pictures as an aid in understanding. Now to your question, which I did answer already before, but anyway: there is nothing wrong with the picture itself, nor if we replace it by a portrait of queen Elizabeth. Why do you ask again/ Nijdam (talk) 19:12, 1 March 2013 (UTC)

So what is the problem? We have words and pictures. The first picture shows a simple fact that the probability the two unchosen doors combined have of hiding the car is 2/3, which is a figure we wish to use, then the words tell us that, with the usual assumptions, this value does not change when a door is opened by the host, finally the second picture shows how this 2/3 probability is transferred to the unopened, unchosen door.

Later on we can use Bayes' rule, or whatever other argument you prefer, to show why the 2/3 probability does not change under the usual assumptions and also why it will change under different assumptions. Martin Hogbin (talk) 22:50, 1 March 2013 (UTC)

The problem is, as I've mentioned several times, that the first picture does NOT contribute to any understanding. That's also what your words imply. The picture shows he figure 2/3, for the wrong reason, that's the problem. What are we missing if we do not show the picture? Nijdam (talk) 09:39, 2 March 2013 (UTC)

The first picture shows a simple and non-contentious fact that the combined probability is 2/3. There is nothing whatsoever wrong with this. It contributes to the understanding of the prior probabilities. You are jumping ahead to the next stage where the posterior combined probability is assumed to be 2/3.

I really cannot understand what you object to, In many explanations a combination of words and pictures is used. Without the words the pictures may not tell the whole story and without the pictures the words may not be too clear. The two together provide the complete and clear explanation.Martin Hogbin (talk) 09:47, 2 March 2013 (UTC)

I thought you'd try to understand my objections, but no, you still are stuck in your way of defending the first picture by saying: nothing wrong. Nijdam (talk) 20:20, 2 March 2013 (UTC)

That is because you have never really explained what your objection is. You say, 'The picture shows he figure 2/3, for the wrong reason...'. That is your assumption. I choose to make a different assumption. Martin Hogbin (talk) 21:04, 2 March 2013 (UTC)

A nice example of a contradictio in terminis. You say I didn't explain my objection, and immediately thereafter you mention my objection. What's wrong with you. Why are you that eager to defend the picture. Even now I told you a lot of people are mislead by it. Maybe you aren't, maybe some of your aquaintances aren't, but the mere fact that you mention that the number 2/3 is shown, makes me wonder. Because in some sense that is the misleading aspect. If the first picture just has to show the initial distribution of the car, then 1/3:1/3:1/3 is the best way. No combining, as there is nothing in the initial distribution that says anything about combining. So you tell me if you object replacing the first picture by one with for each door the chance 1/3. Nijdam (talk) 08:49, 4 March 2013 (UTC)

I do see no point in just giving 1/3 for each door. The purpose of the 2/3 is to lead the reader in a certain direction but not, as you suppose, for the wrong reason. The reason that I want to lead people in the direction is that it is, in fact, that the combined posterior probability does remain 2/3 under the normal assumptions. I see no reason that the reader should be drawn to the belief that this is always the case when the caption makes clear that this is under the normal assumptions. The explanation does provide some insight into the problem and may convince some people who would otherwise not believe the right answer. Any 'wrong reasons' are in your head rather than in the article. Martin Hogbin (talk) 09:51, 4 March 2013 (UTC)

You say: the combined posterior probability does remain 2/3, and there lies part of the problem. The prior combined probability is 1/3+1/3=2/3, whereas the posterior probability also has the value 2/3, but now 2/3=2/3+0. It is not a matter of remaining. it all stems from the prior probability 1/3 for door 1, and the posterior probability also having the value 1/3.Nijdam (talk) 09:59, 6 March 2013 (UTC)

I do not understand your assertion that, 'it all stems from the prior probability 1/3 for door 1, and the posterior probability also having the value 1/3'. We can consider the event that the producer originally placed a car and a goat behind doors 2 and 3 (in any order). This has prior probability 2/3. Using standard methods we can show that the posterior probability of this same event, given that the host has revealed a goat under the standard assumptions, remains 2/3. We have no need to concern ourselves with door 1 or worry that 2/3 = 2/3 + 0. Martin Hogbin (talk) 18:27, 6 March 2013 (UTC)

There is a non-trivial insight in splitting the doors into chosen (1/3) and unchosen (2/3). The idea that the initial probability of 1 that the car is behind one of those three doors is split between two disjoint outcomes by the player's choice leads to the conclusion that, provided the probabilities of the two outcomes don't change when the host opens door 3, door 2 then inherits all the probability of the unchosen pair. It's not a rigorous and complete proof, but it is correct, and can be made rigorous by proving the assumption. The hardest part in getting someone to understand a rigorous solution is shaking free the assumption that the two remaining doors must be equally likely, and a non-rigorous, but correct, argument is worth using for that purpose.

Once someone has an intuition of why the answer might be right, then they can follow the rigorous arguments to prove it. Without that intuition, they're in the position of someone presented with a plausible-looking proof that concludes that 1=0 - they know that something is wrong, even if they can't explain what.Rmsgrey (talk) 14:01, 7 March 2013 (UTC)

A welcome oar

I, of course, agree with you entirely. 'Rigorous' is a comparative term, and making people believe the right answer can be an important step in getting them to accept a correct solution. Now see if you can persuade Nijdam. Martin Hogbin (talk) 17:32, 7 March 2013 (UTC)

Misleading, not wrong

It seems very difficult, not to say almost impossible, for the discussiants to keep permanently in mind that the issue is not the correctness of picture one, THERE IS IN ITSELF NOTHING WRONG with it. My objections, as I repeatedly have to mention, is the misleading effect. My experience is that people are lead to reason: the two unchosen doors have probability 2/3, as the opened door has probability 0, the remaining closed door must have probability 2/3. Got it? Nijdam (talk) 11:34, 11 March 2013 (UTC)

But as I have said before, the misleading effect is in your mind not in the pictures.

The 'false' reasoning you quote is, in the standard case, quite correct. Reasoning,'the two unchosen doors have probability 2/3, as the opened door has probability 0, the remaining closed door must have probability 2/3' is correct and therefore pictures which lead to this line of reasoning are not misleading.

There is a similar line of reasoning which is not correct, which is, 'the two unchosen doors have probability 2/3 before a door is opened and must always have the same probability after a door is opened. As the opened door has probability 0, the remaining closed door must have probability 2/3'. However, I do not accept that the pictures inevitably lead the reader to this particular line of reasoning.

The pictures do involve glossing over a particular step which could, in other circumstances, be critical but, as Rmsgrey points out above, just getting people to accept the correct answer is half of the battle and if we use a dubious short cut to achieve this we have taken the first step in helping our readers to understand. We can always show the other, better, and more general ways to solve the problem once they have accepted that the answer is not 1/2.

It seems to me that we have reached the point where you must agree that this is more a matter of personal taste rather that mathematics and on that subject we could simply agree to disagree.Martin Hogbin (talk) 15:02, 11 March 2013 (UTC)

The reasoning I quoted is not correct, also in the, what you call, standard case. There lies your problem. As long as you think it's correct, you have not understood the MHP. You seem to be mislead too. Nijdam (talk) 13:05, 12 March 2013 (UTC)

Then please explain where the error lies. Martin Hogbin (talk) 23:19, 12 March 2013 (UTC)

Well, we've been ther several times before, so I expect you to either accept it, or come with a rigorous attempt, with proper mathematics, to show why you think it is correct.

With the known formalism of C, X and H for the door numbers of car, first choice and host, we have:

P(C=2 or C=3) = 2/3

and also:

P(C=3|X=1,H=3) = 0.

The quoted wrong way of arguing is: P(C=2 or C=3) = 2/3 and P(C=3|X=1,H=3) = 0, hence (???) P(C=2|X=1,H=3) = 2/3.

Do not attempt to repair the reasoning, nor say that just some steps are left out. It is only about this false way of reasoning.

Nijdam (talk) 17:18, 14 March 2013 (UTC)

The reasoning implied by the images is

1) Obviously, P(C=1|X=1) = P(C=2|X=1) = P(C=3|X=1) = 1/3 and thus P(C=2 or C=3|X=1) = 2/3

2) Hence (???), P(C=1|X=1,H=3) = 1/3 and P(C=2 or C=3|X=1,H=3) = 2/3

3) So, since P(C=3|X=1,H=3) is plainly 0, P(C=2|X=1,H=3) must be 2/3

The issue is step 2. Some sources attempt to defend this step with the reasoning that you know beforehand that the host must open a non-chosen door, making this a probability 1 event. But what this reasoning actually leads to is the following step 2

2) Hence (because we know the host must open door 2 or door 3 if the player initially picks door 1), P(C=1|X=1,(H=2 or H=3)) = 1/3 and P(C=2 or C=3|X=1,(H=2 or H=3)) = 2/3

But this is simply a more cumbersome way of saying exactly the same thing as step 1 (if X=1, H can only be 2 or 3) and does not actually help us determine either P(C=1|X=1,H=3) or P(C=2 or C=3|X=1,H=3). As unintuitive as it may seem, the proper conclusion from this reasoning (which the images distinctly do not show) is that P(C=1|X=1,(H=2 or H=3)) = P(C=1|X=1) = P(C=2|X=1,(H=2 or H=3)) = P(C=2|X=1) = P(C=3|X=1,(H=2 or H=3)) = P(C=3|X=1) = 1/3, so (trivially) the sum of any two of them is 2/3. If we're talking about an event with probability 1, it doesn't change the probability the car is behind any of the doors - even one that has obviously been opened! The converse of this is true as well - i.e. if we're able to say the probability the car is behind door 3 is 0 (as shown in the image), we must be talking about an event that does not have probability 1. The bottom line is that the grouping of doors 2 and 3 is probabilistic nonsense. Misleading, as it were.

With the standard assumptions the probability the host opens door 3 (if the player initially picks door 1) is 1/2 (not 1), this 1/2 being the sum of:

1/3 = the probability the host opens door 3 if the car is behind door 2 (1) times the probability the car is behind door 2 (1/3)

1/6 = the probability the host opens door 3 if the car is behind door 1 (1/2) times the probability the car is behind door 1 (1/3)

The image that clearly shows this is the image currently in the section "Getting the conditional probability by direct calculations". -- Rick Block (talk) 05:01, 16 March 2013 (UTC)

Good to see you back Rick.

The problem with both arguments is that you assume that one particular formal line of (erroneous) reasoning is implied by a simple picture of three doors with the probabilities two doors hiding the car being combined. It is quite possible to propose a perfectly correct line of reasoning that is equally consistent with the pictures given.

As I and several others have said, one of the most important steps in getting people to accept and understand the correct solutions is to convince them that the answer is not 1/2. Once we have done this their minds are opened to more detailed and rigorous arguments. For some readers we can achieve this objective with a quick-and-dirty explanation which, despite its deficiencies, gives the reader insight into the problem and its solutions. Just because you interpret this explanation as being based on an erroneous line of reasoning does not mean that everyone will.

Let me say again, I am perfectly happy to work with you both to fully explain the weaknesses of the 'combining doors' explanation later on in the article. This surely gets us the best of both worlds. Martin Hogbin (talk) 11:35, 16 March 2013 (UTC)

Please enlighten us - what is the "perfectly correct" line of reasoning consistent with the pictures given (and if you're talking about including it in the article you do realize you need some source, or sources, that present this line of reasoning)? -- Rick Block (talk) 00:39, 17 March 2013 (UTC)

Here is a perfectly correct line of reasoning which has been presented in several publications by one R.D. Gill, inspired by Persi Diaconis' writings in support of Vos Savant's informal reasoning. Let us write O for the "other door" (i.e., the one left closed by the host). So X, H, O are, in turn, the door chosen by the player, opened by the host, and left closed by the host, respectively. A permutation of the numbers 1,2,3. Let me assume that C and X are uniform random and independent, and that H given C and X is uniform on the legal possibilities. In other words, all probabilies are subjective probabilities, defined through objective Bayesian reasoning derived from our total lack of information how the car was hidden and how the host opens a door (symmetry of our information with respect to renumberings of the doors). We are interested in P(O=C|X,H,O). By symmetry, the event O=C is independent of the triple (X,H,O) which is itself a uniform random permutation of the numbers 1,2,3. Therefore P(O=C|X,H,O)=P(O=C). But the event O=C is identical to the event X not equal to C hence has probability 2/3.

You may object to my inserting the assumption that X is uniform random, you might prefer to be working with X=1 fixed. But notice that I deduce that P(O=C|X=1,H=3,O=2)=2/3. This conditional probability cannot depend on the marginal distribution of X since it refers to the conditional probabilities given X=1. So mathematically I can "pretend" X is uniform at random, for free. I do this in order to be able to work with a more satisfying symmetry, the permutation group S_3, which corresponds to the intuition "the door numbers don't matter". Richard Gill (talk) 11:09, 17 March 2013 (UTC)

Does this have any connection with the discussed subject?? Nijdam (talk) 11:39, 17 March 2013 (UTC)

Yes. Door numbers are irrelevant. If we realise this in advance then the combining doors argument is completely justified. One can alternatively, if one prefers, patch the combined doors argument in the middle or at the end instead of at the beginning. I have published such a patched version in a Citizendium article and in a StatProb article (both are refereeed) as well as on my homepage. See http://www.math.leidenuniv.nl/~gill/#MHP for all references. For the popular reader it hardly matters: the popular reader needs a verbal argument using familiar concepts. Richard Gill (talk) 13:27, 18 March 2013 (UTC)

Round in circles

One "perfectly correct" line of reasoning consistent with the pictures given is shown just above. Nijdam said about this, 'Your step-by-step explanation is correct, be it that some steps need proof'.

The argument being put is that the pictures somehow lead the reader into assuming the incorrect line of reasoning given by Nijdam rather than one of the possible the correct lines of reasoning. None of us can tell what will actually go through the minds of a reader when they first see these pictures but if it helps them got the answer of 1/2 out of their minds we have achieved a great deal.

Once we have got over that hurdle I have absolutely no objection to explaining later on in the article exactly how the pictures might lead one astray and how one can avoid this. | (talk) 12:48, 17 March 2013 (UTC)

Sorry - missed that. Please explain #5. There seem to be one or more typos and I honestly can't tell what you're intending to say (and to what source would you attribute this line of reasoning?).

I said, We know this (the probability for the two unchosen doors combined is still 2/3) because the revealing of a goat tells us nothing about whether the car was originally placed behind on[e] of the the two unchosen doors.

Let me explain what I meant with door numbers. I am using a Bayesian understanding of probability. We are considering the probability of the event that the car was placed behind either door 2 or door 3. At the start of the game I assume that you agree that this is 2/3. After the player has initially chosen door 1, I assume you agree that the (numerical value of the) probability remains 2/3.

After the host has opened door 3 to reveal a goat, we are given no information that would enable us to revise this probability. We know the host will always reveal a goat and we gain no information from the fact that it is door 3 that he actually opened.

If you would prefer me to repeat this explanation from a frequentist perspective I would be happy to do so. — Preceding unsigned comment added by Martin Hogbin (talk • contribs)

If you're considering the event "the car was placed behind either door 2 or door 3", then after the host has opened door 3 the probability the car behind door 3 is still 1/3. If you're going to use the fact that the probability the car is behind door 3 is 0 after the host has opened door 3, you have gained information that you must use to update all 3 probabilities (not just the probability that the car is behind door 3). Either all 3 probabilities are evaluated given "the car was placed behind either door 2 or door 3", or all 3 probabilities are evaluated given "the host has opened door 3" (and to do this, even a Bayesian then needs to know the probability the host opens door 3 in the case the car is behind door 1). You can't mix and match. And, again, what source uses this reasoning? And what about the suggestion below? -- Rick Block (talk) 15:59, 17 March 2013 (UTC)

I am considering only the event that the car was initially placed behind door 2 or door 3, which initially has probability 2/3. I am not, at this stage considering any other events, and there is no requirement for me do do so.

As the show proceeds we may gain information that that allows us to revise the probability that this event occurred. For example, if the host were to choose randomly but just happens to reveal a goat, the probability of the same event (that the car was initially placed behind door 2 or door 3) would change to 1/2. In the standard problem, however, the probability of the one event that I am considering remains 2/3. Do you not agree? Martin Hogbin (talk) 16:14, 17 March 2013 (UTC)

Regarding references, we have two for 'combining doors', Devlin and Adams. We do not need references for a talk page discussion about what might be in the minds of our readers. Martin Hogbin (talk) 16:26, 17 March 2013 (UTC)

So, specifically, you are considering P(C=2 or C=3) which certainly initially is 2/3, just as P(C=3) initially is 1/3. Relating this back to the images, this is the situation shown in the first image, but after the player has picked door 1, so this image is actually showing P(C=2 or C=3|X=1) and, similarly, P(C=1|X=1), which are still (obviously) 2/3 and 1/3 respectively. What do you think the second image is showing, if not P(C=2 or C=3|X=1, H=3) and P(C=1|X=1, H=3)? -- Rick Block (talk) 22:16, 19 March 2013 (UTC)

I am not sure what you mean exactly by things like P(C=1). If you are referring to my argument above then we must use P(C=1) to mean the probability that the car was originally placed behind door 1. In that basis the second picture does indeed show P(C=2 or C=3|X=1, H=3), in words the probability that the car was originally placed behind door 2 or door 3, given that the player has chosen door 1 and the host has revealed a goat behind door 3. We know that in the standard case this is 2/3. Martin Hogbin (talk) 00:44, 20 March 2013 (UTC)

How do we know that in the standard case this is 2/3? -- Rick Block (talk) 02:46, 20 March 2013 (UTC)

added: It's the role of the host that generates the paradox: the host is fixing the game.
There is only one car behind three doors. Coerciveness:
Initially, any group of two doors out of three *is* to hide [one goat with probability 1] and *may* hide [a second goat with probability 1/3].

The simple *fact* that any two doors [A+B] out of three initially have a chance on the car of [A=0, B=2/3] or [A=2/3, B=0] respectively, should be mentioned explicitly.

The famous paradox is *based* on an unknown host who *intentionally* is showing a goat and never the car, in order to offer a switch to his second door. After the unknown host intentionally has opened one of his two doors in order to show a goat, the chance on the car of his two doors now clearly *are* [0+2/3] resp. [2/3+0], whereas the chance of the door first singled out by the guest remains 1/3 with that unknown host. Find some sources that show this in a way anyone can understand. Gerhardvalentin (talk) 03:04, 20 March 2013 (UTC)

If your primary goal is to initially convince readers that the answer is 2/3 - why not use one of the clearly correct (and sourced) approaches that do just this by first examining a strategy of switching vs. a strategy of not switching, e.g. the approach Carlton says he uses in his classes, or as published by Grinstead and Snell? By explicitly changing the reader's focus from the example situation where the host has opened door 3, to the slightly different problem of whether switching or staying is the best overall strategy (without knowledge of which door the host opens - i.e. moving the decision point to before the host opens a door), the analysis is simple and straightforward (and can be followed by nearly anyone). -- Rick Block (talk) 14:34, 17 March 2013 (UTC)

My objective is to convince readers that the answer is 2/3 to the question exactly as asked. I do not think it is helpful to explicitly mention a slightly different question. On the other hand I do not object to discreetly merging the two questions, as I did in the lead when I changed the wording to, 'Contestants who switch have a 2/3 chance of winning the car, while contestants who stick have only a 1/3 chance'. This actually refers to the 'unconditional' problem but I doubt that a newcomer would notice the difference. I am not sure how we could do this with the 'combining doors' solution. Martin Hogbin (talk) 15:54, 17 March 2013 (UTC)

Martin, there is nothing to gain in the combining, except the misleiding way of arguing. Combining means stating that 1/3+1/3=2/3. Not surprisingly. But to come to the result this 2/3 is also the value for the conditional probability, the combining is of no use. Devlin, who used the combing, gave initially the wrong way of reasoning. There is no correct arguing where the combining plays a significant role. Nijdam (talk) 17:35, 17 March 2013 (UTC)

You agreed that my argument presented above was correct. Combining does play a role in that argument. Remember we also have the captions. Martin Hogbin (talk) 17:51, 17 March 2013 (UTC)

You're constantly missing, or avoiding, my point: COMBINING PLAYS NO ROLE IN THE EXPLANATION, but leads to a wrong way of reasoning. Nijdam (talk) 11:07, 18 March 2013 (UTC)

Nijdam, there is nothing to gain by SHOUTING. Combining does obviously play a role in my explanation, points 2,3,4,5,6 all use a combined probability. You may not use it in your erroneous argument but I do use it in my explanation above, which you agree is correct. Martin Hogbin (talk) 13:35, 18 March 2013 (UTC)

Well, I hope you do hear me. You use (I don't dare to capitalize it) a combined probability, but it is of no help in explanating. And, Martin, it is necessary forthre explanation? And, do you notice the misleading effect? Nijdam (talk) 20:50, 18 March 2013 (UTC)

Moreover, I think Nijdam is wrong. As has been said a hundred times before, Devlin missed one step in his "combining doors" argument, and when a reader pointed that out to him, he admitted that his argument was incomplete. It is easy to fill in the missing step. It has been done by yours truly and no doubt by many others too. The missing step was the fact that the information *which* door is opened by the host is irrelevant to the question, whether or not the car is behind the door initially chosen by the player. The popular reader takes that for granted. The mathematician can confirm that this is true. See http://www.math.leidenuniv.nl/~gill/#MHP Richard Gill (talk) 13:39, 18 March 2013 (UTC)

As I said over and over, it seems very difficult to restrict oneself to the poit of discussion. Nijdam (talk) 20:44, 18 March 2013 (UTC)

Combining the doors is necessary for my explanation, since it is how the figure 2/3 is arrived at. I do see how the picture might lead some people to the wrong conclusion but do you see how this explanation might move some people away from their conviction that the answer is 1/2? Martin Hogbin (talk) 12:17, 19 March 2013 (UTC)

Yes, I do, but they mainly do so on false grounds. Nijdam (talk) 18:53, 19 March 2013 (UTC)

Who is to say what the grounds are. I say the diagram shows the (correct) argument that I gave above, you say it shows the (incorrect) argument that you gave, both are consistent with the pictures and neither of us knows what will be in the reader's mind.

On this specific point we can just agree to disagree. You think the reader will be led towards an erroneous argument, I think they won't.

Luckily there is a simple solution to this problem. Later on in the article, for those readers interested in the details we can give a full explanation of why the probability of the car being behind the originally chosen door does not change in the standard case but may change in other cases. I have suggested that we use Bayes' rule to do this. Martin Hogbin (talk) 00:44, 20 March 2013 (UTC)

There is another solution: If the pictures are not well sourced, they are to be considered as Own Research, and have to be removed. Nijdam (talk) 11:37, 20 March 2013 (UTC)

comment by Rmsgrey

The argument that going in with the decision to always switch does better than going in with the decision to always stick is also incomplete - consider a variation where the host always opens an unchosen door to reveal the car when he has the chance, and your choice is always whether to stick with your door or switch to the host's door. In that game, going in with the decision to always switch does better than going in with the decision to always stick, but you can do better than either by taking account of what the host reveals. For the "decide in advance" argument to be rigorous, you need to show that the host opening the door doesn't make any relevant difference. Of course, that is the key point that underpins the problem - that the host opening a door to reveal a goat (when he would never open it to reveal a car) tells you nothing about your original door. Understanding why that's true and why it matters means you understand the problem; missing that means you don't. The prior decision argument shows experimentally what the outcome is, but avoids even hinting at the explanation; the combined doors argument (implicit in phrasing the problem as stick/switch), while it also glosses over the key point, makes it more obvious. Rmsgrey (talk) 13:32, 21 March 2013 (UTC)

You are right, Rmsgrey. Everyone should be enabled to ponder about variants, variants outside the paradox  (btw the clean paradox exists for hundreds of years).  The article is not presenting the famous paradox, but for years was presenting lessons in maths. Although there is no "mathematical problem" regarding the clean paradox. The famous paradox is not about mathematics, it is all about the correct scenario where the famous paradox arises (unconditional, because its scenario, including the "role of the host", is perfectly fixed).

But if  –  outside the paradox  –  its appropriate strict scenario is distorted by maths experts in adding improper featherbrained additional "conditions" (in droves), all of that has to be treated quite separately quite outside the famous paradox. But editors here seem to prefer their mishmash, prefer to show misunderstandings as an inherent part of the paradox. Misconception.

Since years and years the article does never care a damn about the correct scenario that generates the clean unconditional paradox – never contrasting it clearly to spurious "conditional" deviations. The article is dominated by mathematicians, hence bent towards a mishmash of false and misleading directions, where all those inapplicable "conditions" are overzealously treated in formulas, without saying that such added "conditions" never affect the famous clean paradox. The article should be segmented, clearly separating the scenario of the clean paradox from quite other aberrant scenarios. Gerhardvalentin (talk) 22:39, 24 March 2013 (UTC)

Rmsgrey: you wrote that the host opening a door to reveal a goat (when he would never open it to reveal a car) tells you nothing about your original door. This is true if and only if either host choice of door to open is equally likely when he does have a choice; ie. when your intitial choice is the door hiding the car. But suppose the host's two choices are not equally likely. Then his choice does tell you something. However, it never makes your initial door more likely to hide the car than the other closed door. So you don't have to worry about this.

Gerhard: a reasonable way to approach MHP is to use probability in the subjective sense, and therefore, because we are not given any information to the contrary, to take either host choice of door to open equally likely when he does have a choice. The "combined doors solution" is silently assuming that which door is opened by the host gives you no information as to whether or not your door hides the car. The assumption is warranted, under the usual (subjective) interpretation of probability, and given what we have been told. So the answer is correct. However the argument is incomplete - it silently makes a non-trivial step. Please try to understand the difference between the correctness of an answer, and the correctness of the justification of an answer.

It would help terminate the interminable discussions with Nijdam if also Martin were able to make this distinction. Richard Gill (talk) 17:11, 21 April 2013 (UTC)

I guess you expected a comment from me here Richard. I understand the distinction perfectly well but I agree with Gerhard here. There is no reasonable scenario or interpretation of the standard MHP in which the simple solution are not perfectly mathematically justified. Even you Richard seem to have somehow got is stuck in your mind that the Morgan-style solutions are the 'correct' ones or the ones that mathematicians would 'naturally' use. This is not so, and you can easily prove it to yourself. Martin Hogbin (talk) 17:43, 21 April 2013 (UTC)

I hoped you would respond, Martin! I am *not* saying that Morgan-style solution is the right solution to MHP.

It is a fact that the solutions which mathematicians do naturally use and do naturally prefer are conditional probability solutions (using the natural, symmetric, probabilities), see Selvin's second paper, written in response to criticism from mathematicians that his method of solution was wrong! See Rosenthal. See almost any mathematical text. But nobody is fighting to give those solutions some kind of priority over easy, popular solutions.

I am saying that the standard argument known as the combined doors solution is incomplete. It pretends to deliver a conditional probability but there is one step missing in the chain of deductions/computations. Please try to understand the difference between the correctness of an answer, and the correctness of the justification of an answer.

As you well know, it is an easy task to fill in the missing step. So all this is not a big deal. But if you are going to edit an article which is of interest both to the general public and to mathematicians you have to be conscious of this kind of thing. It is not difficult at all to make the article both correct and accessible and fun. Provided you are aware of distinctions which for some people are extremely important.

Nobody is insisting on the Morgan scenario with a possibly biased host. Everyone is happy with the natural, symmetric, probability assumptions. Only Gerhard keeps fighting those spooks. Some people are interested in logical precision, that's all. Richard Gill (talk) 19:56, 21 April 2013 (UTC)

I perfectly well understand the difference in principle between the correctness of an answer, and the correctness of the justification of an answer but I am sure that you understand that you understand that there is no clear definition of the correctness or completeness of an answer. It is a matter of taste.

The simplest solution to the MHP is, 'The answer is obviously 2/3'. I think we can agree that that is not a very complete or correct solution, even though the answer is correct. The most correct and complete solution would be one using a kind of maths that Boris mentioned a while back. It is exceptionally rigorous and probably completely impenetrable to anyone except specialist mathematicians. In between those two extremes that is a whole range of solutions, the completeness of which depends only on what is accepted as obvious and what is considered to require further explanation. Martin Hogbin (talk) 23:07, 21 April 2013 (UTC)

You say yourself above, 'The missing step was the fact that the information *which* door is opened by the host is irrelevant to the question, whether or not the car is behind the door initially chosen by the player. The popular reader takes that for granted. The mathematician can confirm that this is true'. If we choose to take the fact that the information *which* door is opened by the host is irrelevant to the question as obvious then the combining doors solution is complete.

By the way, you will see above that I have agreed to differ with Nijdam above so there are no more endless arguments. You also say above, 'Some people are interested in logical precision'. What exactly is the logical precision that you are interested in? Martin Hogbin (talk) 23:15, 21 April 2013 (UTC)

In Dutch we have a saying: Paper is patient. Maybe you also will agree to differ with this. Nijdam (talk) 07:11, 22 April 2013 (UTC)

No, see below. 08:48, 22 April 2013 (UTC)

Two interpretational axioms of the MHP

The natural language statement of Whitaker cannot be mathematically or logically solved without some interpretation of what question the statement intends to ask. These interpretations then become axioms of any further mathematical or logical calculations; they cannot be proved from within the mathematics.

The two interpretational axioms (yes, I did make up that name) that I would propose as necessary and sufficient to mathematically or logically solve the standard MHP are:

1) We have no reason to suppose that the originally chosen door is any more likely to hide the car than any other door. Or if you prefer:

1a) The event that the car is behind the initially chosen door is independent of all preceding events.

2) No subsequent events (including the host's choice of door) give us any information as to whether the car is behind the initially chosen door. Or again if you prefer:

2a) The event that the car is behind the initially chosen door is independent of all subsequent events (until, of course, it is revealed whether the player has won the prize).

With both these axioms the problem is easily solved by mathematics. Without either one, the problem is insoluble is not the intended problem and has a different answer.Martin Hogbin (talk) 08:48, 22 April 2013 (UTC)

It's about the article. The article still does NOT pay regard to the intended paradox (staying:switching not 1:1, but 1:2). The clean scenario of this intended (old) paradox tells us without excessive demands that the door first selected by the contestant actually has a chance to hide the car of 1 in 3. And we learn that, in the event that the contestant by luck should have selected the prize in 1 out of 3, both unselected doors actually must hide goats. And we learned that in such assumed actual "Lucky Case", staying will win and switching will loose, whatever door the host opens and whatever goat he shows. And we know that any eventual "host's bias" (MCDD) does regard the host only, but actually is NOTHING to us, as we do not know neither about its existence nor about its degree nor direction (Falk). Any actually applying of such one-sided bias inevitably will return incorrect actual results. Period. Repeat: incorrect actual results. The clean scenario tells us that applying such unknown bias actually is not only senseless, but actually is strictly prohibited. It is of use in teaching and learning conditional probability theory, but – within the given scenario – is a no-no for the actual given situation.

The valid scenario furthermore tells us that the probability NOT to be in the "Lucky Case" is 2 in 3. So having selected not the prize but goat A is 1 in 3, and having selected goat B is 1 in 3. (Teaching and learning conditional probability theory uses to changes those "1/3 : 1/3 : 1/3" to other values. This is correct for teaching and learning conditional probability, but incorrect for the situation actually given in the valid scenario for the famous paradox). The article should clearly say so, but not continue to insidiously cover the tracks.  Gerhardvalentin (talk) 15:19, 22 April 2013 (UTC)

Gerhard, I am not sure how your comments relate to what I am talking about. I am suggesting two, what I call 'interpretational axioms' that clearly define what you call the intended paradox.

The first, and I imagine uncontentious axiom is that the car is equally likely to be behind the originally chosen door as any of the others. This is what most people assume, it is undoubtedly what vos Savant intended, and it is a necessary assumption because, without it the problem is not the intended problem and has a different answer.

The second 'interpretational axiom' is that no subsequent events (including the host's choice of door) give us any information as to whether the car is behind the initially chosen door. This exactly what the intended paradox is intended to imply. Because the host always reveals a goat and the door which he chooses gives us no information about the probability that the car is behind the originally-chosen door, so no new information is revealed. This is again undoubtedly exactly what vos Savant and Whitaker intended to be the case. It is probably though not what most people assume to be the case, which is why they get the answer wrong.

If this axiom is false, which means that some event occurs after the player has chosen a door which changes the probability that the car is behind that door then the problem is again not the intended problem and has a different answer.

I have put things this way in response to Richard's statement that he wants logical precision. Richard insists that we consider which door the host opens but ignore any other ways by which the host might reveal information about whether the car is behind the originally-chosen door. My argument is definitive and logically more precise than his. Martin Hogbin (talk) 15:54, 22 April 2013 (UTC)

Just to underline your correct ascertainment, I have made the above additional demand. It is necessary that the article clearly shows that (for the actual situation that the contestant is in) any distinguishing of "before and after", of "left-hand or right-hand door" is not reasonable, but dangerous. It is beneficial for teaching and learning Bayes only, but never applicable in any given actual situation / question concerning the famous paradox. Gerhardvalentin (talk) 16:33, 22 April 2013 (UTC)

I agree with you. The choice is always between a car and a goat. The doors perform only the function of hiding the choice from view. We need a good source that treats the problem that way. Martin Hogbin (talk) 17:05, 22 April 2013 (UTC)

There is "Stochastik: Einführung in die Wahrscheinlichkeitstheorie und Statistik" by Hans-Otto Georgii. Also available in English translation.See pages 54-56 (dropbox; or Google books) Georgii says right from the start that the door numbers are irrelevant, and goes on to present a simple solution. There is the work of Sasha Gnedin who shows from the decision theoretic point of view that the door numbers can be ignored. There are proofs in several of my articles where I first use symmetry to chuck out the door-numbers from the start. (I just removed the section from the article which reproduced this argument since no-one seems interested in it and since some of my claims could not be supported by references to articles other than my own and because some people think "symmetry" is too difficult a word for a wikipedia article on a popular brain-teaser). So: yes there are sources a plenty for this. Richard Gill (talk) 11:37, 23 April 2013 (UTC)

Gerhard: you are mixing up elements of the problem with elements of the solution. The way the problem is stated we can *deduce* that the door numbers are irrelevant. We don't need anyone to tell us what we are supposed to think. MHP is a puzzle, a problem which can be solved by logical thought. It is not a problem which is solved by psychological analysis of Marilyn vos Savant's brain patterns. Richard Gill (talk) 11:45, 23 April 2013 (UTC)

Nice work! But we do not need interpretational axioms cooked up for this particular problem. We already *have* general principles. One such principle is Laplace's principle of indifference. Because we don't know how the car is hidden, it is equally likely behind each of the three doors. Because we don't know anything about how the host chooses a door to open when he has a choice, each choice is equally likely in that case. It now follows from both these facts combined, by symmetry, that given the door chosen by the player, which door is opened by the host is independent of whether or not the car is behind the door chosen by the player.

In particular, given that the player chose door 1, whether or not there is a car behind door 1 is independent of whether the host opens door 2 or door 3. As Bell (1982) said this is so obvious as to be hardly worth saying out loud. But there's no harm in saying it out loud!

When deciding whether or not to switch, one may safely ignore the specific door numbers in the particular case at hand.

If you like we can invoke symmetry up front and say straight away, that by symmetry the door numbers are irrelevant. Therefore the only choice we have is between staying or switching, independent of the door numbers in the case at hand.

I don't need two ad hoc axioms specially invented in order to justify someone's intuitive quick-thinking answer; I just need one stabdard principle which everyone already automatically agrees to. I use it twice and I make a deduction. Richard Gill (talk) 06:45, 23 April 2013 (UTC)

Richard, you started this conversation by suggesting that I did not understand the difference in principle between the correctness of an answer, and the correctness of the justification of an answer. I do understand this difference perfectly well although I am not sure what I need to do to convince you of this. Now perhaps you could try to understand my point.

The problem with you proposal for a solution above is that it is logically incomplete. You make no mention of goats or the words spoken by the host, both of which are mentioned in the problem statement. Now, of course, I agree that vS/W obviously did not intend these to be part of the problem but nevertheless, in principle, both could affect the answer. We are talking here, therefore, about the correctness of the justification of your answer. As you say above, 'this is so obvious as to be hardly worth saying out loud. But there's no harm in saying it out loud!'. This is my point, there is no such thing as a complete and totally rigorous solution only a range of possible solutions from, 'The answer is obviously 2/3' to the mathematically rigorous solution of the type mentioned by Boris. There is no sharp complete/incomplete dividing line, just increasing levels of rigour.

My other point, which we have discussed before, is that there is no standard way to turn a natural language statement of a probability problem into a precise mathematical one. There are some standard principles, which you quote above, but we still have to make decisions of interpretation.

The best principle, in my opinion, to apply is that expounded by Seymann; we need to ask ourselves what the intent of the questioner was. There is no doubt in my mind that vS/W did not intend us to consider which door the host opened. There is also no doubt that they did not intend us to consider which goat the host revealed or the words that the host used. Our mathematical formulation should therefore take these facts into account. Your solution proposal is logically incomplete; mine is much more complete (although no doubt a sharp mathematician could find some fault in it) it clarifies exactly the way that vS/W wanted us to set up the problem mathematically. Martin Hogbin (talk) 08:30, 23 April 2013 (UTC)

I agree with some but not all of what you say. I agree that we should ask ourselves what the intent of the questioner was. However I think we should focus on their intent as to what the *problem* was, not on what they intended as a solution. From their solution we can deduce what problem they are trying to solve, but we are not obliged to take their solution as the ultimate in logical precision. The water gets very much muddied in the transition from Selvin via Whitaker to vos Savant. The dividing line between problem and solution became much vaguer. Selvin is precise, unambiguous; Whitaker and vos Savant are vague, even sloppy. You will notice in the discussion between Morgan and vos Savant that vos Savant didn't even get the point which Morgan et al wanted to make, which Morgan et al found important. Partly this is probably obstinacy (the cleverest woman on earth obviously couldn't give an inch), I imagine she is smart enough that she could have understood that point if she had wanted to. The point which Selvin also apparently found so important that he completely rewrote his solution in order to take account of it!

It seems to me that you are mixing up ingredients of the problem with ingredients of the solution. Moreover you invent one axiom after another while I point out that there is just one basic principle, which everyone agrees to, and which takes care of both of your so-called axioms.

I don't bother to discuss all of the other non-issues at all, so in that sense my solution is less complete, but I do discuss an issue which comes up again and again in the literature. Devlin even withdrew his combined door solution because he saw that there was a gap in his argument and apparently didn't see how to fix it.

Selvin set up the problem mathematically in a way which is completely unambiguous. vos Savant and Whitaker, trying to reproduce it in the context of a popular magazine, expressed themselves carelessly. Because of their careless description of the problem and imprecise solution, more careless solutions have also flourished. Richard Gill (talk) 11:31, 23 April 2013 (UTC)

You are still missing my point. You talk above of a missing step. By this I presume you mean the step of stating that it is not important which of the two unchosen doors the host chooses when he has a choice.

I agree that this is a missing step, but it is one of several missing steps. We also need to say that the particular goat that the host reveals is not important. I know this is obvious and nobody, as far as I know, thinks it might be important but the fact remains that without stating this assumption a solution is incomplete. With a little perverse imagination it is not hard to think of more missing steps.

I am not proposing my 'axioms' for the article, I just give them because they cover a wide range of possible missing steps. Regarding whether they are part of the problem or the solution, again, there is no clear dividing line. There are several stages in addressing the original problem statement including, deciding what the game rules are, deciding which events could possibly affect the probability of interest (once we have decide what that is) turning these decisions into a mathematical problem and then solving the clearly defined mathematical problem. Some of these steps are, in my opinion, not properly explained in many sources and they are where most of the disagreement lies. Because some steps are often tacitly assumed, mathematics cannot tell us what steps are missing. Martin Hogbin (talk) 12:34, 23 April 2013 (UTC)

Let's not be perverse. Let's restrict ourslves to missing steps which are actually painful. Devlin withdrew his combined doors solution when someone pointed out that his argument was incomplete/unjustified. Selvin rewrote his solution completely. Rosenthal found the simple solutions shaky. Richard Gill (talk) 19:13, 23 April 2013 (UTC)

Yes, I accept all that but my point is that there is a continuum of rigour/perversity in solutions starting with the absurd 'The answer is obviously 2/3', progressing through the 'simple solutions', which address the problem as it was obviously intended (in that we are not to worry about door numbers, goats or the host's words) through the 'Conditional' solutions (where we need take account of the door number opened by the host or at least give an explicit justification for not doing so) through a slightly more rigorous/perverse solution in which we consider which goat the host reveals, to an absurdly rigourous and perverse solution where we try to consider with maximum mathematical rigour all possible events that might affect the outcome.

There is no sharp dividing line between solutions which consider the door opened by the host and solutions which consider the goat revealed by the host. There is a difference, and that is that vos Savant tried to make the problem clearer by numbering the doors but she did not name the goats.

I do, by the way, accept that one of these solutions is mentioned extensively in the literature and, thus far, the other is not. I therefore do not want to make this point within the article. I was responding to Richard's assertion that, 'It would help terminate the interminable discussions with Nijdam if also Martin were able to make this distinction'. I do make this distinction, more so than most. Martin Hogbin (talk) 08:15, 24 April 2013 (UTC)

Naming the doors is part of the trick (building up the mental image which leads the victim astray). It is very interesting, I think, that the father and grandfather of MHP - three prisoners, three boxes - are paradoxes of a different type: everyone can see the correct answer right away, one is presented a formal reasoning which appears logical but which leads to an obviously ludicrous answer. MHP is precisely the other way round! Also interesting is that there is no controversy at all concerning the need for conditional probability in those two ancestral problems. A fascinating, brilliant, shift.

So Martin needs two interpretational axioms to deal with MHP (with door numbers). I only need one (the principle of indifference). And: a solution in Martin's style which does not explicitly invoke axiom 2 is logically incomplete.

You cannot get way with that one Richard. You do indeed have one basic principle, that of indifference (which you said, and I agree, is inherently part of the Bayesian approach) but you have to decide when to apply it. If you do not state explicitly each time you choose to apply the principle of indifference you end up with a simple solution, having tacitly applied the principle throughout your solution. If you do explicitly apply it you need to do so several times and also state why. You should say that you have applied it to the original car placement, the host's door choice, the host's goat choice, the host's choice of words and anything else not mentioned that might be important.

You still seem to be missing my point, which is that the host's door choice is just one of several possible conditions that should, if you want to be complete, be applied. There is no very good reason to single this particular condition as the one essential condition in the stated problem. Martin Hogbin (talk) 10:10, 25 April 2013 (UTC)

Maybe Martin can also give us his opinion on Selvin 1975a solution, in tbe light of Selvin 1975b. In the latter, Selvin deckares prominently that his solution (2/3) depends on the assumption of an unbiased host. Yet his 75a solution did not even mention, let alone use, that assumption. So is the reasoning in 75a wrong (or at best, incomplete)? (Not in your opinion, but according to Selvin's own principles). If incomplete, how to add the missing step?

Selvin missed something that might be argued to be necessary and he later corrected this omission. He had no goats though. I think he sounded a little irritated (maybe with himself, maybe with others) that his simple problem had been perversely complicated. Martin Hogbin (talk) 10:10, 25 April 2013 (UTC)

Morgan

You say that one particular missing step comes up many times in the literature. This is because of the paper by Morgan et al. Do not take my word for that though, it is quite easy to check. You say that any worthy mathematician would take into account the door chosen by the host, I say that this is only considered important because of the Morgan paper.

We can test these hypotheses by looking at the literature before and after the Morgan publication. Let us stick to the literature after the vS/W problem was published. There was very little on Selvin's original version a decade earlier.

As far as I know, there were no published sources considering the door that the host opened published before the Morgan paper, you may know better but I am not aware of any and Morgan themselves assert that there was none. After the Morgan paper many sources mention the question of which door the host opens. I guess you would be able to calculate the probability of this happening by chance. Martin Hogbin (talk) 13:18, 23 April 2013 (UTC)

No published sources considering the door the host opened before Morgan et al.? Martin, you know full well Selvin's second letter does this. And, you also know full well (should know) that Morgan et al. was the first paper that rigorously addressed the problem. Dividing the literature before and after, and concluding that Morgan et al. introduced some perverse interpretation is ludicrous. You admit there's hardly anything in the literature before Morgan. So, how is dividing the literature before and after this paper remotely meaningful? -- Rick Block (talk) 18:48, 23 April 2013 (UTC)

Selvin (1975b) was published as a reaction to letters which Selvin had received from professional statisticians, criticizing (justly) his (1975 a) solution. At least one of them, rather famous (I can find his name for you if you like). The '75b solution is an absolutely standard, routine, trivial calculation of the obvious thing to calculate (if you have any training in probabilty or statistics). Earlier solutions to Martin Gardner's Three prisoners problem, and Bertrand's box paradox, are in exactly the same vein: routine calculation of the conditional probability of winning by switching, given what has happened so far. One of the *less* standard solutions to the Bertrand's box paradox is by symmetry/independence. Beginner mathematicians don't like this so much, since they prefer to "play safe" using standard definitions and blindly computing, rather than using insight in order to short-cut the calcuations. Richard Gill (talk) 19:00, 23 April 2013 (UTC)

The Selvin case tells us that some mathematicians found fault with Selvin's argument and he agreed with them. I do not dispute that but we can make no further deductions from this fact. The question I am addressing is, what proportion of statisticians would naturally include the number of the door opened by the host in their calculations. The Selvin case tells us that some would for sure but little more than that.

Yes he agreed with them. This supports my argument that from a professional's point of view, Selvin's second approach is the natural approach to take. You said that Morgan et al. were responsible for the switch to conditional probabilities. This is simply disproved by Selvin 1975b, it's just wishful thinking on your part, not supported by anything. Morgan et al. are not famous and their paper is not famous. Every professional can solve the problem for themselves routinely, does so automatically using conditional probability, and doesn't know or care about Morgan et al. The novelty of Morgan et al. was to introduce a biased host and show that despite possible bias, whether known or unknown, you should still switch, anyway.

What has always been missing from the literature is any serious attempt to motivate the various probability assumptions. As far as I know I am the first to argue that for a subjectivist, the problem description in itself leads irrevocably to *both* uniformity assumptions (hiding car, opening door). For a frequentist, on the other hand, there is no solution at all unless they are allowed to randomize the choice of door in advance of the player ... or unless the person who states the problem also states how the two choices are done (i.e. by what kind of randomization devices). Richard Gill (talk) 17:14, 24 April 2013 (UTC)

In 1975b, Selvin also writes "The basis to my solution is that Monty Hall knows which box contains the keys and when he can open either of two boxes without exposing the keys, he choose between them at random". Interestingly his 1975a "solution" does not make any use of this assumption. So if it was meant to be the basis of his 1975a solution then the argument he gave then was incomplete. You can't have it both ways. His last words on the subject were 1975b and by implication his 1975a solution is (at best), in his opinion, incomplete. Richard Gill (talk) 17:24, 24 April 2013 (UTC)

Vos Savant, on the other hand, got thousands of letters, many from statisticians and not one, as far as we know mentions the host's door choice. Morgan in their paper give 6 'false solutions', one from Mosteller. They say, 'That we have not previously seen the solution offered in this article...'. We presume from this that they had seen some solutions. Martin Hogbin (talk) 07:47, 24 April 2013 (UTC)

The solution they're referring to not having seen before is not the 1/3:2/3 solution using conditional probability, but rather the 1/(1+p) solution.

This solution takes as givens (because these aspects of the problem were explicitly clarified by vos Savant in her followup columns) that the car is initially placed randomly (uniformly), the host must open a door revealing a goat, and the host must make the offer to switch - but leaves the host's preference between the two goat doors, which was never mentioned by vos Savant, as a (potentially unknown) variable.

The asymmetry between taking the hiding of the car to be random but not the host's choice between goats is their understanding of the "complete" problem conditions, given vos Savant's statement of the problem plus her subsequent clarifications. Note, in particular, her detailed "cup" simulation instructions [1] which explicitly randomizes the initial placement and the player's initial choice, but not the selection of which losing cup to reveal in the case both are losers.

In their response to your letter, they generously concede that this choice is implicitly random as well (even though vos Savant never explicitly mentions it), which (of course) makes the answer 1/3:2/3. However one presumes they still consider the question to be "given you've picked a particular door (e.g. door 1) and the host opens a particular door (e.g. door 3), should you switch" - i.e. what are the conditional probabilities the car is behind your original door (e.g. door 1) and the now only other unopened door (e.g. door 2) - not "what are the probabilities of winning by following an always switch vs. always stay strategy". The latter is the question vos Savant's, and many other solutions, address - note that this question can be paraphrased as "should you decide to switch before seeing which door the host opens". They show these questions have the same answer (iff p=1/2), but also conclusively show these are different questions by introducing variants where p!=1/2 (in which case these questions have different numeric answers, but always switching is still never worse). -- Rick Block (talk) 16:16, 24 April 2013 (UTC)

Morgan et al. hadn't seen Selvin's two letters (in the same journal!) and weren't aware of Nalebuff. They had talked about the problem with students and colleagues and hurriedly written their paper to criticize vos Savant. I don't think we can conclude very much from what they hadn't seen.

The Mosteller solution they refer to is Mosteller's 1965 solution to Martin Gardner's 1959 Three Prisoners problem (itself perhaps derived from Bertrand's box paradox). It's related and in its usual formulation probabilistically equivalent to the standard (symmetric, full) MHP. And it is traditionally, like Bertrand's box before it, solved by a conditional probability calculation! There is a long tradition of problem posing and problem solving behind this! A long and (among mathematicians) very well known tradition. It's been the staple diet of introductory texts, for 100 years, always associated with Bayes. Don't tell me Morgan et al started up a completely new tradition. Far from it.

And on the other hand, vos Savant simply couldn't understand what point Morgan et al were trying to make (or perhaps, obstinately pretended that she didn't see the point, confident in the knowledge that most of her readers would not see the point either). A situation which amusingly prevails to this day. Richard Gill (talk) 17:44, 24 April 2013 (UTC)

At this point, Richard, I think I must bow to your superior knowledge of the literature. I think you should be more explicit when you refer to a 'conditional' solutions though. What exactly do you mean? 'Conditional' seem to me to be a term bandied about by statisticians to try to scare people. In Bayesian probability pretty well every problem is conditional. I have nothing against conditional probability. So, just to be clear, are you saying that there were several published solutions to the W/vS problem which explicitly took as a condition of the problem the fact that the host opened door 3? Martin Hogbin (talk) 09:52, 25 April 2013 (UTC)

When I say "conditional" in this context I always refer to the use of conditional probability, by definition probability of A given B = P(A|B)=P(A and B)/P(B) = probability of A and B simultaneously divided by probability of B. I do not use the word to scare anyone. Everyone on these talk pages has been using the word with this precise, tecnical, meaning for years and years. All the publications about MHP which solve MHP within a formal mathematical context do the same. I say that, completely routinely, Selvin 1975b solved his own MHP by the extraordinarily elementary direct calculation of P( car is behind door 2 | player chose door 1, host opened door 3). There are good reasons for doing this. (a) one hundred years of tradition cf. Bertrand, 3 prisoners (b) the relevant probability for the player about to make his choice *must be* the conditional probability given what he knows at the moment he must make his choice, which is *after* seeing various things happen. You can approach this axiomatically - subjective (Bayesian) "probability" is always in fact a conditional probability given all information available. It's all about what you should believe, given what you know. You can also approach this operationally: it gives you a guarantee of the highest possible overall success rate. Richard Gill (talk) 11:32, 25 April 2013 (UTC)

Yes, I know all that, but when you apply a condition, which you call B above, you must clearly state what the condition that you are applying is. Your condition B is that the player chose door 1 and that the host chose door 3. I know in the general history of the MHP, including talk on WP, the condition is taken to be the player chose door 1 and that the host chose door 3, but in this conversation it would greatly help if you were to make that clear. There are other conditions, which we could, but you chose not to, take into account. The host reveals a specific goat; we know that there are two goats and that he reveals one of them. In our solution we may choose to call this goat X (and the other one goat Y) just as vos Savant chose to number the doors. Why then does a complete solution not also require us to apply the condition that the goat revealed was goat X? Martin Hogbin (talk) 11:59, 25 April 2013 (UTC)

It's a question of systematically working through the events as they happen one by one. But at the same time keeping economical. No superfluous complications for no good reason.

(1) car is hidden behind the left, middle, or righthand door (2) player chooses one of the three doors, (3) host opens one of the other two doors, (4) player decides whether or not to switch. At each step we have to be creative/sensible and figure out what is the chance of the possible outcomes, given the history so far. You propose to complicate the story by adding further refinements (the goats have names...). You'll then have to make further assumptions about Monty's preferences for names of goats, and whether or not the player is able to identity which of the two goats he sees. Quite unwarranted, you agree; perverse, even. You also propose that we could simplify the story by not distinguishing between the doors. But that's stupid, because we can and do distinguish them! The player is able to distinguish between the door he initially chose and the other door which Monty left closed. And you can distinguish the doors too in your solution: you give each door initially probability 1/3. Your approach is inconsistent. Unsystematic. You just happened to be lucky that your shaky solution gets the right numerical answer. You don't even have a clear idea what that number means, what it represents. Sorry. Richard Gill (talk) 12:40, 25 April 2013 (UTC)

You seem to be mixing up logic with opinion. Yes, of course, you may decide that considering which goat is revealed in superfluous or that it is stupid not to consider which door the host opens, but that is just your opinion, it is not mathematics or logic.

You say, 'At each step we have to be creative/sensible and figure out what is the chance of the possible outcomes, given the history so far', I agree completely. We have to consider whether it is possible that the host's door choice could affect the outcome and whether we want to take this into account. We know that it could affect the outcome and that therefore any solution which does not take this into account can be considered incomplete. We also know a host goat preference could affect the outcome and that therefore any solution which does not take this into account can also be considered incomplete.

I shall ignore your concluding remarks. Martin Hogbin (talk) 19:33, 25 April 2013 (UTC)

If you don't distinguish the doors (by number, or right-middle-left position), what does it mean to say you have a 2/3 probability of winning by switching (from the perspective of a player who has picked a door she can see and has the seen the host open a door she can see)? For example, does it mean a player who picks door 1 and sees the host open door 3 has a 2/3 chance of winning by switching to door 2? If not, what (exactly) does it mean? -- Rick Block (talk) 03:53, 26 April 2013 (UTC)

Thanks Rick, that's the polite version of my last sentence. This is the crux of the matter. 2/3 of what? For the rest: Solving the problem systematically, methodologically, seems to me to *necessitate* distinguishing the doors and therefore also modeling both door choices on Monty's side: hiding, opening. It does not, it seems to me, necessitate giving the goats names and discussing whether or not Monty or the contestant know which goat is which. We also don't consider the fact that the contestant might hear or smell a goat. Martin's argument shows indeed that dividing lines are not perfectly sharp. However, for practical purposes they are sharp enough! Richard Gill (talk) 06:05, 26 April 2013 (UTC)

At least now we have attempts at logical reasons to condition on door numbers but not on goat revealed. I will deal with your proposed reasons individually below. — Preceding unsigned comment added by Martin Hogbin (talk • contribs)

Are the following sections meant in some way as a response to the question above, or are you simply ignoring the question? -- Rick Block (talk) 12:57, 26 April 2013 (UTC)

The sections below were intended to be answers to questions that were asked but if there is something that I missed then please let me know and I will address it. Martin Hogbin (talk) 14:08, 27 April 2013 (UTC)

The doors are not necessary

There doors serve only to hide the objects behind them thus enabling the player to choose an object without knowing what it is. Beyond that they play no part in the problem.

In my opinion (and I believe that of many others) it is quite reasonable to dismiss the doors and their numbers, without comment, right at the start of the problem because they obviously are not important. You do exactly the same with the two goats.

The problem can be expressed in terms of the car and two goats, so our sample space becomes just {C,G1,G2}. The problem can now be solved by mathematics. We (obviously) have no need to consider the fact that the door which hid the goat was labelled 3 just as we do not have to concern ourselves with the fact that the host said the word 'pick', even though the problem statement tells us that both events occurred. — Preceding unsigned comment added by Martin Hogbin (talk • contribs)

But what does it mean to model the problem this way? Specifically, if the answer is "you have a 2/3 chance of winning the car by switching" does it mean (in the limit) 2/3 of the players who pick door 1 and see the host open door 3 will win by switching? Or does it mean something else? If something else, what? To make this more concrete, assume an idealized sample of 900 shows where the car is hidden behind each door exactly 1/3 of the time (so 300 times behind each door), and the players (independently) initially pick each door exactly 1/3 of the time (so out of 900 shows, door 1 is the player's initial pick 300 times and of these 300 times the car is behind each door exactly 100 times). What set of these shows does this model pertain to? This is the meaning of Richard's question, "2/3 of what?" -- Rick Block (talk) 13:19, 26 April 2013 (UTC)

Modelling the problem {C,G1,G2} means I consider the door numbers irrelevant. It means that I recognise, right at the start, that even should I want to address the specific case in which the player chooses door 1 and the host opens door 3 (which we know was not actually the question intended by Whitaker/vos Savant or Selvin) I can still ignore the door numbers because they give me no information. It is how the problem must be modelled from a strict Bayesian perspective (well actually that should be {C,G}) because we have no information about the meaning or relevance of the door numbers so we can, indeed must, ignore them, just as you ignore the words spoken by the host; they tell you nothing.

If you want to be a strict frequentist, on the other hand, the answer is simple; the problem is insoluble. Martin Hogbin (talk) 16:12, 26 April 2013 (UTC)

Of course, you may choose to model the problem using door numbers, that is fine, and once you have done that then all you say is true, the host has a choice of two doors and may have a preference for one of them and even if he does not we should reflect the fact that the host has a choice of doors in the calculations.

If we were going to solve one of the variants of the problem in which the host is known to have a door bias then, of course, we must model the problem using door numbers. If, on the other hand we were going to solve a problem in which the host has a known preference for one of the goats then we would need to include the goat IDs in our model. Martin Hogbin (talk) 16:56, 26 April 2013 (UTC)

You're still not answering the question. In a world where doors are distinguishable (if we can distinguish left-middle-right we might as well number them), what does your "numberless" 1/3:2/3 solution mean? How many (and which) of the 900 shows in the idealized sample does such a solution pertain to?

Guessing that you either don't know, or don't want to answer this, I will. Your solution applies only to all 900, and not to the 300 where the player has picked (say) door 1 let alone to the 150 where the player has picked (say) door 1 and then seen the host open (say) door 3. If the doors are dostinguishable, this solution pertains to a problem that might be phrased as follows: Suppose you will be on a show where a car will be hidden behind one of three doors and goats behind the other two. You will pick an initial door and the host, who knows what is behind the doors, will open a different one showing a goat. Should you stay with your initial door, or switch your choice to whichever door the host does not open? Note in particular that in this problem you're deciding whether to switch before you're on the show, without seeing which door the host opens. I think if you want to move the decision point to after the host opens a door you have to consider the doors to be distinguishable. -- Rick Block (talk) 18:15, 28 April 2013 (UTC)

My 'doorless' solution obviously applies to any and every legal combination of doors, otherwise such a solution would not be possible. The point is that we can ignore the doors altogether. Thus my solution would apply equally to all 900 cases, and to 150 where the player has picked door 1 and then seen the host open door 3, and to 150 where the player has picked door 1 and then seen the host open door 2, and to 150 where the player has picked door 2 and then seen the host open door 3 etc.

In other words my 'doorless' solution applies to the case where the player has picked one of the hidden objects (but does not know what it is) and has then had it revealed that at least one of the remaining objects was a goat.

You question makes perfect sense if we frame the solution in terms of doors but having realised at the start that the door numbers, goat ID, and words spoken by the host are irrelevant we can ignore them all in our problem formulation. Martin Hogbin (talk) 21:58, 28 April 2013 (UTC)

Martin, I believe you're incorrect, and perhaps Richard Gill could comment here as well. By ignoring door numbers your solution only applies to samples that were selected by ignoring door numbers. So, it applies to any individual case or any randomly selected subset of cases out of the 900 I'm talking about, which is simply another way of saying it applies only to the entire sample of 900. The conclusion reached by this model does not apply to any subset selected by door number, such as those cases where the player picks door 1 and the host opens door 3. -- Rick Block (talk) 15:24, 29 April 2013 (UTC)

Question: What could be meant in saying: it applies "only" to any individual case out of the 900 you are talking about, and subsequently continuing in saying "but it does not apply to any subset such as player picks #1 and host opens #3 ? – I'm mazed. Gerhardvalentin (talk) 16:30, 29 April 2013 (UTC)

You have the "only" in the wrong place. If your model of the problem ignores door numbers, then if you want to pick a sample of shows (that will, in the limit, show the same proportion of winning by staying and switching as your solution) the sample must be picked in a way that ignores door numbers. Such a solution applies to an individual show (you can say that a player who has picked a door [any door] and sees the host open a different door [either other door] has a 2/3 chance of winning - note that this is a different statement than saying that a player who has picked door 1 and has seen the host open door 3 has a 2/3 chance of winning by switching), or you can say out of any number of randomly selected shows (where you are not specifically selecting these shows based on what door the player initially picks or what door the host opens) as the number of shows in your sample increases the proportion of players who win by switching will become close to 2/3 - but you cannot say out of a sample of 900 shows if you select only the shows where the player picks door 1 and the host opens door 3 that you expect 2/3 of the players to win by switching. Because (and only because) of a model using door numbers, we know that if the host chooses randomly in the case the player initially selects the car (and the car is randomly hidden at the outset, and the host must reveal a goat, and the host must make the offer to switch) we'll end up with a 2/3 chance of winning by switching regardless of which door the player picks and which door the host opens. As Richard has said before - this is a stronger conclusion, requiring "more" effort. A "doorless" model precludes making this conclusion. -- Rick Block (talk) 16:10, 1 May 2013 (UTC)

You are, of course, quite correct if we know the host has a door preference but in the given problem statement we are not told that he has one. As Bayesians we must take it that the door numbers do not matter if we are given no information to the contrary. In other words we must assume that the host has no door preference and therefore that the probability of winning by switch is the same for 150 cases where the player has picked door 1 and then we see the host open door 3 as it is for 150 where the player has picked door 1 and then we see the host open door 2 and for every other legal combination of door numbers (if we are frequentists we can arrive at the same conclusions with some reasonable assumptions or the application of the principle of indifference).

The important point is that the above independence of the door identities and the probability of winning by switching is obvious enough to most people that we can ignore the door identities right at the start, before we start to turn the problem into a clearly defined mathematical one. This removes the need to condition on door IDs later. Martin Hogbin (talk) 15:48, 29 April 2013 (UTC)

I'm fairly certain the conclusion you're making is incorrect. If you're a Bayesian and do not know about a host preference, your conclusion is that out of 900 shows 2/3 of the players will win by switching or in a single instance (where the player has picked a door any door and the host has opened some other door - these will of course be specific doors in a specific instance) the player has a 2/3 chance of winning by switching (because you're unable to distinguish this case from any other). By not distinguishing the cases, you are obligated to aggregate them in any sample you're looking at - because otherwise the "probability" you have determined may not match the observed proportion as the number of samples tends to infinity (the Law of large numbers applies to Bayesians just as much as to Frequentists). This is why many probability problems are posed as urn problems, where what is distinguishable and what is not is made explicit. Again, perhaps Richard can comment. -- Rick Block (talk) 16:10, 1 May 2013 (UTC)

Note that I am not saying that we cannot use the door IDs to formulate the problem just that, once we have recognised that they are unimportant, we can formulate the problem mathematically in a way that does not use them.

If we do choose to formulate the problem mathematically in a way that uses door IDs then, of course, we must strictly speaking, condition on any door numbers that we might consider to be specified, even if we know it makes no difference. Martin Hogbin (talk) 16:09, 29 April 2013 (UTC)

Is it just the numbers then?

It looks as though you are suggesting that we must use as conditions of the problem the fact that the player has chosen door 1 and the host has opened door 3 just because the doors have been given numbers. Can I just confirm exactly what you mean by asking you to say exactly what the conditions that must be considered are in the following hypothetical variants of the problem statement.

A) Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door and the host, who knows what's behind the doors, opens another door, which has a goat. He then says to you, "Do you want to pick door the remaining door" Is it to your advantage to switch your choice?

B) Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick door 1, and the host, who knows what's behind the doors, opens door 3, which has a goat. He then says to you, "Do you want to pick door 2?" Is it to your advantage to switch your choice?

C) Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats, Billy and Nanny. You pick a door and the host, who knows what's behind the doors, opens another door, which reveals Billy. He then says to you, "Do you want to pick door the remaining door" Is it to your advantage to switch your choice?

In each case above could you tell me whether we must condition on 1) Door numbers, 2) Goat names, 3) Neither. — Preceding unsigned comment added by Martin Hogbin (talk • contribs) 08:50, 26 April 2013 (CEST)

It is nothing to do with numbers of the doors. It is all about visualizing the problem and writing down the sequence of actual real world events. There are three doors. We visualize them, we can distinguish them. Look at all the pictures in the article: three doors in a row! Let me call them LeftDoor, MiddleDoor, RightDoor. Suppose we know also in advance that there is a boy goat and a girl goat, and suppose the contestant can tell the difference. The sequence of events is now: quiz team hides car and billy and nanny behind the three doors. There are 6 possible ways to hide these three objects and because we don't know anything else, I take them as all six equally likely. Next the player chooses a door. I don't bother to model how he chooses his door because at the moment (a bit later) that he has to make his choice, he knows his own choice. The doors are distinguishable, remember! Having chosen a door you know which door you chose. Let's say then that he picks LeftDoor because he's a socialist. Next Monty has to open a door. He has either one or two choices. We know nothing about how he does this, so we take the two actual possibilities as equally likely, if there are two; if there is only one choice, it is certain anyway. Suppose Monty opens RightDoor and reveals the boy goat Billy. We have obtained two pieces of information (1) The Door, and (2) the sex of the goat.

Because the contestant knows nothing about how the car and goats are hidden we may as well assume he made his choice (certainly LeftDoor) in advance, and after that one of the six "hidings" is determined, uniformly at random, by the quiz team, next Monty makes his choice, and so on. We can now calculate in any one of a myriad ways, the chance that the car is behind MiddleDoor, given that the player picked LeftDoor and the host opened RightDoor and revealed Billy. We don't have to have a brilliant flash of insight or come up with reasons why door door labels or goat sexes are irrelevant. It is probably safer not to try, since humans' probabilistic intuition is extremely fallible.

Why do we compute what I say we should compute? Because, relative to my mathematical model of the situation at hand, the optimal decision is the one based on the conditional probability given the situation we are in at the last moment we have to make our decision. Richard Gill (talk) 13:23, 26 April 2013 (UTC)

You may visualise things that way but to chose to assign significance to the host's choice of door but not his choice of goat, and then try to say that this is logical precision, is the kind of thing that gives statistics a bad name.

Richard, perhaps you could just answer my question, 'In each case above could you tell me whether we must condition on 1) Door numbers, 2) Goat names, 3) Neither?'. Martin Hogbin (talk) 16:20, 26 April 2013 (UTC)

I did not attach significance to anything. As I told you, I wrote down the sequence of events and gave arguments for how I modelled each step of the developing history. If you include details about individual goats in your problem description, I'm obliged to take account of them, even if only in order to argue later why they are irrelevant. I did not include any door numbers, you are the one who seems obsessed by those numbers. There is a big stage and at the back of the stage there are three large doors: the one on the left, the one in the middle, and the one on the right. At the end of my analysis it turns out that which door is opened by the host is conditionally independent on whether or not there is a car behind the door chosen by the contestant, given the door chosen by the contestant. If I had been smart I could have guessed that in advance and argued for it in advance. I recently deleted from the wikipedia article a big section which contained the mathematician's version of such an argument. People had objected that it wasn't well sourced. Since I was the main source I didn't like to add references to my own publications. You didn't add references and you didn't object when I deleted that section, and now it's gone.

I do not attach significance, I deduce significance. I reason logically and methodically, I apply well understood and well known principles and I reason and I compute with them. No "deus ex machina". No brilliant intuition, justified by inventing ad hoc brand new "axioms". Richard Gill (talk) 18:31, 26 April 2013 (UTC)

It is still not clear what your answer would be to my problem statement A above. Does your logic tell you to condition on the specific door opened by the host, the specific goat revealed by the host, both, or neither. Martin Hogbin (talk) 19:33, 26 April 2013 (UTC)

I explained to you a systematic approach to take to solve problems like this. Now it's up to you to apply my approach to each of your three problems (learn by doing).

The problem you have solved is the three cups problem. There are three apparently identical cups. Your host puts a coin on the table, covers it with one of the cups, puts the other two cups next to it. Then he rapidly slides the cups about so as to shuffle their positions thoroughly. You no longer have any idea which one covers the coin. Of course, your host can feel the coin as he slides the cups about, or the cups are secretly marked, he knows where it is. He asks you to choose one of the cups. He slides it towards you. He then picks up one of the other two cups revealing nothing under it, and asks you if you would like to switch to the remaining upside down cup.

You can *deduce* that the three cups problem and the three doors problem are equivalent, in other words, that the positions of the three doors (Left, Middle, Right) needn't be taken into account: all that counts is the relation between the visibile and invisible *role* of each door. The article had a lengthy mathematical section taken from one of my publications, giving the argument. I think this is the formal companion of the instinctive first step of some good solvers: "the numbers are irrelevant". (It's all about symmetry, a word which was found to long for a wikipedia article on a popular brain teaser). Richard Gill (talk) 06:57, 27 April 2013 (UTC)

Richard, all I am asking you to do is tell me whether in my problem A above you would consider it necessary (If we are to be logically precise) to condition on the specific door opened by the host, the specific goat revealed by the host, both, or neither. I think it is a fairly straightforward question yet, for some reason, you seem reluctant to give me a straight answer.

If you would be kind enough to just answer my question we might be able to make some progress. At present you are addressing your own imaginings of what my position is. I am happy to make my position absolutely clear if you will make yours clear. No learning is needed, just some straight talking. Martin Hogbin (talk) 08:30, 27 April 2013 (UTC)

I can't imagine your position, it makes no sense to me. And I don't answer your question for the same reason: it is meaningless to me. I told you exactly what I would do; and I told you why. I would make a mathematical model of the sequence of events in the story, and use it in order to compute the conditional probability of winning by switching given the information which, in the model, the player has at the moment he makes his choice. In order to build my mathematical model for problems A, B or C I will be obliged to give the doors names (e.g. left, middle, right). Why I condition on "everything the player knows (in the model)":? Because that's the optimal solution for the player (within the given model).

My model for your problems A and B would be essentially the same. For model C, it would be a little more elaborate. For the three cups problem it would be more simple.

The question "is it logically necessary to condition on this that or the other" is (IMHO) a stupid question. It puts the cart before the horse. *First* I make a mathematical model of my problem and *then* I figure out how to solve it. Read my paper "MHP is not a probability puzzle: it's an exercise in mathematical modelling". Richard Gill (talk) 11:56, 27 April 2013 (UTC)

I am not sure why you are telling me that. We have agreed many times that the first stage of solving the MHP is the mathematical modelling of the problem. Once that is done, the mathematical model that you set up determines exactly what conditioning is required. That is what I have been saying for years!

So can you now answer this question, in my statement A above: would the specific door chosen by the host feature in your model? Would the goat revealed by the host feature in your model? Martin Hogbin (talk) 14:17, 27 April 2013 (UTC)

A random thought

Perhaps the reason that this problem is so unintuitive is that the host helps the player to win the prize by not revealing any information as to whether the car is behind the originally chosen door. Of course, the host does reveal other information to the player but it is the fact that the host conspires not to change the probability that the car is behind the originally chosen door that enables the player to gain an advantage by swapping.

The natural assumption is that the more information about the disposition of the car and goats that the host reveals, the better placed the player is to make his choice; this turns out not to be true. Martin Hogbin (talk) 08:41, 23 April 2013 (UTC)

Not to be true? What are you talking about? After the host opens door 3, the relative odds the car is behind door 1 compared to door 2 change from 1:1 to 1:2! Switching to door 2 before the host opens door 3 would be pointless.

Yes you are right. This was just a random thought. It still remains true though that it is important that the host does not do anything that changes the probability that the car is behind the originally chosen door. Martin Hogbin (talk) 07:31, 24 April 2013 (UTC)

It's abundantly clear that the reason the problem is so unintuitive is that the conditional probabilities the car is behind door 1 (the door the player initially picks) and behind door 2 (the unpicked, unopened door) are affected asymmetrically by the host opening door 3. People are used to unknowns being updated symmetrically - as in the case where the host opens door 3 without knowing what is behind it and fortuitously revealing a goat. This is the essence of the "equal probability" assumption. The chances the car is behind door 1 and door 2 before the host opens door 3 are the same (each 1/3). Since they're the same before the host opens a door, they should clearly be the same after the host opens a door (i.e. the conditional probabilities seem like they should be 1/2 and 1/2). Of course, you don't need to speculate about this (or virtually anything else regarding the problem). Read the sources. In this case, Falk. -- Rick Block (talk) 19:06, 23 April 2013 (UTC)

Yes, I agree, the belief that probability must always be equally distributed between all the possible alternatives is a strong one. In the MHP we conspire (curiously by not giving information about whether the car is behind the originally chosen door) to make the distribution non-uniform.

My point was that it is the not giving of information that makes the problem work. I find that paradoxical. Martin Hogbin (talk) 07:31, 24 April 2013 (UTC)

See the punchline of my article in Statistica Neerlandica! Indeed it is paradoxical. "They [a frequentist and a Baysian] will likely disagree about whether the conditional probability that you win, given door chosen and door opened, is also 2/3. I think the frequentist will say that he does not know it since he doesn’t know anything about the two [how the car is hidden and how a door is opened] host actions, while the subjectivist will say that he does know the conditional probability (and it’s 2/3) for the very same reason." Richard Gill (talk) 10:24, 24 April 2013 (UTC)

I agree with what you say but are we talking about the same thing there? I am referring to the difference between 'Forgetful Monty' and the standard MHP. In the former case, the host's action of revealing a goat does tell the player (frequentist or Bayesian) something about the probability that he has initially chosen the car, in the standard problem the host's revealing of a goat tells the player nothing about his original choice. Martin Hogbin (talk) 10:53, 24 April 2013 (UTC)

I am talking about the paradoxical difference between conclusions of a frequentist and a subjectivist approach to standard MHP. Richard Gill (talk) 15:24, 24 April 2013 (UTC)

As I thought. You are taking about something different from me. Regarding your point, this goes to the heart of the meaning of the word 'probability' although the differences are easily reconcilable.Martin Hogbin (talk) 09:56, 25 April 2013 (UTC)

What if it was reformulated? You still get 3 options but 2 choices, and you win if one of them is the good one. That's a 2/3 chance right away and it eliminates all the misleading game show shenanigans. 95.33.20.177 (talk) 02:42, 6 May 2013 (UTC)

Superset [math] and Subset [math]   –   the missing step

The famous paradox (stick:swap not 1:1 but 1:2) tells us that the actual location of the objects is unknown to the contestant and unknown to us, nor do we have knowledge about "how" the contestant made his decision to select "his door". But it is well known that the host, after the contestant selected his door, is to open one of the two unchosen doors in order to reveal a goat and to offer the contestant a switch to the second still closed door. That's all we know.

Indeed. That's all we know, and this is exactly what Selvin (1975b) uses. It's exactly what I used, and exactky what Devlin (patched) uses. Please read Selvin 1975b and some other good maths papers carefully (e.g. Rosenthal). While you're at it, take a look at Gardner's Three Prisoner problem, and Bertrand's three box problem. Forget about Morgan et al. Completely irrelevant.t Richard Gill (talk) 19:06, 23 April 2013 (UTC)

So there are 3 possible scenes (superset 1:1:1):

Lucky Guess scene If in 1 out of 3 the contestant by luck selected the prize and switching will lose, the host is free to open anyone of his two doors.
Wrong Guess scene A If in 1 out of 3 the contestant selected goat A, or
Wrong Guess scene B if in 1 out of 3 the contestant selected goat B, then the host never will "randomly" open an unchosen door, but is constricted to deliberately "choose" the second goat to show, but never the car. All the contestant knows and all that we can know is that altogether, in the superset of 1 out of 3 he is in the Lucky Guess scene, and that altogether in 2 out of 3 he is in the Wrong Guess scene, The decisive failure of MCDD was to forget about "altogether" (1/3 : 1/3 : 1/3) and, by dividing the superset into subsets without saying so, treating one special "bias-subset" as if it was the full superset, deliberately forgetting about the "subsets" they blanked-out. Reducing the full scene to an examined subset only is suitable for teaching and learning conditional probability, but never represents the famous paradox.

Morgan et al. presented a correct solution of a possible though uncommon formalization of vos Savant's problem. They contributed something original and modestly notable to the field of Monty Hall studies. One can complain about various features of their paper but one can't alter the mathematical truth of their main result. You can't abolish true mathematical facts just because you don't like them. Richard Gill (talk) 16:45, 25 April 2013 (UTC)

You are right, mathematically correct. But to show that in one subset the result will be 1:1, in another subset the result will be 1:2 and in another subset the result will be 0:1, imo is treating subsets though, without saying that all of this does not apply to the superset. Mathematically correct and very notable, but hiding that this does not concern the superset of the famous paradox but only subsets thereof. Conditional probability is useful and valuable, especially if it does not conceal which mere subset it focuses. Gerhardvalentin (talk) 18:37, 25 April 2013 (UTC)

And if the host in one Wrong Guess scene was not constricted to "choose" the second goat to show but, in randomly opening of a door would have shown the second goat just by luck, then this again would be a subset only. And in that subset staying:switching is 1:1, as per the common intuitive wrong answer.

It would help the article if those subsets (Morgan et al. of 1/2:1/2 to 0:1, the forgetful host of 1:1 etc.) are clearly pointed out as subsets, never confusingly mistaken as the "superset" of the intended paradox. The "missing step" is to take a subset to being the superset, without saying so, and to forget about the blanking out. Gerhardvalentin (talk) 16:19, 23 April 2013 (UTC)

Richard, thank you for your reply. It is on "superset versus subsets". Please read carefully what I tried to say, in fact it is this: From the well defined clean scenario where the paradox arises (not 1:1 but exactly 1:2), as a valid irreversible "superset", mandatorily follows that those three scenes (Lucky Gess : Wrong Guess : Wrong Guess) unalterably are and forever will remain 1/3 : 1/3 : 1/3. Regardless which door the contestant selected, and regardless which unchosen door the host opened to show a goat and to offer a switch. No way out, if this superset is not inadmissibly reduced to any "subset" thereof, by blanking out integral aspects (e.g. by by means of inadmissibly adding any "additional assumptions" that will "blank out" parts of the given sample space). This means that no mathematical analysis (frequentist nor Bayesian) – if serious and not ludicrous – will ever be capable to change this obvious [fact of 1:1:1], but remain a cumbersome way to proof that given fact. I suggest to use that aspect [obvious fact] in your papers. I guess that's not your way of approach, just give it a try. Regards, Gerhardvalentin (talk) 11:21, 24 April 2013 (UTC)

I don't know what "mandatorily" means. This is an appeal to authority. It has no place in a logical or mathematical deduction. We solve a brain-teaser by logical deduction, not by papal decree. The fact that the probability that door 1 hides the car is not changed by seeing door 3 opened by the host follows by a very easy logical or mathematical deduction. I do not think that the argument is cumbersome. Some mathematicians find it is so easy to prove that they do not find it worth the trouble to write down any mathematical detail.

For mathematicians it is sufficient to remark: given the player chose door 1, it follows easily by symmetry that whether or not it hides the car is independent of whether the host opens door 2 or door 3.

In a probability theory class-room one might want to write out a formal proof. But among professionals it is superfluous.

If we are talking about a popular discussion of the Monty Hall problem in an internet encyclopedia, it seems to me sufficient to state "in the absense of any knowledge of how the host chooses to open a door, the information which door he opens has no impact on the probability the car is behind door 1". Richard Gill (talk) 12:52, 24 April 2013 (UTC)

mandatory? necessaryly, inevitable... – The article is full of references to experts in maths, all treating the effect (and the problem) of reducing the superset by blanking out integral parts of that superset. No jota of "shaky" if blanking out (without saying so) is not "trendy" to those experts like a duck takes to water. And the "conditionalists" still think that understanding the paradox needs conditional probability theory, "otherwise shaky". The article should show that that does not apply. You "can" use maths, but you do not "need" it. Just be careful not to "blank out" what is given in the superset. Gerhardvalentin (talk) 13:31, 24 April 2013 (UTC)

Gerhard: perhaps you should try to understand *why* all those experts (Selvin, Devlin, Rosenthal, ...) think that the Monty Hall problem should be solved by conditional probability, rather than trying to enforce a ban on careful thinking and precise argumentation. The Wikipedia article, I think in a balanced way, now first presents a selection of simple, verbal, intuitive arguments; and later presents a selection of more refined arguments. The reader can choose for themselves. You however want to forbid the general reader from finding out what the experts think. While you're at it, take a look at Monty Hall's ancestors: Martin Gardner's Three prisoners problem and Bertrand's box paradox. Richard Gill (talk) 17:50, 24 April 2013 (UTC)

Thank you. It's about the experts. It is about mechanically "blanking out", without explicitly saying so, and about consequently not treating the full superset of the famous paradox, but mechanically treating subsets only by using conditional probability. The excessive maths literature is full of "overall 2/3" quite contrary to "actually". Your proper efforts helped. But, within the undamaged superset of the paradox, such antagonism can never exist. Utterly impossible. Please try to read what I said above. Gerhardvalentin (talk) 11:21, 25 April 2013 (UTC)

I don't understand you at all. I see no antagonism. Neither in the article, nor in the literature. I get the impression that you have a rather strong personal point of view which you want to impose on the article as a whole. Sure, maybe your point of view is the same as that of some reliable sources. But is it different from that of yet other reliable sources. Some readers of this wikipedia article will come from mathematics, others come from the general public. Like it or not, you have to cater to readers and to sources of both categories. That's a challenge. Richard Gill (talk) 11:42, 25 April 2013 (UTC)

I see that you don't understand me at all and don't see the antagonism between "door 3 opened" and strategy, but the literature is full of this antagonism: "players who always swap have a 2/3 chance of getting the car" – you see the antagonism? And "the player is never worse off switching".
Rick above did clearly specify this antagonism in saying to me on 17 March:

Why not use one of the clearly correct (and sourced) approaches that do just this by first examining a strategy of switching vs. a strategy of not switching, e.g. the approach Carlton says he uses in his classes, or as published by Grinstead and Snell? By explicitly changing the reader's focus from the example situation where the host has opened door 3, to the slightly different problem of whether switching or staying is the best overall strategy (without knowledge of which door the host opens - i.e. moving the decision point to before the host opens a door), the analysis is simple and straightforward ...

Blanking out and treating "subsets" of the paradox, combining them with another "subset" of the paradox using probability theory may be fine if experts would say what they do. But blanking out, and forgetting about the superset of the paradox imho is not the wisest of options nor the ideal solution. We should not lose sight of the full superset of the scenario, where that antagonism never did exist. Gerhardvalentin (talk) 13:21, 25 April 2013 (UTC)

I still have no dea at all what you're talking about, sorry. Richard Gill (talk) 17:22, 25 April 2013 (UTC)

"Before and After"

Gerhard - Is it a paradox that by choosing one door out of three you have a 1/3 chance of picking the car and win by switching (with a 2/3 chance) if it is behind either of the other two doors, i.e. by committing to switch before the host opens a door? Or is the paradox the situation after the host opens a door, where you're looking at only 2 closed doors (say, door 1 which you've initially picked and door 2 given that the host opened door 3), and the chance the car is behind the door you originally picked remains 1/3 while the chance it is behind the one (not two!) other unopened door has (magically) become 2/3?

The "after" situation requires that we talk about a subset. If you restrict yourself to the superset, you're talking about the "before" situation - which (IMO) is hardly paradoxical at all. Talking about the "before" situation is what makes the "simple" solutions simple. It completely avoids the paradox. It changes the problem into an exercise in mathematical obfuscation, i.e. a complex way of asking whether you would prefer to go home with the car if it's behind only one door or both of the other doors. Well, duh!

IMO, the paradox is after the host opens a door (say door 3), by switching (to, say, door 2) you still double your chances even though the car is now behind only one of two doors. If the host opens door 3 the chance the car is behind door 3 is plainly 0. If the car has a 1/3 chance of being behind door 2, where does the other 1/3 chance of winning by switching (after the host opens door 3) come from? It's certainly no longer the case that you win the car with a 2/3 chance because there's a 1/3 chance it's behind either door 2 or door 3. Something else is going on. What is it, exactly? This is the paradox. -- Rick Block (talk) 16:00, 25 April 2013 (UTC)

Rick, you are seeking the permanent old blue-eyed naive amazement: how can the other unopened door have (magically) become 2/3? – Did not both unselected doors originally have a probability of only 1/3 each to hide the prize?

So let's go back to the scenario of the clean paradox: It says that there are three doors but only one car, okay? And did you ever bother to imagine that - as soon as any two of those three doors (say #2 and #3 e.g.) have been singled out - they will hide:

door #2   door #3

  goat       goat

  car         goat

  goat       car

 

There evidently is the "either-or-chance" to hide the car of

door #2 door #3

  2/3        zero          no magic,  there evidently is asymmetry,  and "still 1/3 + 1/3" is myopic

  zero      2/3            no magic,  there evidently is asymmetry,  and "still 1/3 + 1/3" is myopic

This is valid for any two armed doors as soon as they randomly will be singled out of three, if there is only one prize and no further info is given. Evident and no magic at all.

In 1 out of three, both doors will hide goats. But in 2 out of 3, one of both doors definitively hides the car, we just still don't know which one. In other words: as soon as any two ( ! ) of three doors have been singled out, one of them (we just still don't know which one) does hide the prize with probability 2/3. Once more: we still don't know which one, but we know that definitely just one of them hides the car with probability 2/3. Nijdam used to say "no, both singled-out doors have a probability of 1/3 each to hide the prize".

Now that we know for sure that at least one of those two singled-out doors will hide a goat (unknown which one), and that we know that the "other one" (unknown which one) hides the prize with probability of 2/3, there is no "magic" any more after we have been told "which one" of both singled-out doors "is which one". No more magic at all, we know it in advance that one of those two doors has a 2/3 chance to hide the prize, while any "second one" will be empty for sure.

Before and after? – Let's be serious. What I said is known to apply to "any" two doors that are singled out of 3 doors. Okay? We know that even before the contestant has made his choice. And we know that this applies to any (!) of two unchosen doors. We know that beforehand, if we regard the superset of the paradox, and not only subsets.

Can you agree with the following:

The answer "1/3 : 2/3" depends on the rules of the world-famous paradox, requiring the host to always reveal a goat. This means that in the 2 out of 3 cases, when the player originally has chosen a goat and switching wins, the host must reveal the remaining second goat, thus ensuring that a player who swaps wins the car in every such case. However if, contradictory to the rule of the standard paradox, the host just opened a door at random and revealed a goat simply by chance, the overall outcome staying:switching would have been reduced from 1:2 of the standard paradox, to only 1:1 as per the common intuitive wrong answer, because half of the potential winning cases are wasted when the host can accidentally reveal also the car.

"Before and After" is an illusion, imo induced by seeing and treating subsets (host's preference e.g.) but forgetting about the superset of all given results / outcome. Gerhardvalentin (talk) 17:45, 25 April 2013 (UTC)

This explanation (that one of the two doors has probability 2/3, you just don't know which one) is presumably your OR, and it's simply wrong. The two doors together, before the host opens one, indeed have a total probability of 2/3 (the probability of each is 1/3, so two together adds up to 2/3) but this doesn't mean that either individually has a 2/3 chance after the host opens the other one. Indeed, by focusing on the superset and saying that the host must reveal a goat and therefore you have a 2/3 chance of winning the car by switching, you're only talking about the combined probability (1/3 + 1/3) - which applies only before the host opens a door. If you commit to switching at this point (before), you don't know which door the host will open or which door you'll be switching to, and because of this you have a 2/3 chance of going home with the car, i.e. you're still in a situation where you go home with the car regardless of whether it's behind door 2 or door 3.

But this doesn't mean that the chance the car is behind door 2 after the host opens door 3 is 2/3! The biased host variant amply demonstrates this.

The magic that turns door 2's probability from 1/3 to 2/3 when the host opens door 3 is conditional probability - and this magic only works (ends up 1/3:2/3) if the host is not biased! In fact, unless you're talking about conditional probability, when the host opens door 3 the probability the car is behind door 2 does not change at all. It remains 1/3! If the car is behind door 2 the host must open door 3 (probability 1), so the combined probability of the car being behind door 2 and the host opening door 3 is 1/3 * 1 = 1/3. What actually changes is the probability the car is behind door 1! What is the probability the host opens door 3 if the car is behind door 1? 1/2, right (because he's unbiased)? So the combined probability the car is behind door 1 and the host opens door 3 is 1/3 * 1/2 = 1/6 (note that this comes out differently if the host is biased, but it's never greater than 1/3). This is only half as much as the probability the car is behind door 2 and the host opens door 3. So if the host opens door 3 you double your chances of winning the car (from 1/6 to 1/3) by switching (and with a biased host, you're never worse off by switching). Only if you express these as conditional probabilities (talking about the subset) do they scale up to 1/3 and 2/3. Another way to say this is if you're going to say the probability the car is behind door 2 is 2/3, you have to be talking about conditional probability. If you're not talking about conditional probability, the probability the car is behind door 2 starts 1/3 and either stays 1/3 (if the host opens door 3), or drops to 0 (if the host opens door 2 - the host has a 0% chance of opening door 2 if the car is behind door 2, so the combined chance the car is there and the host opens this door is 1/3 * 0 = 0).

The "superset" starts as

1=1/3 + 1/3 + 1/3

The host opening one of door 2 or door 3 splits this into

1=(1/6 + 1/3 + 0) + (1/6 + 0 + 1/3)

where the term on the left are the probabilities if the host opens door 3 while the term on the right are the probabilities if the host opens door 2. There's never any 2/3 in the superset (other than the sum of two of the 1/3's). In particular, it doesn't start (as you're suggesting) as

1=1/3 + ((2/3 + 0) or (0 + 2/3))

The 2/3 comes only from either of the subsets, by expressing the probabilities in one of the subsets as conditional probabilities. -- Rick Block (talk) 01:20, 26 April 2013 (UTC

Thank you. No, it's not OR. It's the definition of the paradox that says that there are three doors but only one car. This is not OR. And after any two of those three doors have been singled out, then those two doors cannot hide two cars. This again is no OR. They can hide two goats, yes, but they will hide at least one goat. No OR, no, that these two doors are to hide at least one goat is no OR. You say above that the two doors together, before the host opens one, indeed have a total probability of 2/3. That's right. And you know in advance that at least one of them must hide a goat. At least one of them, but you don't know which one. What is the probability to hide the car of that door, of that one (unknown which one it is) door that must hide a goat? You still stick on "before and after", still stick on MCDD. Didn't you notice the huge mistake they made? It is not much to see their huge mistake, it's easy to see because it's so oversized. I'll be back after a break. Thank you once more. Gerhardvalentin (talk) 02:13, 26 April 2013 (UTC)

Yes any two together have a probability of 2/3. Yes there must be at least one goat behind any two. But this does not ensure the probability the car is behind one is 2/3 when the other is opened showing a goat! Per Falk: "The host can always open a door revealing a goat [equivalent to your assertion about 2 doors and 1 car] and the probability that the car is behind the initially chosen door does not change [same as your assertion that the one door ends up with probability 2/3], but it is not because of the former that the latter is true." The biased host variant proves this (by example!) - in this variant the host always opens one of the two unpicked doors, but the probability does not end up 1/3:2/3. If all that is required is 1 car behind 2 doors with a total probability of 2/3 and a host forced to show a goat, then in this variant as well the probability would have to be 1/3:2/3.

But it's not!

Only in the standard version (with an unbiased host) is this true. If the host is (or assumed to be, which is perhaps the natural assumption if you don't know about any bias) unbiased, it's true. If the host is biased unevenly, it's not true. So, an unbiased host (or an appeal to indifference because of lack of knowledge about any bias) is a critical addition to the scenario. Just as important as the host knowing what's behind the doors (and always making the offer to switch).

You might say "but the question is, should you switch?" Indeed it is. But the situation you're being asked about is after the host has opened a door (not before). In this situation, with an extremely biased host, the answer might be "it doesn't matter whether you switch or not". This is not the answer you want (assuming you want the answer to be 1/3:2/3), so just add in as a condition that the host is unbiased. It's very simple. Pretty much everyone does it these days. And then the conditional probability in any of the 6 specific cases must end up the same, and the (conditional) probability the car is behind door 2 (given the host opens door 3) really is 2/3 - but still not because of the reason you're saying. -- Rick Block (talk) 03:45, 26 April 2013 (UTC)

Exactly. Gerhard's arguments are OR and (in my opinion) they're wrong. Forget about Morgan. Read Bell's comment on Morgan. Read Rosenthal, Gill, or any of a hundred other professional sources. Read the second part of our wonderful Wikipedia article! Learn some elementary probability theory. Richard Gill (talk) 05:55, 26 April 2013 (UTC)

Rick and Richard, both of you are wrong, see the literature. Probability theory unnecessary. As soon as  "any 2 armed doors" (out of 3 armed doors) will be singled out, they "always" must hide at least 1 goat (yet in 1 out of 3 even both armed doors hide goats), and "in 2 out of 3" they can hide never more than only one car.

So, long before a contestant made his first selection, and long before any host's action, "at least one" of those any two armed doors (later on, in advance only the host knows which one) did not hide the car. No probability theory. This is known not only to the host, but is known to the contestant also. We have to account for both views, and so does the literature. Probability theory unnecessary in this respect. Item: No "miracle". Anyone who ever believed in "mysterious jumping"? Gerhardvalentin (talk) 22:10, 30 May 2013 (UTC)

Richard, admonitions like 'Learn some elementary probability theory' do nothing to help resolve this dispute. I am sure that Gerhard, like myself, is fully aware of all the probability theory necessary to solve the MHP, it is not that hard.

The way you use the word "condition" appears to contradict this. I also get the impression neither of you have read the second part of the wikipedia article. Sorry. We wouldn't be having this discussion if you were familiar with the arguments presented there.

Richard Gill (talk) 12:58, 28 April 2013 (UTC)

A condition is just an event, any event. Any probability can be conditioned on the occurrence of any event. Conditioning may or may not change the value of the probability of another event occurring depending on whether the events are independent or not. What is not to understand?

On the other hand you, many times, have used the term 'conditional' about a probability without making clear what event you are conditioning on. That is meaningless. Martin Hogbin (talk) 16:07, 28 April 2013 (UTC)

The difference in understanding comes elsewhere. As you have pointed out above, before we can attempt to solve the problem with probability theory, we have to first construct a mathematical model of the problem. Mathematics or even strict logic cannot help us with this. We can only consider what events might affect the probability (as yet formally undefined) of interest.

There is no reason, after consideration of all the relevant facts in the problem statement and consideration of what our audience will find obvious and therefore not worth mentioning, that we should not, for example, model the problem in terms of a sample space of {C,G}, without comment or explanation. Some may prefer to model the problem differently, taking account of the goat revealed by the host for example, but there is no right or wrong way or even a 'logically precise' way to model the problem. Once you start to understand that other people may, with complete justification, model the problem differently from you you will begin to understand what Gerhard, myself, and many others are saying. We cannot be proved wrong with probability theory because this argument is beyond its scope. Martin Hogbin (talk) 15:34, 27 April 2013 (UTC)

Rick you are wrong with your "biased host", see Falk. If your "biased host" opens his "strictly avoided door" to show a goat and his unopened second door hides the prize, then this is not because the host was biased. If that event occurs, it is a wrong conclusion to say "I knew that beforehand because the host is extremely biased." See Mlodinow who tells us about a candidate who "knew" that his number 48 will win, "because 7 times 7 is 48".

And you are wrong with the other unopened door that "magically" did become 2/3. You knew before the host showed the goat, that behind the door opened there already WAS a goat. He always has at least one goat. No magic, the paradox does not provide for two cars. No wrong conclusions, no matter of probability theory. No magic!

That in a specific situation of the clean paradox, after the host opened a door, the chance to win by switching is "not 2/3" but with a biased host could be 1:1,  1:2,  0:1 or anything in between, is incorrect and wrong. See Falk. The probability to win by switching of 2/3 is not merely on the fact that there always will be at least one goat that can be shown, but also is firmly based on the scenario of the paradox that DOES NOT GUARANTEE any biased host. Falk does NOT SAY that it is necessary to know that the host is not biased, but in contrast she very clearly says that, if you should like to treat any host's bias, then this is possible only if the host IS BIASED and you "KNOW ABOUT HIS EXISTING BIAS." But you don't. The famous paradox does not guarantee any biased host. Addressing any host's possible bias without saying that this is contrary to the valid paradox is devious an insincere.

You and maths and anyone else can "assume" what you like, and blank out 1/3 of possible outcomes, treating only a subset thereof, but then you and any others should say so. If someone treats such subset only, then it does not reflect the paradox. They should not cheat but finally say so.

In one rounded per mille of 100 millions of cases, switching will win in 66'667 cases and staying will win in 33'333 cases.

If the host should be extremely biased to open the door with the highest number, the result is identical. If he should be biased to open the door with the smallest number only, the result is identical also, if you do not blank out 1/3 of all possible events.

As the paradox does not guarantee a biased host, it is simply a wrong conclusion to say "I knew that result beforehand, because 7 times 7 is 48". You must say that, by inventing your biased host, you are blanking out 1/3 of possible events, and you must say that you are treating only a subset of events. And never the "paradox". Saying this addresses the paradox is insincere.

For the famous paradox, there is no "before and after", because "after" (!) the host of the famous paradox opened a door to show a goat, the probability to win by switching is 2/3, as per Falk. Gerhardvalentin (talk) 15:44, 27 April 2013 (UTC)

Gerhard, you are confusing assumptions from deductions. Because we know nothing about any possible host bias, the two choices of the host are equally likely, when he has a choice. Because of this, given the player chose door 1, whether or not the car is behind door 1 is independent of whether the host opened door 2 or door 3. Therefore the probability of winning by switching is 2/3: the same before as after.

We do not need hundreds of words of polemics. We just need a few clear words of logical deduction. To which you should add, on wikipedia, suitable sources which confirm that deduction. Understand probabilistic reasoning, read the sources. Richard Gill (talk) 06:22, 29 April 2013 (UTC)

To address the weak arguments of experts in conditional probability theory is no polemic. If there are two cups and you know that those two cups have a joint chance of 2/3 to hide the only coin, knowing at the same time that (at least) one of them definitely MUST BE EMPTY and consequently cannot comprise any chance, then it is beside the point to predicate that this "no chance cup" has a chance of 1/3 though to hide the only price. You may say that this "no chance cup" has a chance of 1/3 though, and you can proof this by methods of conditional probability theory. IMHO not authoritative but absurd.

And that the famous intended paradox says that, under its clean scenario, in each and every case swapping will double the chance to win is correct. The intended paradox does NOT provide for any foisted "might be possible" additional information that never may be planted in, belated. The intended clean paradox is what it is, it does NOT guarantee that any inapplicable "additional information" perfidiously might be planted in as an "integral addition". Either you understand and respect the obvious conclusiveness of the intended paradox, or you don't. I prefer to bid farewell to all that attempts to distort and to falsify. As per Falk we have to accept that the intended paradox does not provide for falsifications. We have to bid farewell to "before and after" with regard to the clean paradox.

Of course you can show that famous mathematicians presented dissenting scenarios with subsets of the probability space and with mathematically correct conclusions. Not having understood the conclusiveness of the intended paradox, and hiding that they did not even intend to address the existing clean paradox. As a mathematician you can try to mend that misconception, to mend and to treat absurd falsifications hereof by methods of probability theory. To remigrate and converge again to the superset of the clean paradox. I welcome your effort, but I prefer to follow the sources that are unmasking such attempts to falsifying. Gerhardvalentin (talk) 11:36, 29 April 2013 (UTC)

Three cups problem

There are three apparently identical cups. Your host puts a coin on the table, covers it with one of the cups (bottom up), puts the other two cups next to it (also bottom up). Then he rapidly slides the cups about so as to shuffle their positions thoroughly. You no longer have any idea which one covers the coin. Of course, your host can feel the coin as he slides the cups about - or perhaps the cups are secretly marked, he can see which one covers the coin. He asks you to choose one of the cups. He slides the one you chose towards you. He then picks up one of the other two cups revealing nothing under it, and asks you if you would like to exchange the two cups whose contents are still concealed: yours and his.

Marilyn vos Savant clearly thinks that the three doors problem and the three cups problem are the same. The "simple solutions", in particular the "combined doors solution" are excellent solutions to the three cups problem. Monty Hall demonstrated the cup problem to Luke Tierney (NYT).

I think one can solve the three door problem by solving the more simple three cups problem and then arguing (whether formally or informally, briefly or at length) that the door numbers are irrelevant. vos Savant and many others found any argument superfluous. Devlin, Rosenthal, Selvin and many others found it necessary. Bell thought it a matter of taste.

This strategy for solving the problem - realizing it is the same as a more simple problem - is one of the strategies proposed by Kraus and Wang. They call it "less is more". You have to get away from the static image of the three huge doors at the back of the stage. It misleads us (two closed doors - must be 50-50!). The three cups, moving about on a table, allow us to see our choice, vividly, as the choice between one randomly chosen cup versus the other two. Richard Gill (talk) 12:13, 27 April 2013 (UTC)

Could you briefly give your reasons why the 'The "simple solutions", in particular the "combined doors solution" are excellent solutions to the three cups problem'.Martin Hogbin (talk) 08:35, 28 April 2013 (UTC)

Because the cups are indistinguishable to the player. The first move is not "the quiz team hides the car behind one of the three doors". No. The first move is: the player picks one of the three cups. The host reveals that there is nothing behind one of the other two. The only randomness in this story (viewed from the point of view of the player) is whether or not the player initially picked a cup or not. However with three distinguishable doors we start with the quiz team hiding the car behind one of the three doors, completely at random. The player picks the left hand door because he's a socialist. The host chooses (or is forced) to open the middle or the right hand door. The player is then offered the opportunity to switch to the right hand or the middle door accordingly.

Because the player can't distinguish between the cups, a natural choice of outome space for the three cups game is the space of two elements describing what is hidden under the cup chosen by the player: {coin, nothing} with probabilities 1/3, 2/3. The host reveals nothing under one of the other cups. The player can choose whether or not to switch, the other cup contains nothing, or the coin, according to whether or not the initially chosen cup had a coin or nothing.

With the three doors problem, a natural outcome space has at least four elements. Read the article for the details - the tree diagrams have four end points with probabilities 1/6, 1/6, 1/3, 1/3.

It's also educative to look at Bertrand's three boxes problem. Actually three cabinets, each of which with two drawers, each containing a gold or a silver coin. The natural probability space has six elements: which cabinet was first selected, which drawer of that cabinet. Kraus and Wang, and the article, present two main ways to solve three doors problem: one by merging outcomes and reducing to a two element outcome space, one by splitting and expanding to a six outcome space. You'll find in the wikipedia article a solution of Bertrand's problem "by symmetry" as well as the conventional solution with big outcome space.

It's also educative to look at The Economist solution: Door 1 is the door hiding the car, the randomness is exclusively in the initial choice of the player. An outcome space of three elements of equal probabilities 1/3. See also Georgii's textbook solution. Same approach. Door 1 is the door hiding the car.

For the three cups problem, the player obtains no information what ever before he has to make his choice. In the three doors problem he does (middle or right door is opened). You can ignore this information without saying anything, or you can argue why it should be ignored. I'm saying that if you build up your probability model in a systematic way you can't avoid saying something. Richard Gill (talk) 12:53, 28 April 2013 (UTC)

I only wanted your first sentence, 'Because the cups are indistinguishable to the player'.

At the time that player makes his original choice the cups are sitting on the table, where the player can see them. The player can distinguish between them in exactly the same way that he might distinguish between three identical doors, left, middle or right (or whatever other pattern they might be arranged in). Martin Hogbin (talk) 13:36, 28 April 2013 (UTC)

Read the other sentences. You misunderstand the first sentence (I had better delete it). It was a too brief summary of my explanation. Richard Gill (talk) 06:14, 29 April 2013 (UTC)

So are the cups indistinguishable to the player or not? To put it in a better way, does your problem statement define the cups to be indistinguishable to the player? Martin Hogbin (talk) 08:39, 29 April 2013 (UTC)

The intelligent and constructively inclined reader understands perfectly well the idea of my puzzle: the player can't distinguish between the cups, the host can. Of course anyone who wants to be difficult can come up with objections and alternatives. Richard Gill (talk) 09:04, 29 April 2013 (UTC)

There is no need to reply to a simple questions with a thinly veiled insult, a simple, 'Yes, I do define the cups to be indistinguishable to the player' would have sufficed.

The intelligent and constructively inclined reader understands perfectly well the idea of the MHP and understands right at the start that door numbers or other door identities were not intended to be part of the problem. The problem is about three objects, all hidden from view. One is picked by the player and another is revealed by the host. The doors are obviously irrelevant.

All I am doing is insisting on a consistent standard of logic throughout our discussion. Either we take a problem as it is obviously intended (which I fully support) or we look for weaknesses and loopholes in the arguments. Which would you like to do? Martin Hogbin (talk) 09:24, 29 April 2013 (UTC)

Neither. I proposed a systematic approach to building a probability model for problems of this kind. You haven't commented on it yet. I applied the approach to several variant problem statements. That brought out a clear difference between the three cups problem and the three doors problem.

You stated about MHP: "the doors are obviously irrelevant". That is your personal opinion. It's not everyone's personal opinion. It was Marilyn's opinion, but it wasn't Selvin's opinion.

I showed how you can *deduce* that the doors are irrelevant, using only well accepted principles of probabilistic reasoning.

I offer the three doors cups variant as a pedagogical device in order to illuminate the issues which we are talking about here but if you continue to insist on "the doors are obviously irrelevant" then discussion is terminated. (We know that vos Savant thought they were irrelevant). But she is not the only person who has ideas about how the problem should be solved). Richard Gill (talk) 15:51, 29 April 2013 (UTC)

Richard, I do not know why you get so agitated. All I wanted to know was why you considered the three cups problem to be different from the MHP. All you had to say was that you defined the cups to be indistinguishable. There is no need for your illuminating device though because I perfectly well understand that the doors are distinguishable in the MHP and that, if you choose to make them part of your mathematical description of the problem and if you understand the question to be referring to specific doors then we must use conditional probability, even if it makes no difference to the value of the answer. Goats, cars, doors, and words are all distinguishable and, unless the problem statement defines them otherwise, we must treat them as so..

I am not the one being dogmatic. I chose to address the problem differently from the start and, as I am a Bayesian and I have no information to the contrary, I take the player's choice of door, the host's choice of goat to reveal, legal door to open, and words to say to be unimportant. I therefore set up the problem mathematically in a way that does not involve any of these unimportant choices. A frequentist might chose to apply the principle of indifference to the above choices an come to the same conclusion.

I do not claim that my approach is the only one or even the best one but it is as logically precise as any other. I do not even object if you say that I must give my rationale for ignoring these choices at the start of the problem. I would only insist that you do the same and explain why you have not included the host's choice of goat in your solution. Martin Hogbin (talk) 17:22, 29 April 2013 (UTC)

I started a new thread in order to introduce a new topic (I'd said everything I wanted to about choosing goats in a previous thread.) I hope that the difference between the three cups and the three doors problems might be useful to editors wanting to understand the rationale behind conditional solutions to MHP. However if like vos Savant you are immediately instinctively certain that the problems are identical, then there is nothing for you to think about. You already made the creative problem reduction ("less is more") which is one of the smart ways to solve MHP, and you made this step without noticing it. Like Marilyn herself, in fact. Richard Gill (talk) 19:41, 29 April 2013 (UTC)

Richard, you go from harsh words to kind words; I guess I cannot complain about both, but I do not think any special talent or insight is required see that the door numbers do not matter; most people seem to understand that perfectly well.

You still have not quite got my point though and as long as you continue to sweep goats under your carpet you will never get it. The solutions which condition on the door number chosen by the host are better than the ones which do not (but not by much); the solutions that also condition on the specific goat revealed are better still (by even less), and solutions which also condition on the words spoken by the host are even better (by less still). You pays your money and you takes your choice but nothing, in my opinion, makes the 'conditional' (on the door opened by the host) the 'correct', or 'the logically precise' solution. It has an arbitrary level of rigour that is neither fish nor fowl. Martin Hogbin (talk) 15:48, 30 April 2013 (UTC)

I argued that, in the three doors problem, modelling the sequence of actions of quiz-team, player, and host can’t be done without keeping track of door identies. It thereby necessitates modelling the host’s choice and thereby automatically necessitates us to probabilistically condition on what the player has got to see at the moment he must make his decision. I explained how a similar modelling process in the three cups problem leads to the player making his decision with no modelling of the host’s choice and nothing random in the history to condition on.

I therefore vehemently dispute that the standard way in which every trained probabilist or statistician models the three doors problem is neither fish nor fowl. It certainly does not have a level of rigour which is completely arbitrary. On the contrary. It has a level of rigour which is entirely natural and appropriate.

You could argue that in this particular case it leads to more work being done than is necessary. Having got to the 2/3 : 1/3 answer, the same as in the much simpler three cups problem, you could ask yourself if it would not have been cleverer still to argue the irrelevance of the door identities in advance, after which the much simpler three cups analysis is justified. Richard Gill (talk) 07:19, 2 May 2013 (UTC)

The 3 cups problem does not introduce a problem in which the "door numbers" do not play a role. Furthermore, I'm getting rather tired of hearing: "the door numbers do not matter". In fact: the host does not matter, the wheather does not matter, the player does not matter, etc. It's easy to write down such statemants, their meaning is not clear, until you specify what "does not matter" actually means. I said so many times before, and still has to repeat it here. Martin, show me a sane analysis, in mathematical correct terminology, and you will see, there is nothing to gain in saying the door numbers do not matter, nor will it justify vos Savant's mistake.Nijdam (talk) 14:49, 1 May 2013 (UTC)

Why the door ID that the host opens does not matter

Let us start at the beginning with the W/vS problem statement. Do you agree that before we can produce a 'a sane analysis, in mathematical correct terminology' we must first decide what the question is asking? Martin Hogbin (talk) 15:08, 1 May 2013 (UTC)

I think we must first form a mathematical model of the situation we are talking about. I have already proposed an answer to this, or rather, I proposed that our model should be structured according to the causal sequence of events, so as to take account of causality: causality works forwards in time, not backwards. First a car was hidden behind one of three doors. Then a player picked a door. Then the host who knows where the car is, picked and opened a different door. Can we agree on that? From here on, everything is plain sailing, I think. Richard Gill (talk) 10:01, 2 May 2013 (UTC)

From the recent discussion and from Rick's patient responses I hope to see that we are getting a step forward, maybe. I want to put a question to Martin and to anyone here, it concerns the known standard paradox (chance staying:switching said to be 1:2 and not 1:1).

1. Three doors, behind one door is the prize, behind the other two doors are goats, location of the car is unknown, and the host did not yet arrive because of a traffic jam.
Is it correct to say that, without host, as soon as 'any' 2 doors out of 3 (imaginary or real) should have been singled out, they will hide the car with either a chance of "0 : 2/3" resp. of "2/3 : 0". – And is it correct to say that the probability for these two doors to hide the car is either "0 : 2/3" or alternatively "2/3 : 0" ?

2. Let's assume that the host arrived meanwhile, the contestant has already selected his first door and the host has already opened another door to reveal a goat and to offer a switch to the second closed door, but the contestant declined the offer, so his last choice was still his door that he initially did select. And now the host is just going to open it.

2a: The contestant and the audience can see that there is goat A. What is the conditional probability now that the door opened is a loser and he is in the Wrong Guess scene?

2b: The contestant and the audience can see that there is goat B. What is the conditional probability now that the door opened is a loser and he is in the Wrong Guess scene?

2c: The contestant and the audience can see that there is the only car. What is the conditional probability now that the door opened wins and he is in the Lucky Guess scene?

Because Mlodinow says that there is only one Lucky Guess scene, the contestant having first selected the car and switching will lose (probability 1/3), and there are two Wrong Guess scenes, the contestant having first selected goat A or goat B and in both cases switching will win (probability to be in a Wrong Guess scene 2/3). Obviously the actual scene that the contestant actually is in, is unchangeable. Overall, the rate of  1/3 : 2/3  therefore is constant. Now to the host's interaction:

As to questions 2 – 2c: Suppose the player already made his last decision (not to swap but to stay), but before the host opens that door, he says before in 2a and in 2b: "behind your door is a goat", and he says before he opens the door in 2c: "behind your door is the car". What is the conditional probability in those three cases, after the host has told the contents of the actual door, but before he opened it, that the contestant wins by staying (resp. would have won by switching, but he didn't switch)? That he is in the Wrong Guess scene resp. is in the Lucky Guess scene? Imagine that only thereafter the host opens the actual door to show what he said was correct.

If you could answer my stupid questions, you would help me to understand, and if I'm wrong please help me with sources so I can see differences. Regards, Gerhardvalentin (talk) 21:48, 1 May 2013 (UTC)

I have no idea what you mean by 0 : 2/3, Gerhard. And I don't know what you mean by Goat A and Goat B. Do you want to suppose that we all (host, player, audience) know in advance that there is a boy goat, called Billy, who wears a blue ribbon, and a girl goat, called Nanny, who wears a pink ribbon; and that the host knows not only where the car is hidden but also where the boy and the girl goats are hidden? And that when a door is opened revealing a goat, everyone can see which goat is there? Why make the problem more complicated when we already have difficulty discussing the simpler version where no one cares about the identities of the two goats?

However, if you insist, it's a simple exercise to mathematically model this extended three doors problem, by focussing on the sequence of actions of quiz-team, player, and host, and using Laplace's principle of indifference to assign probabilities (representing the player's knowledge or lack of knowledge) whenever the various actors have to make choices. Having done this, one can compute the player's conditional probability of winning by switching given the player's information at the moment he must make his choice. One will be able to deduce that the specific door numbers and goat names are irrelevant. In probability language: whether or not the car is behind the door initially chosen by the player is statistically independent of the number of the door opened by the host and the sex of the goat behind that door. In retrospect, this statistical independence is obvious, and it could have been argued for right at the start, rather than "discovered" by modelling everything and just calculating away.

I wrote down natural probability models, following the principles I have laid out, both for standard three doors problem and for standard three cups problem and I showed that the natural mathematical models for the two problems were different. However, in the three doors problem we can *deduce* that door identities are irrelevant, and therefore can be ignored. Having done that, the three doors problem *reduces* to the three cups problem.

Probability theory gives us a well established and well understood framework for systematically, methodologically, solving problems like these. Probability theory provides us with a language which allows us to make necessary distinctions, to work at an adequate level of precision. Probability theory allows us to solve problems in a routine way; no need to have a brilliant flash of insight. No need to introduce ad hoc axioms in order to deal with each new problem. Of course, you might occasionally in this way discover that the answer to some problem is striking or simple in some respect, and that might lead you to figure out an alternative, short cut route to getting the right answer. That might finally offer a more insightful and intuitive analysis of a given problem. Richard Gill (talk) 07:41, 2 May 2013 (UTC)

Thank you. As to goats A and B, I didn't intend to complicate, just to express that I am speaking "of one" of the two goats, and not of the "other second goat".

All 3 doors "together" hide the prize with probability 1.0 okay?

I am perplexed. I mentioned "2 doors" that have been singled out of those 3 doors, and I said that with a chance of 2/3  "they"  will hide one car. More precisely: only one of them can hide one car while "the other door" will be empty in that case. My view: Just ONE of them can hide the car while - if one of them really should hide the car, in that case the other one will be empty of course (so "1 : 0" or "0 : 1"). But you say "I have no idea what you mean by 0 : 2/3". (It is on the mystery).

You still haven't given me a clue what you mean by 0 : 2/3. Richard Gill (talk) 09:54, 2 May 2013 (UTC)

At first it is BEFORE any host's action. Rick above correctly said that both unchosen doors have a chance of 1/3 each to hide the prize and that both together have a 2/3 chance before any host's action. This is correct. And I understand him saying that, after any host has opened one of those two doors to show a goat, the second unselected door still has 1/3 chance on the car. I supposed an unbiased host and I do not believe that it would be a mystery if the second still closed unselected door suddenly would have changed from 1/3 chance to 2/3 chance. I understood without a mysterious miracle the chance of the second unselected still closed door cannot have changed from 1/3 to 2/3.

So I try to say that, although both doors together have a 2/3 chance, at least one of them must hide a goat for sure but no car. I say that at least one of them will hide a goat with probability 1.0 and with a chance on the car of 0.0. So IMHO the chance of 1/3 + 1/3 of those two doors could be specified to be 0 + 2/3 resp. 2/3 + 0, just from the beginning. I said "no mystery". You see what I meant? I tried to say that no miracle is necessary to see that at least one of any two singled out doors (out of three) must hide a goat. Please try to understand my jolty argument. Do we need any miracle or not? Of course you can show by Bayes that the second unchosen still closed doors has a chance of 2/3. But is this really a mystery if we say so without using Bayes? Gerhardvalentin (talk) 11:18, 2 May 2013 (UTC)

And please try to tell me what is the probability that the contestant actually is in the Lucky Guess scene, 3 seconds before the host opens the door finally selected by the contestant, that the contestant stayed on, but after the host truthfully announced that the door he is just going to open hides the prize (additional info was given before the host actually opens the door). Please help. Gerhardvalentin (talk) 08:34, 2 May 2013 (UTC)

You mean, after the host says he's going to open the door hiding the car, but before we find out which door that is? And whose probability are you talking about? (The player and the host have different information so their subjective probabilities are not the same). Richard Gill (talk) 09:54, 2 May 2013 (UTC)

Okay, only the host is omniscient, not the player nor the audience nor me. Here I tried to address the knowledge of the player. Initially the player only knew that in 2 out of 3 he first will have selected one of the two goats, and in 1 out of 3 he first will have selected the car. That's all he can now at this stage. After the host opened one of the two unchosen doors to reveal a goat and offered a swap, let's assume that the player did not worry about the chances but decided to stay with his first door based on his "good feeling". Just let's assume that the host now truthfully said "your door hides the car indeed!". What is now the conditional probability (player view) to have made a Lucky Guess? BEFORE the host, some seconds later, will be opening the player's door? – And what is the conditional probability (player's view) to be in in the Lucky Guess scene, AFTER the host really showed the new car behind the player's door? That's what I tried to ask. Regards, Gerhardvalentin (talk) 11:18, 2 May 2013 (UTC)

I don't see any point in you asking these questions and no point in my answering them so I am going to quit for a while! Regards, Richard Gill (talk) 11:45, 2 May 2013 (UTC)

Richard, you say, 'it's a simple exercise to mathematically model this extended three doors problem, by focussing on the sequence of actions of quiz-team, player, and host, and using Laplace's principle of indifference to assign probabilities (representing the player's knowledge or lack of knowledge) whenever the various actors have to make choices'. This is where we disagree. I agree that we must model the problem mathematically before we can use mathematics to solve it but the modelling process, from a natural language statement to a precise mathematical description, is not trivial, as you seem to suggest. Mathematics, by definition, cannot help us in this endeavour. Throughout this process we must use our judgement and knowledge to decide exactly what we must and must not include in our model.

At one extreme we can simplify the problem as much possible before putting it into mathematical terms. Using simple logic we can reduce the sample space to just {C,G}.

At the other extreme we can model the problem mathematically by including every single event given in the problem statement (I think we can agree to consider only the events mentioned in the question). We can then apply probabilities to these events, based for example on Laplace's principle, and condition our sample space based on our understanding of the events that actually did occur. With the normal assumptions, this will show that many things, such as the door originally chosen by the player, the door opened by the host, the goat revealed by the host, the words spoken by the host do not affect the value of the probability that a player will win the car by swapping.

I have no objection to the two methods described above, one relies heavily on intuition and the other slogs laboriously through every possibility. You may decide that you want to use a method somewhere in between these two, where you deduce logically before modelling the problem that some of the events can be ignored and include others in the model, to later show they do not affect the result. That is fine but it is your personal choice. There is nothing mathematically or logically superior or more precise about your model and there is no reason why solutions based on your model should be considered the logically precise ones.

Again, just to make my point absolutely clear, I am not objecting to mathematical or logical precision, or to intuition, just to treating your preferred arbitrary combination of these methods as the 'logically precise' way of solving the problem. Martin Hogbin (talk) 08:52, 2 May 2013 (UTC)

What you call my preferred arbitrary combination of methods is not arbitrary at all, it is a routine application of a well-tried and well-motivated methodology, which the best mathematicians and scientists in the world have been working on and refining for 400 years. There are two fundamental yet common-sense principles at work in the model-building: (a) consider the sequence of events in our story in their temporal sequence, so that the model will respect fundamental principles of causality (b) apply Laplace's principle (symmetry) where needed to specify subjective probabilities. The two principles take care of the deterministic and the random parts of the problem, respectively.

In the model-solving (advising the decision maker) there is just one principle: (c) given what the decision maker knows at the moment he must make a decision (inside the previously constructed mathematical model), find out what are the (so-called conditional) probabilities of the things that he doesn't know.

Are you telling me that you have no use for principles (a), (b) and (c) and prefer to solve every new problem by inventing your own principles for each new occasions?

Three doors: first the quiz-team hides the car (three possibilities). Then the player picks a door. Then the host picks and opens another door (two possibilities). Three cups: the host puts a coin under one of three cups and shuffles the cups. The player picks one (two possibilities, but not equally likely). The host (who did the shuffling and still knows where the coin is) turns over another cup revealing nothing underneath. I could go on to analyse Bertrand's three boxes, Gardners' three prisoners in the same way.

I do not see any arbitrary combination of methods. I see *one* method followed consistently which enables us to *routinely* and uncontroversially solve problem after problem, old and new.

If you insist on complicating the three doors problem by supposing that the two goats on the show are individually known to host, audience, and player - no problem. First the goats and car are hidden by the quiz-team (six possibilities now, not three), etc etc. Richard Gill (talk) 09:46, 2 May 2013 (UTC)

Please read what I have written above. I have nothing against the standard principles of probability and I have made no mention of any of my own principles in this ection.

The principles you expound above, if applied at the start of the problem, clearly show that we can treat the problem as having the sample space {C,G}. It the time the player chooses whether to swap or not he has no knowledge of the relevance of the specific door open by the host, or the goat revealed by the host, or the door he originally chose. By applying the principles you suggest we can discard all these choices as unimportant.

Of course we can do it that way round too, if we are clever/creative enough to realize that it is applicable. I recently deleted from the article a section which did exactly that, in mathematical language, so that those who prefer a formal mathematical solution can see that there is a formal mathematical version of the "symmetry implies irrelevance" argument. Some editors thought it not well enough sourced, others thought that I used too expensive words ("symmetry"). You did not complain or put it back. You say "clearly"; your use of that word betrays the fact that you don't actually have an argument - you just jump to the conclusion. You "know" intuitively it is irrelevant. Just like Marilyn vos Savant. Mathematicians learn that people who say "clearly" are often jumping to conclusions and occasionally making a fatal error.

The plodding way (systematic approach, working with the principles I laid out) gets us to the same final result and is fool-proof. In general, being more informal and intuitive might lead us to making mututally contradictory assumptions and hence generating nonsense (one can easily jump to wrong conclusions especially regarding probability puzzles). Cf. the two envelopes problem. Intuition leads us, in that problem, to make assumptions which are actually mutually incompatible. That's exactly why the most brilliant mathematicians and scientists have spent the last 400 years getting probability theory into its present refined state. Because people kept getting wrong answers, when doing it by intuition. Richard Gill (talk) 11:32, 2 May 2013 (UTC)

If you want to apply the principles after you have mathematically defined the problem that is fine too. You are arbitrarily applying some at the start (with you argument that the player's original door choice is unimportant) but leaving some until later, where you use the same principles to show that the host's door choice is unimportant after you have defined the problem in mathematical terms. Why? Martin Hogbin (talk) 10:22, 2 May 2013 (UTC)

When did I say that the player's original choice is unimportant? It is made without any information of where the car is hidden. So after it is made, the car is still equally likely behind each of the three doors. At the moment the player must make his choice of whether or not to switch the player still knows which door he initially chose. So it is part of the information on which he conditions. Having conditioned on the outcome of this choice, the way it was made becomes irrelevant. There are no arbitrary choices in my approach, once I have determined on the sequence of events which the story is all about. A car is hidden. A player (not knowing how the car is hidden) chooses a door. A host knowing where the car is hidden and which door was chosen by the player opens a door. Richard Gill (talk) 11:38, 2 May 2013 (UTC)

Martin, you said that by simple logic you can reduce the problem to a sample space consisting of just two elements "car" and "goat", with (I suppose) probabilities 1/3 and 2/3, and representing the initial choice of the player. I have yet to see that simple logic. But anyway, you claim that by some simple logic at the start, you can argue that the three doors problem may be simplified to the three cups problem. My own more elaborate but entirely principled approach leads to the same conclusion since with it I *deduce* that which particular doors were chosen by player and opened by the host are conditionally independent of whether or not the player's choice of door hides a car, given the player's initial choice. Richard Gill (talk) 10:19, 2 May 2013 (UTC)

I use the very same principles that you expound above. We have no information that allows us to distinguish between, the doors we originally choose from, the two doors the host might then open, and the two goats the host might reveal. The solution can be considered symmetrical with respect to these choices. Martin Hogbin (talk) 10:25, 2 May 2013 (UTC)

Well consider my solution(s) a formalization of yours, and placing it within an existing scientific tradition and existing language and existing methodology. Richard Gill (talk) 11:21, 2 May 2013 (UTC)

Three doors, three cups, it's all the same. Do the cups have colors, are they numbered? Are they presented to audience and player as left, middle and right? You need to specify this. Nijdam (talk) 17:29, 2 May 2013 (UTC)

I agree, having three cups is not helping, I was talking about the standard MHP anyway.

Nijdam, I made it absolutely clear that the player cannot see any difference at all between the cups. A coin is placed under one of them, the host shuffles them about on the table. The player has no idea any more where the coin is. The host however does know - maybe he can feel the difference as he moves the cups around. Maybe there is a tiny mark on one of the cups, the one hiding the coin, which the player can't see. Does it matter how the host knows? Richard Gill (talk) 05:19, 3 May 2013 (UTC)

Richard, I do not have a solution. I am discussing a family of solutions that all use exactly the same principles that you expound.

Richard, the cups may be unidentifiable, their positions at the moment the game starts are not. How do you present the shuffled cups to the player? Nijdam (talk) 09:59, 3 May 2013 (UTC)

Huh? The player can't distinguish between the cups except by their relative positions, and their relative positions get shuffled by the host, so the player can't say which cup (as identified by initial position) became which cup (as identified by final position). Use your imagination and/or take a peek at my solution (model) for the three cups problem. Richard Gill (talk)

Okay, in that case you may be right. In fact it looks as if you're trying to find a model for our well-known rv's X(first choice), H(host opened door) and C(car), but then as outcomes. I can't get hold of it at the moment, and have difficulty in describing a sample space. Nijdam (talk) 12:06, 4 May 2013 (UTC)

My model for the three cups problem (as seen from the player's point of view) is rather simple. The player chooses a cup. Probabilities 1/3, 2/3 that it hides the coin (and these are objective probabilities, coming from the physical act of randomization by the host, which preceded the player's initial choice). The host turns over one of the other two cups, shows there's nothing underneath it, and offers a swap. That other cup hides the coin if and only if the player's initial choice of cup does not hide the coin.

In other words: the natural probability model for the three indistinguishable cups problem corresponds with the simple solutions, including the "combining doors solution". Richard Gill (talk) 15:17, 5 May 2013 (UTC)

Richard, please give me the sample space for your 3 cups. Nijdam (talk) 08:35, 8 May 2013 (UTC)

I give below what I believe to be the standard 'conditional' solution. Do you agree? Martin Hogbin (talk) 19:58, 2 May 2013 (UTC)

Just to be clear, I am not looking for a long discussion here, I just want to know whether what I have written reasonably represents the logic of the standard 'conditional' solution or whether I have misunderstood something or missed something out. Martin Hogbin (talk) 20:05, 2 May 2013 (UTC)

The standard 'conditional' solution

0) We decide that the question requires us to consider the case where the player initially chooses door 1 and the host opens door 3.

1) By applying the principle of indifference (or by noting that we have no information about the initial car placement) to we take initial car placement to be uniform at random. The player's initial door choice is therefore independent of the initial car placement. We can therefore fix the player's initial door choice as door 1.

2) By applying the principle of indifference (or by noting that we have no information about the initial goat placement) we can ignore the specific goat that is revealed by the host.

3) Having done the above, we can set up a sample space showing all the legal options of car placement and host door choice. In order to assigns probabilities to the events in our sample space we also need to apply the PoI to the host's door choice. He has a choice of two doors and we have no reason to suppose that he prefers either door so we assign a probability of 1/2 to each door.

4) We now remember that we are to consider only the case where the host opens door 3 so we condition are sample space accordingly, removing all events which do not meet our condition.

5) From the conditioned sample space we now calculate the probability that a player who swaps (given that the player has chosen door 1 and the host has opened door 3) will win the car. The answer is 2/3. Martin Hogbin (talk) 19:58, 2 May 2013 (UTC)

Delete lines 0, 2. 4. Rewrite the others (I have given solutions earIier on this page, you seem to have ignored them). For instance, line 3: PoI: because we don't know anything about the host's choice, either choice would be equally likely for us, when he does have the opportunity to choose. Line 5: condition on what the player knows: he knows his own choice and he knows the choice of the host. For instance, the left and the right hand doors, respectively. The player needs to know the chances on the possible locations of the car given his information at the moment he must choose. Richard Gill (talk) 05:31, 3 May 2013 (UTC)

0) Richard you are just being perverse now. The argument for the 'conditional' solutions has always been that question requires us to consider the case where the player initially chooses door 1 and the host opens door 3.

I am not being perverse! What we should probabilistically condition on is determined by the model, once the model is written down. The player wants to know what to do, and therefore he sensibly - indeed, optimally - conditions on the information which is in his possession at the moment he must make his decision. As I argued before, the solution would look exactly the same, whether or not specific door numbers or names are mentioned in the question. Richard Gill (talk) 11:57, 3 May 2013 (UTC)

Your argument that the solution would look exactly the same, whether or not specific door numbers or names are mentioned in the question is interesting but in the problem, as many choose to understand it, the doors are numbered and if we are going to refer to specific doors we might as well use the numbers. Martin Hogbin (talk) 23:47, 3 May 2013 (UTC)

It's more than important, it's crucial. Richard Gill (talk) 06:52, 4 May 2013 (UTC)

2) There are two goats (goats are always distinguishable) and only one of them is revealed. We must at least give rationale for choosing not to distinguish between them. Remember your admonition against logical imprecision.

Doors, cups, and goats are all in principle distinguishable: this is not quantum physics. What we are talking about here is a kind of relative distinguishability, relative to a particular person and a particular context. No doubt the philosophers have some good language for making these distinctions. Instead of using philosophers' technical terms, let me give examples. Three cups: after they have been shuffled, the player can no longer associate their present position on the table with their original positions. That's what I mean by the cups being indistinguishable for the player. That's what we mean by the word "shuffle". For the host they can be distinguished, it doesn't matter how (secret marks on the cups, or by feeling the resistance of the coin as he shuffles?). Two goats: at some point in the game the player gets to see a goat but does not recognise it individually. The host possibly does know the two goats on tonight's show personally, but so what? At some point he opens one of the two doors. We don't know how he does it. Maybe he prefers to open Nanny's door on those occasions when Nanny and Billy are on the show. So what? Because we know nothing about how he makes his choice, either choice is equally likely. Richard Gill (talk) 11:53, 3 May 2013 (UTC)

The would be the principle of indifference then. We have nothing to give either goat any special significance so we take it that either is equally likely. You do not want to delete 2) just maybe change the wording a little. I have no objection to that.

4)The doors are numbered! We are not even told that there are right or left hand doors. Let us use the door numbers. Martin Hogbin (talk) 08:20, 3 May 2013 (UTC)

There are three huge doors on the back of a stage. Marilyn gives them names "door 1", etc, but the only thing we can be sure of is that everyone can see three huge doors and people would tend to number them 1 to 3 from left to right. The important thing is that the doors stay fixed throughout the story and the story is prefaced by the quiz-team placing a car behind one of the three doors, goats behind the other two. Compare with the three cups. The cups are shuffled and the player no longer knows which one hides the coin. The doors aren't shuffled. The car is placed behind one of the doors before the show starts and it stays there all the time. The doors stay there in fixed positions all the time. Richard Gill (talk) 11:57, 3 May 2013 (UTC)

Can we just forget the three cups. Martin Hogbin (talk) 23:47, 3 May 2013 (UTC)

The section is entitled "The standard 'conditional' solution", I am not discussing your preferred solution or what you might think to be my preferred solution to be but the common 'conditional' solution as given in the article and many sources. Martin Hogbin (talk) 08:24, 3 May 2013 (UTC)

You entitled this section "the standard conditional solution" but I don't recognise it at all. If you want to discuss a standard conditional solution please choose one from the literature or from the article. Richard Gill (talk) 12:18, 3 May 2013 (UTC)

Nijdam's comment: 0) okay. 1) The formulation of the problem is the normal formulation that leads ordinary people to interprete as random placement of the car, and the assumption (not logical consequence, as Martin suggests) of independence of the player's choice and the position of the car. 2) Okay, or just take undistinguishable goats, or compare with empty doors 3) Okay. 4) Do not remove anything, just calculate the appropriate conditional probability. 5) I would not speak of "answer", that's what grammar school pupils do. The outcome 2/3 is the value of the desired conditional probability. Nijdam (talk) 10:15, 3 May 2013 (UTC)

I prefer to use the term 'answer' rather than 'conditional probability' because some of this discussion is about exactly what conditional probability we aim to calculate. If you prefer, we could call it the probability of winning by switching (given what we are told in the problem statement)

Martin, it is not about the conditional probability of winning by switching given what we are told in the problem statement. The player needs to know the conditional probability of winning by switching given what he knows at the moment he must decide. Richard Gill (talk) 06:51, 4 May 2013 (UTC)

Regarding 4) would you prefer to say something like, 'consider only those events that meet the given conditions', that would be fine with me? Martin Hogbin (talk) 23:47, 3 May 2013 (UTC)

I disagree. We cannot assume that the quiz-team uses some randomization mechanism to choose a location for the car uniformly at random. We don't know anything about how it is done. Therefore (in the sense of subjective probability) all three locations are equally likely, both for us, and for the player. Therefore (in the sense of subjective probability) the player's initial choice is independent of the location of the car. After making his choice, the car is still equally likely behind each of the three doors. Therefore, it is, for us and the player, *as if* the location of the car had been chosen by a fair randomization machine, and as if the player's choice is made independently of the outcome thereof. Once done it's done and the player knows it, so we condition on it (we are not in the "little green woman from Mars" problem).

I don't *assume* anything. I *deduce* everything. My total ignorance is symmetric under permutations of the doors, so I deduce a uniform distribution and I deduce that the player's choice is independent of the location of the car.

This is important. Because probability assignments are *deduced* from symmetry it is clear that we might argue in a different sequence. Instead of first getting probabilities from symmetries, and then computing, and then observing that the solution of the three doors problem has some similarity to that of the three cups problem, we could use symmetry in advance to argue that the three doors problem reduces to the three cups problem, and only then explicitly assign the probabilities which need to be assigned in the smaller problem. Richard Gill (talk) 12:09, 3 May 2013 (UTC)

Richard, you say, 'My total ignorance is symmetric under permutations of the doors, so I deduce a uniform distribution and I deduce that the player's choice is independent of the location of the car'. That is absolutely fine but please be consistent, follow that statement with, 'My total ignorance is symmetric under permutations of the doors, so I deduce that the host's choice is independent of the location of the car, in fact we can forget about the doors altogether.'. Martin Hogbin (talk) 23:47, 3 May 2013 (UTC)

Please read the last sentence I wrote yesterday: "we could use symmetry in advance to argue that the three doors problem reduces to the three cups problem, and only then explicitly assign the probabilities which need to be assigned in the smaller problem". In fact, I first wrote but later deleted a section of the article which advanced this point of view. I deleted it because no one liked it. I did not hear you complain about the deletion. Richard Gill (talk) 06:46, 4 May 2013 (UTC)

We agree then, we can do our application of well-established principles of probability at the start, before we produce a mathematical description of the problem or after we have formulated the problem mathematically. The only choice we have is what to do at the start and what to do later.

It is hardly my fault that you deleted something! I do not check through all your edits. Martin Hogbin (talk) 21:59, 4 May 2013 (UTC)

I proposed on the talk page to delete certain sections. I waited a week, there was no comment. Then I did it and reported on the talk page that I'd done it. There was no comment. Richard Gill (talk) 09:50, 5 May 2013 (UTC)

A proposition for Nijdam

It looks like we do not disagree on anything substantial regarding my description of the standard 'conditional' solution.

Suppose now that somebody questions step 1) above, saying that it is not clear why we can fix the players original choice of door.

Would you agree that a reasonable response to that question might be to consider all possible initial player door choices and include them in our sample space. Later on, we can condition our sample space on the given initial choice of door 1 and end up with essentially the same problem. Martin Hogbin (talk) 22:00, 4 May 2013 (UTC)

It *is* clear why we condition probabilitistically on the player's choice of door. We seem to have agreed to solve this problem from the Bayesian (subjective probability) perspective. That is the approach which comes naturally to many (most?) people. The approach entails modelling the information stream as received by the player in order to determine the player's optimal decision in the light of the information at his disposal at the moment he must make his choice whether to switch or stay. Once he's made his initial choice of door, it's made and it's known to the player. He is not the green woman from Mars who turns up later, unaware of the unfolding history of events as seen by the player!

Whether we update probability distributions step by step or all in one go doesn't make any difference to the final result. Richard Gill (talk) 09:57, 5 May 2013 (UTC)

Richard, you seem to keep telling me things I already know and challenging things that I have not said.

Regarding, 'Whether we update probability distributions step by step or all in one go doesn't make any difference to the final result', Yes!. That is what I have been saying for years. Whether we consider the host's door choice and goat choice at the start of the problem and rule them both out as unimportant factors in the calculation using some well-accepted principle of probability, or whether we include everything we can think of in the mathematical formulation and then consider the host's door and goat choice formally, in mathematical form, later does not affect the result. My general feeling is that more events we include in our mathematical formulation, the more formal and rigorous our solution is. Martin Hogbin (talk) 11:33, 5 May 2013 (UTC)

How to experimentally verify the "doorless" solution?

Tying this into the thread above - if you reason you know nothing about how the car was hidden or how the host chooses between doors, and (by ignoring door numbers) arrive at an answer of "2/3", 2/3 of what? In particular, if you want to experimentally observe this 2/3 answer with a large number of repetitions (per law of large numbers) how do you do it? Lets say the show was on long enough that there are 9000 recordings. Within this sample we have occasions where the players pick either door 1, door 2, or door 3, and in each of these cases there are some number of occasions where the host opens either remaining door (always showing a goat). Which cases of this sample is your 2/3 solution talking about (how many, and which, do we count)? Is it

a. All of them, i.e. in about 6000 times switching wins and 3000 times it loses

b. Only those where the player initially picks door 1 and the host opens door 3, i.e. 2/3 of these players will win by switching

c. Any randomly selected subset up to and including all, which will include a mix of all 6 possible combinations of initial door pick and door the open opens

d. All of the above

Got your answer?

OK, now, to make it more interesting I'll tell you (you didn't know this - the answer is 2/3 for you because you didn't know, right?) that the car was hidden behind a door chosen by rolling a (fair) die (door 1 if the roll is 1 or 4, door 2 if the roll is 2 or 5, and door 3 if the roll is 3 or 6) and the host strongly (but not totally) preferred to open lower numbered doors. Under these conditions nearly all (way more than 2/3) of the sample where the player picks door 1 and the host opens door 3 win by switching. Is the law of large numbers wrong? Or, perhaps, does your 2/3 answer not apply to this sample? If you approach the problem this way, what does your 2/3 answer actually mean - i.e. 2/3 of what? -- Rick Block (talk) 17:18, 4 May 2013 (UTC)

This is just a slight obfuscation of a simpler (non)paradox. A car is hidden behind one of three doors. What is the probability that it is behind door 3?

The answer is 1/3.

Aha, you say, but the car is, in fact always placed behind door 1 so the real answer is 0. So it is, if you know that fact, but if I do not the probability remains 1/3. Probability is a state of knowledge.

Another well known example is: A card is taken at random from a standard pack of cards. Person A is told that it is a five, B that it is a spade and C that it is a black card. It is in fact the five of spades. What is the probability that the card is the five of spades. It depends on who you are to A it is 1/4, to B it is 1/13 and to C it is 1/26.

Interesting, but nothing specifically about the MHP. Martin Hogbin (talk) 22:18, 4 May 2013 (UTC)

Yes, probability is a state of knowledge - I'm asking what is the meaning of the knowledge. If you say a car hidden behind one of three doors has a probability of 1/3 being behind door 3 because you know nothing about how it was hidden, you're NOT saying that in any given experiment where a car is hidden behind one of three doors the proportion of times it will show up behind door 3 approaches 1/3 (as the number of samples tends to ∞). You're saying something else. In particular, you're saying if you randomly (uniformly) pick door 3 (and the other times pick door 1 or door 2), your choice will be correct about 1/3 of the time. This statement is true, and can be experimentally verified, regardless of how the car is hidden. Subjectivist probabilities definitely follow the law of large numbers. On the other hand, you have to be aware of what they actually mean before you try to experimentally verify anything.

This definitely relates to the MHP. In particular, a subjectivist conclusion that a player has a 2/3 chance of winning the car by switching does NOT mean in a sample of 9000 shows where we consider only those cases where the player has picked, say, door 1 and the host has opened, say, door 3 that the car is behind door 2 in 2/3 of these shows. I think most people would say 9000 shows (or the subset selected from 9000 shows where the player picks door 1 and the host opens door 3) is enough to show a "true" probability. In which case, the subjectivist's expectation must be met. It will be, as long as you understand the "2/3 of what" the subjectivist's answer is talking about. -- Rick Block (talk) 00:34, 5 May 2013 (UTC)

Yes, of course this relates to the MHP so I see no problem about discussing it here but it is not specific to the MHP.

The Bayesian concept of probability has nothing to do with picking a door at random. If I come across 3 doors and, the only thing I know is that there is a car behind one of the doors then the probability that the car is behind door 1 is 1/3.

So what if I repeat the experiment many times? How does this relate to the law of large numbers? The first, and critical, question is, 'What exactly is the experiment?'. The Bayesian probability of 1/3 for door 1 (and doors 2 and 3) is based on coming across 3 doors for the first time and knowing only that there is a car behind one door. Ther are two ways of repeating this experiment, the first way is to try to contrive to repeat the first time that I come across the 3 doors. Each time I know nothing about how the car might be placed and must therefore assume the probability is uniformly distributed. Each time, the way by which the car is placed might be different so, after a large number of repeats, the proportion of times the car is behind door 3 is 1/3.

Alternatively I can repeat the experiment in which I know that the car is placed by the same mechanism each time. Suppose that, the car is, in fact always put behind door 1. The first time the probability is 1/3 but when I win the car, because it is always behind door 1, I have more knowledge and I revise my probability in favour of its being behind door 1. Each time the car proves to be behind door 1 the probability that it is behind door 1 moves closer to 1. After may tries the Bayesian probability matches the law of large numbers.

Frequentist and Bayesian probabilities always match, so long as we do the right experiment. Martin Hogbin (talk) 09:25, 5 May 2013 (UTC)

So let's agree on a particular experiment. Location of the car is determined completely at random. Player chooses door 1. Host opens door 2 or door 3. If he has a choice, he does it by tossing a fair coin. We can now see whether the player wins by staying or by switching. Let's repeat this whole thing 9000 times, and keep track of the outcomes.

I think we all agree that in about 6000 of those 9000 repetitions, the player wins by switching. I think we also agree that in about 4500 of those 9000 repetitions, the host opened door 3; and of those 4500 repetitions, the player wins by switching about 3000 times.

3000/4500 = 2/3 = 6000/9000. The long run relative frequency of winning by switching given the player chose door 1 equals the long run relative frequency of winning by switching given the player chose door 1 and the host opened door 3.

Does everyone agree so far? Richard Gill (talk) 10:13, 5 May 2013 (UTC)

I agree. Martin Hogbin (talk)

Then we must also agree, that a correct proof of (A): the long run relative frequency of winning by switching given the player chose door 1 is 2/3; and a correct proof of (B): the long run relative frequency of winning by switching given the player chose door 1 and the host opened door 3; are two different things.

It's a matter of fact that when approaching MHP, some sources seem to be happy with A (most popular sources), other sources insist on B (most mathematical sources). Some sources think the difference is rather important, other sources don't see a difference (e.g. Marilyn vos Savant reacting to Morgan et al.). This major split between the way popular writers and mathematical writers approach MHP is a nice challenge to wikipedia editors but I hope that wikipedia editors at least can agree on the facts. Richard Gill (talk) 15:33, 5 May 2013 (UTC)

I have no idea why you are saying all this. Do you really think that after all the years of discussing the MHP I do not know the difference between the probability of winning by switching given the player chose door 1 and the probability of winning by switching given the player chose door 1 and the host opened door 3???

The first thing we must ask when responding to the MHP is what question does the questioner intend to ask. It is quite clear to me, from the context of the question, a column in a popular general interest magazine, that the intended question was neither of the above but that the intended question was about probability of winning by switching given the player chose any door and the host opened door another one, which hid a goat. We know that this was indeed the intended question, so assuming the question to be about specific door numbers is perverse.

Now the thing that even after all this discussion you seem not to have grasped is that, even if we take the question to be asking for the probability of winning by switching given the player chose door 1 and the host opened door 3, the difference between what you call the the way popular writers and the way that mathematical writers talk about the problem is not that different. If you really need me to explain why one more time I will. Martin Hogbin (talk) 16:54, 5 May 2013 (UTC)

I was not addressing you personally Martin, just trying to get some facts out on the table in plain view of everyone. But what you say is good. Like me, you think that there is not so much difference between the popular and the mathematical solutions. If that's the case, you cannot claim that the standard mathematician's solution makes absurd arbitrary choices. I've tried to argue above that the mathematician's solution is not arbitrary at all but is the straightforward result of applying a couple of sensible and time-honoured basic principles. I've also argued that the mathematician's solutions (i.e., both following the same basic principles) to the three doors problem and to the three cups problem are interestingly different. I believe that the popular solution to the three doors problem is based on an intuitive grasp that by symmetry, the three doors problem can be reduced "in advance" to the three cups problem. Intuition and mathematical analysis agree here. Intuition can be so strong that clever, insightful, people do not see there is anything to discuss. Vos Savant is an example of someone with such "intelligent-intuition". Richard Gill (talk) 09:36, 6 May 2013 (UTC)

Maybe you misunderstand what I meant when I referred to arbitrary choices. I am only referring to the decision over which events to deal with, using standard principles of probability, at the start of the problem before it is formulated mathematically and which events to deal with more formally after the problem has been put into mathematical form.

Most 'mathematical' solutions deal with the player's initial door choice and the host's choice of goat right at the start of the problem, often tacitly. They then go on to deal with the host's door choice more formally using conditional probability. There is nothing wrong with this but it is a somewhat inconsistent approach. One consistent, but long-winded, approach would be to include both the player's initial door choice and the host's goat choice in the formal mathematical formulation of the problem and then deal with them all using conditional probability. Another consistent approach would be to deal with the the player's initial door choice and the host's choice of goat or door at the start of the problem, ideally stating your reasoning, and then solve the very simple mathematical formulation {C,G} that remains.

Can you just confirm that you understand the point that I am making. Martin Hogbin (talk) 14:52, 6 May 2013 (UTC)

I have no idea what point you're making. You don't seem to have understood mine. I've spelt out a completely standard mathematical approach to these problems. Focus on the sequence of events. 3 doors: quiz team hides car; player chooses door; host opens door. 3 cup: player picks one from three (for him) indistinguishable cups, host eliminates one of remaining two. Remember the tree-diagram in the article? The tree-diagram represents all possible histories. Each possible history is a point in the sample space, if you like to think in terms of sample spaces. I don't see why you keep mentioning goats, and a host's choice of goat. The host opens a door. That's what he does, that's what we're told he does. That's what goes into the model. Try the three prisoners and the three boxes. You can solve each problem by making a brilliant intuitive guess, or you can solve each problem by carefully analysing the sequence of events. Richard Gill (talk) 18:15, 6 May 2013 (UTC)

What tree diagram? How does it represent all possible histories? See below. Martin Hogbin (talk) 18:42, 6 May 2013 (UTC)

Goats again

The host has to decide which door to open when he has a choice. We are told that the host opens a door but we are not told how he chooses that door. There are four possibilities: door 1 which hides goat A, door 2 which hides goat B, door which hides goat B, and door 2 which hides goat A. There are four possible histories which we must include in our sample space, not two. We are told that the host reveals one goat and we know that there are two goats. Maybe it matters which one he reveals, maybe he has a goat preference but, even if he has no goat preference, we must still include the two goats in our sample space just as we must include the two doors he might open even if he has no door preference. Martin Hogbin (talk) 18:51, 6 May 2013 (UTC)

Listen very carefully, I will say this only once (since I already said it a million times). As you say, maybe this, maybe that. We (you, me and the player) don't know. Therefore, for us, when the car happens to be behind our door, either of the other two doors is equally likely to get opened. And you asked: what tree diagram? The answer to that question is RTFA (A for Article instead of the usual M for Manual). The systematic standard method to approach this kind of problem is to list the historical sequence of possible actions or events. Model each one in turn, taking account in each case of the history so far. The aim is to advise the player what to do, so we focus especially on the stream of information reaching the player as the sequence of events unfolds. Forget about sample spaces. We can easily write down a suitable sample space to match a given tree diagram, after we have the diagram. There is one elementary outcome for each "leaf" of the tree. Each possible complete history. If we want to know the probability of each outcome we simply multiply the sequence of conditional probabilities of taking each branch, given the route taken so far, as we trace a particular history from root of the tree to leaf. Richard Gill (talk) 07:23, 7 May 2013 (UTC)

I think we may as well give up. Fairly obviously I understand what you are saying and I completely agree with it. Your explanation is perfectly clear, although not needed. I know what a tree diagram is and how it is used to solve the MHP, you surely must realise this.

On the other hand you still seem to have no idea what I am talking about. You mention, 'Each possible complete history', yet you do not do this. In the case (leaf if you like) where the player has chosen door 1 and the host has revealed a goat behind door 3 there are two possible histories, that the goat revealed is goat A, that the goat revealed is goat B. I understand that this 'information' is of no value to us but it still should be represented on our tree. Martin Hogbin (talk) 07:54, 7 May 2013 (UTC)

My analysis focusses on a sequence of events: team hides car, player chooses door, host opens door. It might be considered a matter of taste to focus on those three and only those three steps in the game, but I think it is a very natural, reasonable choice. When I talk about complete history I refer to the complete history in those terms. I do not add the day of the week, or what the player had for breakfast. Monty's relationship with his goats is taken into account in my model. The player doesn't have any personal relationship with those goats. So no: I definitely should not add goat identities into my model. Richard Gill (talk) 11:40, 7 May 2013 (UTC)

Everything you have said above applies equally to the door ID opened by the host. Considering only what is stated in the question:

We know that there are two doors that the host might open; we know that there are two goats the host might reveal. The host opens one of two possible doors; the host reveals one of two possible goats; the player sees the door that is opened and also sees the goat that is revealed. We are not told how the host chooses how to open a door to reveal a a goat, maybe he has a door preference, maybe he has a goat preference, maybe he secretly tosses a coin to decide which door to choose, maybe he secretly tosses a coin to decide which goat to reveal, we have no way of telling. The player gains no information as to whether the car is behind the originally chosen door from the door number opened by the host or from the specific goat revealed by the host. The player has no personal relationship with either the goats or the doors.

Please tell me as clearly and succinctly as you can exactly what is the reason to consider the door opened but not the goat revealed. You have already stated that the door numbers are not relevant. Martin Hogbin (talk) 13:32, 7 May 2013 (UTC)

I already did this as clearly and succinctly as I can. See if you can find someone else who understood me, to rewrite what I wrote in yet other words, if that is what you need. Richard Gill (talk) 08:46, 9 May 2013 (UTC)

Quit if you like but please do not try to tell me that you have given me any valid reason for taking into account the door opened by the host but not the goat revealed by the host. Martin Hogbin (talk) 09:20, 9 May 2013 (UTC)

I won't try to tell you this any more, but the fact remains that I believe that I have given you some very good reasons. Richard Gill (talk) 07:32, 29 May 2013 (UTC)

Perhaps you could briefly restate the reasons in bullet form for me they seem to have got lost in the many words. Martin Hogbin (talk) 11:27, 2 June 2013 (UTC)

Similar versus equivalent

Regarding three cups versus three doors: see the Bertrand box paradox talk page for a similar discussion on whether problems like these are equivalent or merely similar: http://en.wikipedia.org/wiki/Talk:Bertrand%27s_box_paradox#Similar.2C_not_equivalent. See also the solution to the Bertrand box paradox by symmetry. See also Gardner's three prisoners problem. Richard Gill (talk) 08:40, 9 May 2013 (UTC)

Richard, your 3 cups problem is difficult to describe with a sample space, however it seems to be an effort of modelling the 3 balls problem people come and came up with, like Martin did in the past. Put a black and 2 white balls in a bag and let the player take one out, unseen and put aside unseen. Then the host lookes in the bag and produces a white ball. The player then is offered the choice of sticking to his original choice, or swapping to the remaining ball in the bag. This should be considered equivalent to the 3 cups, although I have my doubts. It looks (is) similar to the MHP, but is not equivalewnt to it. Nijdam (talk) 05:48, 10 May 2013 (UTC)

I think the 1 white, 2 black balls in an urn version was my suggestion - and I don't think it's equivalent to either MHP or Richard's 3 cups problem. Re 3 cups: what is the fundamental difference between hiding a car behind one of three doors vs. randomizing the position of a coin under one of three cups by moving them around? If you number the cups beforehand, then you're left with a randomized arrangement which makes numbering them beforehand pointless - but the cups are still distinguishable (left-middle-right) in a manner accessible both to the player and to the host (indeed, the host must be able to distinguish the "coin" cup from the others in order to be able to reveal a loser) and the host can still express a left to right preference in the case the player initially picks the winner.

If you want to simplify the problem, then simplify it! Turn it into an urn problem where neither the player or the host can distinguish the two losing choices - or (equivalently) move the player's switch/stay decision to before the losing choice is revealed (which is the approach actually taken by most of the "simple" solutions). Richard's 3 cups problem is equivalent to an urn problem with 3 different colored balls (winning white, losing red and green). The two (different) losers can be distinguished (in the same way) both by the player and the host, meaning the host may exhibit a preference for which one to show the player (if he has a choice). Perhaps Richard is thinking his version is equivalent to saying the player must be completely red/green colorblind. To be equivalent to the simplified urn problem (1 winner, 2 identical losers) two randomization steps are needed, perhaps as follows. 1) Put the coin under one of the cups. 2) Shuffle (randomize). 3) Player picks one. 4) Shuffle (randomize) the remaining 2 cups. 5) Host reveals an empty one. But this version creates a pragmatic difficulty - how does the host know where the coin is after step 4?

The main point here is that as long as the two losers are distinguishable (in the same way) by the player and the host, then the host may exhibit a preference for revealing one rather than the other (in the case the player initially picks the winner), and knowing which loser is revealed therefore gives the player additional information. This leads to the "Morgan" answer of 1/(1+p). -- Rick Block (talk) 18:02, 11 May 2013 (UTC)

First of all, some history: the three cups (three shells) goes back to Martin Gardner some years before MHP became famous. Marilyn refers to the three cups (three shells) in one of her Parade articles, it is obvious (to me) that she got it from Gardner. She seems to think that the two problems are equivalent. I think that lots of people would agree with her. I think that the simple solutions to MHP are basically the result of intuitively seeing that the door numbers are irrelevant, hence MHP can be reduced to the three shells problem. So in my opinion, the problems are not identical; one is a bit more complicated than the other; but it can be solved by realizing that certain aspects of the problem are irrelevant, and once they have been discarded, the problems have become "the same".

I assume that, as far as the player is concerned, the three shells are completely indistinguishable or "apparently identical". The host puts a pea under one of them, and after he has shuffled them about on the table, the player has no idea any more which shell has the pea under it. The player picks one shell. I assume that the host can distinguish between the shells (maybe he can feel the pea as he moves them about, or maybe there are some tiny distinguishing marks on the shells which he can see but which the player is not aware of). Now the host reveals that one of the other shells doesn't have the pea underneath. Should the player switch?

I think that the natural probability space for this problem has just two outcomes: chosen shell hides pea, chosen shell doesn't hide pea. They have probabilities 1/3 and 2/3 and these probabilities come from the combination of the physical randomization, the shuffling and the player's arbitrary choice of one of the three. The point is that because the shells are apparently indistinguishable and because I naturally do not incorporate the positions of the shells on the table into my mathematical model, I do not distinguish in any way, in the model, between which of the two other shells is lifted up by the host. As far as the player is concerned, there is only one thing the host can do, and what he does is certain: he picks up a shell and show there's nothing underneath it. In the Monty Hall problem the car is hidden in advance and throughout the whole story the three doors and the car remain in fixed positions. As far as the player is converned, the host does one of two different possible things: either he opens door 2, or he opens door 3.

We can do this with balls in an urn, one red and two blue, and a player who is wearing special glasses which make him colour-blind: he cannot see the difference between red and blue. The player picks one ball out of the urn, but of course he doesn't know which one he has. The host takes a blue ball from the remaining two balls in the urn, puts it aside, and asks the player if he wants to stay or switch.

I notice here an ambiguity in the meaning of the English word a (the indefinite article)! When you mathematically formalize these problems, you have to resolve the ambiguity. In probability theory, there is a difference between the events "host opens door 2 or door 3", "host opens door 2" and "host opens door 3". So in probability theory, "host opens a door" is ambiguous.

Notice that the difference between these problems lies in whether certain objects can be recognized individually (tracked in time) by particular persons in the story, or not. It's all about information, and about the difference in information available to different parties. Three shells: the player can't track the shells in time as they are moved about the table. The host can. Three doors: the three doors stay fixed throughout, the player can track them; at the moment when one has been opened and he may choose anew, he knows which one he initially chose and which one was left closed by the host. The difference in information in this problem concerns the location of the car: the host knows it all the time, the player not.

It's all about information; about who has what information. In particular, the player has incomplete information but in the course of the game he gains some information and uses it to update his beliefs about what he doesn't know. Richard Gill (talk) 10:46, 30 May 2013 (UTC)