Nijdam

User talk:Nijdam/Discussion

Welcome!

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And don't forget, the edit summary is your friend. :) Oleg Alexandrov (talk) 01:34, 24 January 2006 (UTC)

Monty Hall

Hello Rick. The 'discussion' about this subject gets completely out of hand. It's almost everywhere the same. Schools in different countries use the problem to illustrate something about probability, or use it as a task for their pupils, always in the wrong way. Most of the discussiants have no idea what they are talking about. I make an exception for Martin Hogbin, but I'm surprised he doesn't understand the conditional nature?

In my opinion the article needs to be changed. And no comprimising, that does'n lead to a better understanding. Let's look for an easy understandable formulation of the right solution. I also think it will be necessary to mention the wrong explanation, and show why. I'll try to think about the wording, but hope you'll be willing to improve my text if (and it will be) necessary. Nijdam (talk) 21:33, 14 February 2009 (UTC)

Hi. Thanks for the encouragement. I've asked folks who presumably do understand this to weigh in at Wikipedia talk:WikiProject Mathematics#Proposed addition to Monty Hall problem. I suspect where this is headed is that the article will take on a slightly different structure, effectively with a "popular solution" section (referenced to some particular popular explanation, perhaps vos Savant's) and then a "rigorous mathematical solution" section (which I think should be referenced to Morgan et al. and Gillman - as the first two to publish rigorous treatments). The focus needs to be on what the sources have to say about the problem. Non-academic sources nearly uniformly present unconditional "solutions". The article should as well (sourced to a non-academic source). Wikipedia doesn't (shouldn't) "take sides" here, but instead should use wording like "vos Savant's solution is <this>" and then "Morgan and Gillman say "<that>". It usually isn't this hard in a math related article - but for whatever reason this particular problem generates a tremendous amount of deeply entrenched (usually wrong) opinions. I suspect any inherently conditional problem does as well, see for example Boy or Girl paradox. Bedankt (is that idiomatically correct?). -- Rick Block (talk) 22:16, 14 February 2009 (UTC)

Okay, I'll have a look at the mentioned page. BTW: your Dutch, allthough short, was absolutely correct. Where did you pick up your Dutch?Nijdam (talk) 22:40, 14 February 2009 (UTC)

I don't speak Dutch at all, just noticed you were (and looked up how to say thank you). -- Rick Block (talk) 02:44, 15 February 2009 (UTC)

Martin reverted your change, which is what I thought would happen. I'm not sure, but I think some progress is being made on the talk page. Fighting over the lead at this point is probably not that productive. It may be best to just leave it alone until things settle down a bit. -- Rick Block (talk) 01:45, 16 February 2009 (UTC)

Regarding the German Wikipedia, I think the approach you're suggesting is roughly how you originally found the article in the English Wikipedia - i.e. carefully worded (not incorrect) unconditional solution first, followed with an explanation and a conditional solution. As I recollect you didn't like this very much. Another approach is to present, say, vos Savant's solution clearly identified as such (with a citation) followed by a conditional solution (perhaps some choice words from Morgan et al.). At this point I'm actually leaning to a fully rigorous solution (like Morgan's), followed with various other "solutions" (with critical commentary, meticulously sourced). -- Rick Block (talk) 03:03, 3 March 2009 (UTC)

BTW - Regarding this edit aren't we generally assigning the odds to the doors, not the players? The reason the little green men have a 50/50 chance is because they must randomly select between the doors, so 50% stay and 50% switch. If the probabilities of the doors are p and 1-p this results in 50% win regardless of p. -- Rick Block (talk) 03:53, 3 March 2009 (UTC)

Of course, but I made the remark, because I read somewhere in the discussion, that even the "green men" would win in 2/3 of the cases when switching. The meaning of my remark is that a different type of player may have different odds. So the odds regarding the doors are assigned to the type of player.Nijdam (talk) 09:41, 3 March 2009 (UTC)

Ah. The comment was that of the green men who switch (as opposed to all green men), 2/3 would win - i.e. the 2/3 applies to the door, not the player. This is exactly where I'm going with user 216. -- Rick Block (talk) 13:10, 3 March 2009 (UTC)

Sorry, may be my English fails here. Anyway I meant to say that green men cannot switch, because they did not choose. They just may open one of the still closed doors, and on the average win the car in 1/2 of the cases. So for them 1/2 is applicable for each door. Note that a green man, finding door 3 open, doesn't know which of the other has been originally picked, and hence the average is taken over all situations with door 3 open, the ones with door 1 picked and the ones with door 2 picked. Nijdam (talk) 15:13, 3 March 2009 (UTC)

But of the green men choosing door 1, 1/3 will win the car (and similarly, 2/3 who choose door 2). BTW - I'm "watching" this page so you don't need to ping me when you reply. -- Rick Block (talk) 15:24, 3 March 2009 (UTC)

No no, be carefull, because in repeating, not only green men are involved where door 1 has originally be picked by the earthly player, but also the ones where door 2 has been picked. That's precisely the meaning of introducing the green men! They form a conditioning on the event "door 3 is open" in the complete sample space. And that's why I said chances (regarding the doors) are assigned to the play(er).

Yes, it is important to be precise. I mean in cases where the player has initially picked door 1 and the host has opened door 3. -- Rick Block (talk) 16:00, 3 March 2009 (UTC)

But no green man ever knows which door is picked. Nijdam (talk) 18:19, 3 March 2009 (UTC)

Yes, but whether they know or not, in cases where the player has picked door 1 and the host has opened door 3, if they pick door 1 they have a 1/3 chance of picking the car. I'm conditioning on the event "player picked door 1 and door 3 is open", not just "door 3 is open". -- Rick Block (talk) 00:07, 4 March 2009 (UTC)

Your remark puzzls me. The trick of the green man is his ignorance of the choice of the player. I come back to it later. Nijdam (talk) 22:50, 4 March 2009 (UTC)

Yes, but any given green man is trading places with a player who has initially picked a door. Whether the green man knows the player's choice or not, any green man's choice wins with probability 1/3 if that green man chooses the door that was the player's choice and wins with probability 2/3 if that green man chooses the door the player did not pick. The green man's choice wins with probability 1/2 but it's 50%(1/3) + 50%(2/3). This is the exact issue I'm discussing with 216. -- Rick Block (talk) 01:20, 5 March 2009 (UTC)

Back

I'm back, and have been considering green men. In all of the problem the wording with payers and host and green men are not relevant, but the probabalistic formulation is. Then the "normal situation" is covered by the event: chosen door 1, opened door 3. The green man is an imaginative representation of the event: door 3 opened, that's all. So it is no use to ask about the probability of the green man to find the car behind door 1 in the "normal situation"; simply because he is not in that situation. A problem with the green man is the distribution of the initial choice of the player. Because it is not necessarily random, the green man has not necessarily a 50/50 chance.Nijdam (talk) 00:32, 16 March 2009 (UTC)

The green man (who has swapped places with a player at the point the player is asked whether to switch) has a 50/50 chance, but it is because his choice is random, not because the distribution of the car behind the doors is 50/50. At the point the green man is choosing we might as well have rolled a 3-sided die and put the car behind one door if the die was 1 and behind the other door if the die was 2 or 3. p(D1) = 1/3, p(D2) = 2/3 - but a random choice between these ends up with a 50/50 chance of winning the car. We can also ask what is the probability the green man wins if he happens to choose the player's initial door (D1), which is the same question as what is the probability that the car is behind this door. And the answer is 1/3. We can make the distribution of the car behind the doors as uneven as we'd like, and a random choice always has a 50/50 chance of finding it, since (1/2)p + (1/2)(1-p) = 1/2. According to 216 (in a discussion on the arguments page), Massimo Piattelli-Palmarini (a cognitive psychologist) presents a version of the MHP where the player "guesses" at the end. This strikes me as extremely bizarre - why go to all the trouble of establishing a non-equal probability distribution between two unknowns and then turn it into a 50/50 chance because the player guesses? -- Rick Block (talk) 04:10, 16 March 2009 (UTC)

Fair coins

BTW: in your discussion with 216 a fair coin itself as a coin doesn't have the 50/50 property. As a fair coin it is physical symmetric, so in a right flipping experiment the total experiment has the 50/50 property. A deck of 52 cards only has the property of cardinality 52. The experiment of drawing one at random shows the property of 1/52 chance for each individual card. Nijdam (talk) 15:37, 3 March 2009 (UTC)

Agreed, although I'm not sure this distinction is particularly relevant. -- Rick Block (talk) 16:00, 3 March 2009 (UTC)

Well, I think in the MHP the doors have no intrinsic probability properties, but the rules of the game determine the probs. Nijdam (talk) 18:19, 3 March 2009 (UTC)

Or, put another way, the probabilities of the doors are due to the rules of the game. Like the probability of heads/tails is due to the physical properties of the coin. -- Rick Block (talk) 00:07, 4 March 2009 (UTC)

Well, we like to believe this, but the way the coin is flipped is part of the fairness. I even may argue that the flipping to some extent is more important than the physical symmetry. Nijdam (talk) 22:35, 4 March 2009 (UTC)

Yes - there are two issue, the fairness of the coin and the fairness of the experiment. But again I think not relevant at 216's level of understanding. -- Rick Block (talk) 01:20, 5 March 2009 (UTC)

If you think that this notice was placed here in error, you may contest the deletion by adding {{hangon}} to the

Proposed new math formulation for the MHP.

Hello, I have responded to your comments on the proposed Math. Form. for the MHP. Thanks for taking your time to reply, if you care to. glopk (talk) 17:19, 1 July 2010 (UTC)

My Monty Hall analysis page

Thanks for your input to my page. I have started again near the end with the player's initial choice being shown explicitly under the main heading 'Full calculations including the players initial car choice'. I think the player's choice is important in some cases. I have then worked the most general case through to provide a solution for every possible formulation of the MHP with the standard rules I think this is a better way of looking at the problem. We now have a general solution to which can be applied the various distributions and conditions to get a specific answer. My aim is to separate the maths, about which there can be no disagreement, from the philosophy which we have all been arguing about.

Although there are no new concepts in the complete solution it has not been published anywhere. Maybe it could be. If you are interested I suggest we discuss it further on the analysis talk page. Martin Hogbin (talk) 16:36, 24 July 2010 (UTC)

I am grateful for your help in getting my analysis page right. I want to try to separate plain mathematical fact from philosophical argument. I would like to discuss the best way to present the information clearly with you, either here or on the analysis talk page. Martin Hogbin (talk) 11:05, 29 July 2010 (UTC)

MHP reference

• Martin Hogbin, W. Nijdam. [1] The American Statistician. May 1, 2010, 64(2): 193-194. doi:10.1198/tast.2010.09227.

Mediation resumes

The mediation of the MHP case has re-started. If you wish to participate, would you be willing to check in on the case talk page here? Note that the mediators have asked that participants agree to certain groundrules. Sunray (talk) 06:57, 11 August 2010 (UTC)

Game theoretic issue

You put up:

Nijdam: MHP has to be treated with game theory

Most people consider the MHP a psychological game, and hence it may best be treated with simple game theory.

I respond as follows.

@Nijdam, isn't this quite a straw man you're putting up here? I'm not aware of anyone in the world that thinks this, and I cannot imagine how you could possibly think they exist. Suggestion for rephrasing (but of course, it's your perceived issue of a perceived opponent):

Nijdam: Game Theory gives insight to MHP

Some people (perhaps many, at first exposure to MHP) start instinctively, it seems, to think in terms of psychology. Is Monty Hall trying to trick me? If I do that he might... Does he know that I know that he knows...? Correspondingly, one might say, it has also been treated mathematically with elementary game theory in the economics / decision theory / game theory world. Correspondingly, vos Savant did not ask 'What is the conditional probability?' but 'Should you switch?'

Vos Savant asked for a decision, please note, not a probability!

Now go on to try to mention the useful insights. Let me give you a little help

From game theory we learn that the wise player would randomize his initial choice secretly at home in advance, in order to get the best of all possible worlds; to be totally free of any mathematical "assumptions", how academically natural or conventional they might seem to be.

You're supposed to present a perceived view of a perceived opponent with as much empathy as you can muster.

It's about being on a game show. But we're not nuts. You're not allowed to ask advice when you stand there on the stage! You'ld better ask advice in advance.

Maybe vos Savant should have written "You're going to be on a game show. If you answer the silly questions correctly, you'll be faced with a choice of three closed doors..."

I'm not trying to put words into your mouth here, just trying to help things along. Gill110951 (talk) 12:43, 15 August 2010 (UTC)

Reading this message could be to your advantage

Here's something else it would be good to discuss over a beer one of these days, when I get out of this hell-hole (Athens in August).

Please take a look at [[2]]

There you will find out why the answer is "2/3" or "switch", as you like, and the method is unconditional, but the assumptions are NOT what some people like to call the standard assumptions, but the result is much much better, since much more useful in practice, much more often applicable, and just as easy to argue, whether formally or informally. And if you like you can refine or complete it, if you like, by proving that 2/3 is the best answer i.e. this is the best strategy. But I don't think anyone will ask you to prove that, since if one could do better than 2/3, we should have heard about it by now, agreed?

Hellpimp showed us the way but no-one noticed. Now the economists and game theorists know all this too and have frequently published on it so there would be no problem backing it up with super reliable sources. (Nalebuff is a bigger guy in the big scheme of things than Selvin or Morgan et al, sorry). But anyway it is such a clean and different simple argument that is "out there" that I think it deserves some consideration. But really I am not talking about pushing it onto the MHP page in order to push my POV; I am talking about something which I think is worth thinking about, if one likes to think of oneself as an authority on MHP. Which seems to be something we both have in common, right? Gill110951 (talk) 18:46, 15 August 2010 (UTC)

test

$Y_{0}={\frac {1}{Z_{0}}}\,$

$Y_{0}={\frac {1}{Z_{0}}}\,$

$zzz\,$

In case you missed this (I missed your comment)

You wrote:

If you mean to say that when the car is placed randomly (what used to mean uniformly), the conditional probs are at least 1/2, okay. But what do you mean by equivalent? If we do not make assumptions about the distributions, the conditionals being at least 1/2 is not equivalent to the overall being 2/3.Nijdam (talk) 15:19, 19 August 2010 (UTC)

Sorry, let me spell it out. Sometimes I'm too fast, often I'm too slow. You naturally agree that

Prob(switching wins) = sum (over 6 pairs of possible door chosen, door opened)

Prob( switching wins | door chosen, door opened ) times Prob( door chosen, door opened) .

Meaning:

$P(C\neq X)=$

$P(C=1|X=2,H=3)P(X=2,H=3)+$

$P(C=1|X=3,H=2)P(X=3,H=2)+$

$P(C=2|X=1,H=3)P(X=1,H=3)+$

$P(C=2|X=3,H=1)P(X=3,H=1)+$

$P(C=3|X=1,H=2)P(X=1,H=2)+$

$P(C=3|X=2,H=1)P(X=2,H=1)=??$

=2/3 if you make an assumption about the distribution. So what do you want to proof? Nijdam (talk) 23:40, 20 September 2010 (UTC)

We know the left hand side equals 2/3 if and only if the initial choice had 1/3 chance to hit the car. The only way we could have a strategy with a larger hit-chance than than 2/3 would be by not switching for some configuration which we know satisfies Prob( switching wins | door chosen, door opened) < 1/2. Because then we would replace a term on the right hand side by a larger term, Prob( staying wins | door chosen, door opened), resulting in a better probability on the left hand side for Prob ( this modified and mixed strategy wins). Assuming Prob( door chosen, door opened) > 0, which is the case in non degenerate situations - every door can be chosen, every door can hide the car.

We know that if all cars are initially equally likely, then all conditional probabilities are at least 1/2. Proving 2/3 (unconditional) can't be beaten is equivalent to proving that all the conditional probabilities are known to be at least 1/2. So yes, the conditionals being provably at least 1/2, under assumptions known by the player to be true, is mathematically equivalent to it being impossible to beat unconditional 2/3. QED.

BTW, I like your math! Must learn this myself. Gill110951 (talk) 17:27, 20 September 2010 (UTC)

Another try at explaining. One can make more or less assumptions and get more or less conclusions. I am telling you about the equivalence of two of the conclusions. They are mathematically equivalent.

There are six configurations (door chosen, door opened). Hence there are 2^6=64 non-randomized strategies for the player and a 64 dimensional continuum of randomized strategies: for each of the 64 configurations one can stay or switch or randomize between them.

Under no assumptions at all, the law of total probability tells us that the strategy "always switch" is the best strategy of all (with respect to unconditional hit-chance), if and only if no conditional probability disfavours switching.

If we only know the players' initial choice is right with probability 1/3, then we know that "always switch" beats "always stay" (unconditional 2/3 versus 1/3), but we don't know anything about the hit chance of mixed strategies.

If however we also know that the host is constrained to hide the cars uniformly at random, then one can show with Bayes that no conditional probability disfavours switching, hence "always switch" is optimal - beats all possible strategies. But one can also prove via a different route, in that situation, that "always switch" is optimal (2/3 can't be beaten). By the mathematical equivalence which I've explained to you, you've then also proven that no conditional probability disfavours switching.

The conditional solution completes the unconditional solution by showing that "always switch" is globally optimal, unconditional 2/3 can't be beaten. At the cost of a further condition on the host's behaviour. While the weaker assumption can be engineered to be true by the player.

In summary:

"always switch" is the optimal player strategy in terms of optimal unconditional hit-chance

if and only if

all conditional probabilities are known by the player to favour switching (or are neutral about it)

if and only if

the host is constrained to hide the cars uniformly at random.

But if you only know that your initial choice hits the car with probability 1/3, you do at least know that "always switch" beats "always stay" (2/3 versus 1/3). My conclusion is that all the approaches give you complementary and important information. The good solution to MHP is to give the results of all the approaches.

Gill110951 (talk) 05:49, 21 September 2010 (UTC)

confidence interval

Hi, please use the talk page of confidence interval if you want to include the word, likely. Your last edit incorrectly implied that a realized confidence interval has some probability of containing the true value. This is exactly the implication you want to avoid.

Generally, since you didn't respond to my last comment and another editor RVed you on this edit, I think you should start with the talk page. 0¹⁸ (talk) 15:42, 11 October 2010 (UTC)

friendly reminder

Hi, I just wanted to remind you that RVing an good faith edit (such as this edit) that is not vandalism or a policy violation, is a little jarring for the other editor. If you do want to take this extreme action for a good faith edit, please, at a minimum start a talk page discussion on the topic. Everyone forgets these things from time to time, so it is not a huge deal, but I just wanted to remind you that it is not taken as a kindness by other editors. Thanks, 0¹⁸ (talk) 17:13, 19 November 2010 (UTC)

I do not doubt the good faith of the edit. On the other hand, discussion doesn't just hold for the revert, but likewise for the edit.Nijdam (talk) 10:22, 20 November 2010 (UTC)

I don't understand what you are saying. I edited a page, you RVed it. I'm saying, please don't RV good faith edits, and when you do, please go to the talk to start a discussion. What are you saying? 0¹⁸ (talk) 03:14, 21 November 2010 (UTC)

Well if you want to make a major edit, first discuss it. Nijdam (talk) 16:21, 21 November 2010 (UTC)

Nah, better to just be bold. 0¹⁸ (talk) 00:53, 22 November 2010 (UTC)

arbitration case

You are involved in a recently-filed request for arbitration. Please review the request at Wikipedia:Arbitration/Requests#Monty Hall problem and, if you wish to do so, enter your statement and any other material you wish to submit to the Arbitration Committee. Additionally, the following resources may be of use—

• Wikipedia:Arbitration/Requests#Requests for Arbitration;
• Wikipedia:Arbitration guide.

Thanks, Rick Block (talk) 06:38, 9 February 2011 (UTC)

Monty Hall problem opened

An Arbitration case involving you has been opened, and is located here. Please add any evidence you may wish the Arbitrators to consider to the evidence sub-page, Wikipedia:Arbitration/Requests/Case/Monty Hall problem/Evidence. Please submit your evidence within one week, if possible. You may also contribute to the case on the workshop sub-page, Wikipedia:Arbitration/Requests/Case/Monty Hall problem/Workshop.

On behalf of the Arbitration Committee, (X! · talk)  · @144  ·  02:27, 12 February 2011 (UTC)

Timeline for evidence in Monty Hall case

Please see Wikipedia talk:Arbitration/Requests/Case/Monty Hall problem/Evidence#Timeline for Evidence, Proposed Decision. On behalf of the Arbitration Committee, Dougweller (talk) 16:41, 21 February 2011 (UTC)

Your comment on the arbitration conclusion talk page

You say I admit that the simple solutions are wrong: this is not true. Of course I agree that it is "wrong" (incomplete) as a full answer to what you see as the "right question". But I do not agree with you that what you call the right question, is the right question.

Please correct your summary of my point of view. Richard Gill (talk) 09:11, 17 March 2011 (UTC)

Done, not a correction, but an clarification. Nijdam (talk) 10:11, 17 March 2011 (UTC)

Wikipedia:Arbitration/Requests/Case/Monty Hall problem closed

An arbitration case regarding Monty Hall problem has now closed and the final decision is viewable at the link above. The following is a summary of the sanctions that were enacted:

• Standard discretionary sanctions are enacted for all articles related to Monty Hall problem (broadly interpreted)
• Glkanter is banned from Wikipedia for one year, and is further subject to an indefinite topic ban on subjects related to the Monty Hall Problem.
• Nijdam is topic banned from the subject of the Monty Hall problem for a period of one year.
• Rick Block is restricted to 1RR on the Monty Hall article for a period of one year.

For the Arbitration Committee, NW (Talk) 00:47, 25 March 2011 (UTC)

Discuss this

About the simple solution

From Richards talk page

Richard, Why are you that keen in stating that the simple solution is correct be it together with symmetry arguments, where it's just the simple solution without any further arguments we, the other editors are fighting, and that is not correct, as you yourself have said, as a solution to the full MHP. You seem to spread some smoke curtain as if you want people to accept the simple solution (in its simple form), because your authority guarantees it may be extended to be correct. It never will be, because as you also have agreed, the simple solution does not set out to consider the needed conditional probability. I should not have to ask you this. And I hope you remember, calculating (or determining) the conditional probability by symmetry arguments is one of the solving methods I showed, and I also like to promote in the article. But why do you want to refer to this as "simple solution", while you definitely know, it is something completely different than THE simple solution. WHY? Have you yourself in the past accepted and defended the simple solution (the simple one) as solution to the full MHP? And do you try to justify this error??? Nijdam (talk) 10:13, 22 February 2011 (UTC)

Of course the simple solution alone does not solve the "full problem". But I do not agree that you *have* to solve MHP with the "full problem". Simple solution plus symmetry, and symmetry plus simple solution, do solve the full problem. Actually, symmetry first tells you that you do not have to solve the full problem: you need not condition on stuff which is independent. From the mathematical point of view the full conditional solution is just one way of showing that the simple solution is optimal (in the sense of achieving the highest possible overall success-rate), as well as good. Mathematics does not have moral or legal authority. Mathematics can't ever tell you that you *must* act in a certain way. It can only tell you that it is wise to act in a certain way. The applied mathematician must explain to his client why it is wise. In the real world there are many other issues, and maybe it is wise not to be wise in some respects. Richard Gill (talk) 11:27, 22 February 2011 (UTC)

Conclusion

1. Full MHP = the player is asked to decide after the host has opened the goat door
2. Simple MHP = the audience is asked to decide even before the player made her first choice
3. Simple solution = decision based on the unconditional probability
4. Bayes solution = decision based on the conditional probability, calculated with the use of Bayes' law
5. Symmetry solution = decision based on the conditional probability, calculated with the use of symmetry
6. Simple solution solves the simple MHP
7. Simple solution does not solve the full MHP
8. Bayes solution solves the full MHP
9. Bayes solution is equivalent to the symmetry solution
10. Simple solution <> Bayes solution
11. There are a number of logically incorrect reasonings, presented as solutions

Situation

Involved editors (before Arbcom decision): On one side:

• Rick Block
• Nijdam
• Kmhkmh
• glopk

all more or less experts on the matter and aware of the above mentioned points; hence strongly advocates of the clear mentioning of the problem with the simple solution. As they are completely right (confirmed by several reliable sources) the criticism that they are not flexible is unjust. Rick Block, more or less the spokesman, is completely right in sticking to his idea of presenting the article. BTW sources that consider the simple solution satisfying, cannot be reliable.

On the other side:

• Martin Hogbin (originally)
• Glkanter
• Gerhardvalentin (as far as I know, still does not understand the shortcoming of the simple solution)

all laymen at the start with only a basic knowledge of probability; strongly opposed to point 7

More or less in between:

• Martin Hogbin (recently),

has changed his position concerning the simple solution; now admitting (in an email to me) it is deficient as a solution to the full MHP, but considers the defect of minor importance, and just like Gill opposes the clear mentioning of the criticism

• Richard Gill

an expert on the topic, and also aware of the mentioned points, but for some strange reasons taking side of the second group as it comes to the presentation of the MHP in the article

Question

Why are the editors in the second group that strongly opposed to the clear mentioning of the criticism on the simple solution????

My concern

I did take part in this discussion because the MHP is widely used for educational purposes, And by just accepting the simple solution, which is no solution at all, students and teachers get the wrong idea about probability.

From Gill's talk page

Let me spell it out for you: the simple solution S0 reads: the car is with probability 1/3 behind door 1. As the opened door 3 does not show the car, it will be with probability 2/3 behind the remaining door 2. Another simple solution S1 reads: the initial choice of door 1 hits the car with probability 1/3, Hence switching gives the car with probability 2/3. My simple (!) question is: do you mention any of these solutions as a solution to the MHP in your courses? Just answer with no or yes, and the number of the solution. Nijdam (talk) 05:25, 18 July 2011 (UTC)

Nijdam, I do understand what you're getting at. I am careful to say things which are correct. So of course I don't give S0 as a solution. I do give S1 as a solution. But on wikipedia the rules are that you can only write what has been published in reliable sources, and the definition of reliable sources had got nothing to do with the truth or falsity of what is written in them. Wikipedia summarizes what people write. Also things which are incorrect. If you would like to correct wikipedia the only option you have is to write reliable sources yourself, and hope that others will write about what you have written. Then maybe in ten years or so wikipedia will cite you, too. Richard Gill (talk) 15:00, 18 July 2011 (UTC)

The Guy Macon Solution

No doubt you have been watching as I try to resolve the longstanding conflict. I will also be watching your talk page and will take any comments or suggestions you have under consideration. I also encouraged Glkanter to email me if he had any comments or suggestions, but he started insulting me so I blocked his emails. Guy Macon (talk) 21:16, 28 May 2011 (UTC)

Thank you for your trust. Nijdam (talk) 09:44, 29 May 2011 (UTC)

Suggestions

[keep] Vos Savant explained that the contestant should always switch to the other door. Initially, the contestant's chance of winning was 1/3. The chance the car was behind one of the other two doors was 2/3. When the host uncovered door No. 3, the chance that the car is behind door No. 2 became 2/3. Consequently, if the contestant switches, he has double the chance of winning the car.

Many people refuse to believe that switching is beneficial. After the Monty Hall problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine claiming that switching was wrong. (Tierney 1991) Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy.

[change; Savant's solution is not the simple solution] Some sources criticise the simple solutions because they because they fail for a variant of the problem in which the car is still placed randomly but the host is taken not to randomly choose between which of the two doors he may open under the game rules. In this variant the player still cannot lose by switching and will probably gain, having a 2/3 chance of winning on average if he switches.

[into; (I need help with my English)] Some sources point to the fact that Vos Savant seems to automatically assume the chance of the first door to hide the car to be fixed. However, after the host opened door 3 a new situation arises with new chances. It turns out the new chance for door 1 has the same value 1/3 as before, and hence one may conclude that door 2 has now a chance of 2/3. When Vos Savant was confronted with this criticism she tried to find a way out and finally said she considered the overall chance of winning the car when switching. This is often called a simple solution, as it does not count for the specific situation the contestant is in. This simple solution does not solve the full Monty Hall problem, in which the specific situation of the contestant is known.

On 2nd thoughts

As Savant's "solution" is not correct, it may be better to replace the explanation in the lead with the following:

[into (English??)] Most readers automatically assume the doors to equally likely hide the car, the contestant to be unaware of its position. and furthermore symmetry as to the role of the doors. Then the choice of the player and the opening of a door by the host does not influence the original 1/3 chance for the chosen door No. 1 on the car. When the contestant is offered to swap doors, the chance for door No. 1 is also 1/3, and hence the remaining door No. 2 has then chance 2/3 on the car. Consequently, if the contestant switches, he has double the chance of winning the car.

Incredibly: now write this out in proper math

Read the recent eprints by A.V. Gnedin on arXiv.org (maths section) if you want to see things written out completely and rather formally. But I would expect that for a smart person it is enough to notice that a player who is going not to switch in some situation, has at least a 1/3 chance of ending up with a goat. For instance, consider a player who chooses Door 1 and intends not to switch if the host opens Door 3. There is a 1/3 chance that the car is actually behind Door 2. Player chooses Door 1. Host is forced to open Door 3. Player doesn't switch. Player gets a goat.

We know that a player who always switches gets the car 2/3 of the time. So such a player has only a 1/3 chance of ending up with a goat. Conclusion: however you play, you'll end up with a goat with probability at least 1/3. "Always switching" is therefore optimal. Therefore all conditional probabilities of getting a car by switching must be at least 1/2 (none of them can favour staying).

Note that we only assumed here that the car is initially equally likely behind any of the three doors. We didn't assume anything about how the host chooses a door, when he has a choice.

Thus this short and elementary analysis covers the biased host case and makes computation of conditional probabilities superfluous. Remember: Vos Savant asks for a decision, not for a probability! Insights from decision theory (or game theory) can be useful when solving decision problems. Probability is not the only game in town. Richard Gill (talk) 10:21, 5 July 2011 (UTC)

Reliable source

Copied from Wikipedia:Requests for comment/Request board:

Can a source that is clearly mistaken, still be considered a reliable source for reference? Nijdam (talk) 19:07, 18 July 2011 (UTC)

Thank you for your question. If a source makes an obvious error of fact, it is clearly unreliable. Were you thinking of a specific source that makes such an error? hare j 22:25, 18 July 2011 (UTC)

The dispute

Then main difference in the dispute is about the "simple solutiion". This solution reads: the probability to hit the car in the initial choice of door is 1/3, hence switching gives the car with probability 2/3. This simple solution is NOT a solution for the full MHP, i.e. the version in which the contestant is offered to switch after the chosen door and the opened door are known (MartinHogbin and Gill have mentioned to agree on this). Yet this is the version most people consider to be the MHP. One side, MartinHogbin, Gerardvalentin, Gill as the main participants, are unwilling either to accept this, or to make this clear to the readers. The other side, Rick Block, Kmhkmh, glopk, myself and others, want to structure the article in such a way that from the start this difference is clear to every reader. 09:45, 23 July 2011 (UTC)

Nijdam, I guess you are talking here about a famous paradox, not about teaching and learning conditional probability theory in the mathematics class room, being quite another thing. What you call here a "difference" makes no difference whatsoever to the decision asked for, because this is completely irrelevant to the decision asked for. Even the greatest difference is completely irrelevant, because to switch will never ever be of disadvantage, and by switching now and here you produce the maximum benefit, just read the sources. Gerhardvalentin (talk) 10:08, 23 July 2011 (UTC)

Sorry Gerhard, it turned out to be pointless to try to make you understand the problem and its solution. You just demonstrated this again. Nijdam (talk) 16:33, 23 July 2011 (UTC)

@Gerhard: instead of wildly kicking around and commenting, just write down your own vision on the content dispute. That may help others to understand where you stand. Nijdam (talk) 11:32, 24 July 2011 (UTC)

Comment on Martin's proposal for a content resolution question

Martin Hogbin is circumventing the actual content dispute. He formulates in one long sentence several statements, some of which are not true. (1) The MHP is indeed a simple probability puzzle, but just because of this, used as an undergraduate exercise in conditional probability. (2) It is part of the dispute whether VosSavant and others solve (correctly) this simple puzzle. Nijdam (talk) 14:03, 25 July 2011 (UTC)

Richard's truth

Interesting to notice Richard's plea for the truth, seemingly as long as it is his truth. As he himself has admitted (see above) the simple solution does not solve the full MHP. Yet the full version is what most people consider the MHP. And for good reasons! So, let's present the truth, i.e. the correct (conditional) solution (maybe without being too technical and not explicily mentioning the term conditional) to the full MHP. Nijdam (talk) 21:26, 26 July 2011 (UTC)

Maybe Richard can be more specific about: The simplists should be aware of the concepts of conditional probability and take care not to write statements which are mathematically speaking false. The conditionalists could think about presenting conditional solutions which build on the simple solutions. Or is this also a red herring? Nijdam (talk) 15:38, 28 July 2011 (UTC)

Covariance

Hi there. I am new to Wikipedia. May I know what's your objection with my edit on "Geometrical Interpretation" of the covariance?

Tfkhang (talk) 07:55, 4 December 2011 (UTC)

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From Martin's talk page

Nijdam, you continue to try to teach me what I already know. Let me start by giving you a brief moment of victory. Using modern probability theory and with the sample space that you prefer and seem to assert is the only possible one to use, what you ask cannot be done. You are quite right, on the basis that you prefer, the "combined doors solution" makes no sense. However, there are other ways of tackling this problem, even within modern probability theory. Please explain to me why you select the sample space that you do. Martin Hogbin (talk) 09:20, 22 February 2011 (UTC)

Dispute resolution survey

Dispute Resolution – Survey Invite Hello Nijdam. I am currently conducting a study on the dispute resolution processes on the English Wikipedia, in the hope that the results will help improve these processes in the future. Whether you have used dispute resolution a little or a lot, now we need to know about your experience. The survey takes around five minutes, and the information you provide will not be shared with third parties other than to assist in analyzing the results of the survey. No personally identifiable information will be released. Please click HERE to participate. Many thanks in advance for your comments and thoughts. You are receiving this invitation because you have had some activity in dispute resolution over the past year. For more information, please see the associated research page. Steven Zhang DR goes to Wikimania! 22:53, 5 April 2012 (UTC)

Invitation to comment at Monty Hall problem RfC

Because of your previous participation at Monty Hall problem, I am inviting you to comment on the following RfC:

Talk:Monty Hall problem#Conditional or Simple solutions for the Monty Hall problem?

--Guy Macon (talk) 22:23, 6 September 2012 (UTC)

Apology

It seems I misunderstood your intention. I have struck my comment. I do believe you have some uber-solution in your head that you are going to reveal eventually (or coax Martin into demonstrating), I apologise for believing that you would try to put it unsourced into the article. --Elen of the Roads (talk) 12:58, 17 September 2012 (UTC)

Accepted, no hard feelings.Nijdam (talk) 18:34, 17 September 2012 (UTC)

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Regarding your recent reversion in Statistical Hypothesis Testing.

Nijdam, You recently reverted an edit of mine in Statistical hypothesis testing with the comment "Why the remove?". I don't know whether it is better to reply here or in a talk section. The short answer is that opinions by Fisher properly belong in the prior Origins and early controversy section which is already lengthy. I was hoping that the Null hypothesis statistical significant testing vs hypothesis testing section would focus on more recent facts and (perhaps) opinions rather than repeating the classical disagreement. The reverted edit also deleted the "opinion" that the two formulations are complimentary while retaining the supporting facts. What is your view?159.83.196.1 (talk) 23:35, 1 May 2013 (UTC)

I do not care much where the remark of the clash properly belongs. Of course it fits into the opinion section, but the section from which you removed it, also even needs the mentioning of it. Nijdam (talk) 16:07, 2 May 2013 (UTC)

I disagree and may remove it again, but not before I have well-sourced material to replace it. I just got a recent book by Lehmann that discusses the contributions of Fisher & Neyman including the dispute. Lehmann seems balanced to me, but reflects Neyman's opinion on the mathematical unity of the two formulations. Do you have any more recent references that support Fisher's position that they are different?172.249.254.111 (talk) 18:12, 2 May 2013 (UTC)

Confidence intervals

Why did you undo my '(or less)' edit? Thanks. Mmitchell10 (talk) 14:30, 26 December 2013 (UTC)

Sorry, I didn't well look at the given definition. It all depends on the definition. As the article mentions, some authors, quite a lot actually, and for good reasons, use as a definition

${\Pr }(U<\theta <V)\geq \gamma$

Then the "(or less)", is necessary. Maybe this has to be explained better. Also in some discrete cases, the given definition may not lead to a CI with exact the desired level. Nijdam (talk) 12:34, 30 December 2013 (UTC)

OK, but the whole of the article up to that point assumes that a 90% CI (for example) means exactly 90%, not at least 90%, in which case it is exactly 10%. As we can't have the probability of the CI covering the parameter value + the probability of the CI not covering the parameter value not equalling 100%, surely the sentence should either be a '90% (or more)' CI, to go with a '10% (or less)' chance, or else it's a 90% CI with a 10% chance.

The fact that a calculated CI may not end up giving exactly the expected or required level of confidence sounds to me like an important issue which merits a section of its own, rather than appending this fact onto the end of a section with a different aim. What do you think? Mmitchell10 (talk) 11:54, 31 December 2013 (UTC)

I agree; maybe we even mention this already in the definition. Nijdam (talk) 12:14, 1 January 2014 (UTC)

You are boxing yourself into a corner

In the MHP article talk page you seem to be asserting that the answer to this question:

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 change to the unchosen door?" Is it to your advantage to switch your choice?

Requires the use of conditional probability in which the specific door chosen by the host is a condition.

Firstly, I am sure that you will agree that the door initially chosen by the player must also be considered as a condition.

How, though, do you explain why the specific goat (we know two exist) revealed must not also be conditioned upon? Martin Hogbin (talk) 08:38, 20 June 2014 (UTC)

I mentioned the first chosen door and the door opened by the host as known, and hence as a condition. And now for something completely different: the exposed goat. If the problem states the black goat is shown (the other one being white), the goat would definitely be part of the condition. If on the other hand we would understand 'door with goat' is equivalent to 'empty door', there is no extra information, and the goat will be no part of the condition. Now what if you would be more catholic than the pope (as we say), and stipulate that you see a goat, although you do not know which one of the two. As you do not know which goat you see, your condition would be: {X=1, H=3, G=1 or G=2} (Iin the usul notation, with G the exposed goat). But this event is the same as {X=1, H=3} . Seeing a goat is not informative. Nijdam (talk) 18:26, 20 June 2014 (UTC)

In what way is an unnumbered door informative? Martin Hogbin (talk) 18:31, 20 June 2014 (UTC)

Which door is unnumbered? BTW: do you understand what I explained?Nijdam (talk) 18:34, 20 June 2014 (UTC)

Maybe I have misunderstood you. I thought that you were saying that the question stated above, in which the doors are not numbered, still requires a 'conditional' solution. Martin Hogbin (talk) 18:48, 20 June 2014 (UTC)

Actually I don't know what you mean by unnumbered doors. We discussed part of this before, if I recollect well. When I can see the doors, they are "numbered", i.e. distinguishable. Please explain to me what you exactly mean. Nijdam (talk) 11:11, 21 June 2014 (UTC)

In the problem, exactly as written above, do you think it is necessary in the solution to condition on the specific door opened by the host? If so, what are your reasons for your choice and what notation/terminology would you use for describing the doors in the solution. Martin Hogbin (talk) 11:37, 21 June 2014 (UTC)

Of course, as soon as a door plays a specific role, we have to conditon on it. As we discussed: label the doors, the simplest way is numbering them 1,2 and 3. Introduce the random variables C (car), X (first choice) and H (host), each taking the possible values 1, 2 and 3. At the moment of the decision to switch or not. we know: X=x and H=h; x<>h. Etc. Notice the difference between X and x, and between H and h. Both X and x may be described as the door first chosen by the player (and analogous H and h). Yet they are different, and in just plain text the difference is often not noticed, and hence leads to misunderstanding. Nijdam (talk) 09:06, 22 June 2014 (UTC)

The question tells us nothing about how the doors are arranged or what they look like. I agree that doors are not bosons; they are not fundamentally indistinguishable. So, you are suggesting that we label the doors ourselves. On what basis do we do that? Do we assume that the problem statement refers to a single specific play of the game in which doors can be identified? If so, then we can label the doors on the basis of their role on that single game. We could label the door that the player initially choose I, the door that the host opened H, and the remaining door R. If you prefer, as the the labels we use are a matter of our free choice, could use 1, 3, and 2 respectively. Do we agree so far.

We can then introduce the random variables as you suggest but before we can do anything else we need to consider what might have happened or, what might happen if the game is replayed a number of times, otherwise we have no basis for any probability calculations. Agreed? Martin Hogbin (talk) 13:38, 22 June 2014 (UTC)

I didn't make any assumption about the arrangement of the doors or their looks. hy should I? I do not need any base to label the doors. We may think as a frequentist, but also as a Bayesian. If you prefer to name the doors after their role in this specific instance of the game, no problem. But If you would like to repeat the game, you might be forced to mention for instance the host opens the door which was chosen first in the first instance of the game. BTW: why dod you use symbels I, H and R, as I already used x and h? Nijdam (talk) 16:33, 22 June 2014 (UTC)

I cannot see what basis there is for labelling the doors except by their role in this specific instance of the game. What, otherwise, does door 1 mean. How can we ever say which door it is? Martin Hogbin (talk) 23:34, 22 June 2014 (UTC)

Why should we label the doors?

Just look at it, it's the door you've just painted a number 1 on. Nijdam (talk) 07:57, 23 June 2014 (UTC)

OK, I will accept that it is, in principle, possible to uniquely label the doors, because they are, by their very nature, distinguishable. My next question is, why should we do this?

Would you say that it is absolutely necessary to label the doors to answer the problem above or would you say that it it just gives us a better and more complete solution if we label the doors? Martin Hogbin (talk) 09:19, 23 June 2014 (UTC)

Well, it is not absolutely necessary, but it makes it easier to talk about the problem and to formulate sotlutions. We could as well say "given the door chosen and the door opened by the host", as long as we keep in mind these are specific doors. For the solution it makes no difference. Nijdam (talk) 15:09, 23 June 2014 (UTC)

I rather meant: "Is it necessary to distinguish the doors, in some way or other, to answer the problem", or would you accept that we could decide at the start that the door identities are unimportant and just ignore them as an alternative, but less rigorous, way of solving the problem.Martin Hogbin (talk) 15:46, 23 June 2014 (UTC)

We were here before. The problem as stated asks for conditional probability given the door chosen and the door opened. This is forced as these doors are known to the player (us, the audience). This also is the interesting formulation, as at he moment of decision to switch or not, the player is left with two closed doors, and hence tends (is this good english?) to consider the two as equal in chance. Actually noone doubt this formulation, but some, being mistaken by the simple explanation, then are trying to find another formulation to justify there mistake.Nijdam (talk) 15:42, 24 June 2014 (UTC)

The problem does not mention any conditional probability, it asks whether you should swap or not, although I agree that most people consider it to ask for the probability that you will win the car if you swap. I agree that this probability is, trivially, a conditional probability in so far as a series of events take place after the car has been placed and we are to calculate the probability after these events have taken place. There is nothing in the statement, however, to tell us that the events 'a door is chosen by the player' and 'a door is opened by the host' must be conditioned upon and that these are the only events which must be conditioned upon. These are decisions to be made by the solver of the problem.

You justify your choice of events to condition upon by saying 'This is forced as these doors are known to the player (us, the audience)'. If you mean by this that we should condition upon all events which are specifically mentioned in the problem statement then I would agree that this is a logical position to take. I would also agree that it is reasonable to condition only on events in which there is a clearly stated or implied choice of action. So, although the host says "Do you want to change to the unchosen door?" we are not told, or lead to believe, that he might have said anything else so we do not need to condition on this event. These would seem a reasonable and objective rules by which we can decide what to condition upon.

Although the statement above only says 'You pick a door' it can be argued that we know that there are three doors and the player picks one of them so we should condition on that choice.

The host though, knows what is behind the doors and uses this information to make his choice, specifically he does not choose the door hiding the car. It is clear therefore that it is not the door that is the basis for the host's choice of action but what is behind the doors. Behind the doors, we are told, is one car and two goats and the positions of these items are the basis for the host's decision. The host must not choose the car but he can choose either of the goats (if the player has initially chosen the door hiding the car). We must therefore by the rules that you have stated (and I have expanded) also condition on the host's choice of goat. Martin Hogbin (talk) 08:51, 25 June 2014 (UTC)

The host does not choose a goat, at least nothing of this kind is mentioned. In the case he has a choice, it is widely accepted he flips a coin. Of course you may study situations in which the host behaves different, if you like. Nijdam (talk) 09:17, 25 June 2014 (UTC)

On the contrary, the host must choose a goat; that is the rules of the game. How the host chooses between available goats is another matter. He may choose randomly between them or may choose randomly between the doors hiding them. Alternatively, he may have a preference for a particular goat (people often have favourite animals) or a preference for a particular goat-hiding door (although I have never known anyone with a favourite door), we are not told. I agree though that from a Bayesian perspective we should consider all eligible goats to be equally likely to be chosen, and for that matter the host equally likely to choose by goat rather than by door. Martin Hogbin (talk) 12:15, 25 June 2014 (UTC)

No contradiction. Nijdam (talk) 07:11, 26 June 2014 (UTC)

I certainly agree in the sense that, with the standard Bayesian assumptions, we always calculate the probability that the player will win the car, given the events described in the problem statement, to be 2/3. So, in that respect, there is no contradiction between the simple solutions, Morgan's solution, and my proposed solution. With the standard Bayesian assumptions, all the solutions give the answer (to probability that the player will win the car by switching, given the events described in the problem statement) of 2/3.

The argument in favour of the Morgan solution is that the simple solutions use the wrong method to get the right answer, in particular that the sample space used is incomplete. This is demonstrated by considering a, somewhat contrived, situation in which we apply standard Bayesian inference to the original car and goat placement but we assume that the host's choice of legal door is not uniform and this is therefore parameterised. This variation of the problem is used to show that the answer (as previously described) is not 2/3 but a value between 1/2 and 1 depending on the host-door-choice parameter. It is argued from this that, even though the simple solutions give the correct answer with Bayesian assumptions for the host door choice, they are incomplete or use the wrong sample space. It is certainly true that the Morgan solution is more general than the simple solutions in the sense that it allow for a non-uniform host door choice and I accept that in this sense it is better than the simple solutions.

My argument in favour of the my proposed solution solution is similar. I say that the Morgan solution uses the wrong method to get the right answer, in particular that the sample space used is incomplete. This can be demonstrated by considering an, equally contrived, situation in which we apply standard Bayesian inference to the original car and goat placement but not to the host's subsequent actions in general, and these are therefore parameterised. We need three parameters, one showing the probability that the host's action will be based on his door preference rather than his goat preference, another showing his potential door preference, and a third showing his potential goat preference.

Suppose that the host has no door preference but does have a favourite goat that he always reveals where possible. The probability that the host will open a given one of the two unchosen doors is still 1/2 but, if the non-favourite goat is revealed, the player is certain to win the car by swapping.

This variation of the problem shows that the answer (as previously described) is not 2/3 (even though the host is equally likely to choose either of the two unchosen doors) but a value between 1/2 and 1 depending on the host-goat-choice parameter. Thus, even though the Morgan solution does give the correct answer with Bayesian assumptions for the host goat choice, it is incomplete or uses the wrong sample space. My proposed solution is therefore more general than the Morgan solution in the sense that it allows for a non-uniform goat choice by the host and, on that basis, I assert that it is better than Morgan's solution. Martin Hogbin (talk) 10:23, 26 June 2014 (UTC)

Let me first say that the simple solution(s) are not correct. Then your comparison between Morgan's host with a possible preference for a door, and your host with a possible preference for a goat. Morgan's host ,as you recalled, leads to a different answer of the value of the conditional probability, lying between 1/2 and 1. Now you state your host also leads to such values. Please demonstrate this. Nijdam (talk) 15:08, 26 June 2014 (UTC)

Is that not obvious?

Let us consider the case where the host cares nothing about the doors but has a favourite goat (G2) which he always reveals where possible.

The player chooses from C G1 G2. The host then reveals G1. The player must have chosen G2 or else the host would have revealed it, so the player wins with certainty by switching. If the host reveals G2 then the player might, with equal probability have chosen C or G1 and wins with probability 1/2 by switching. Martin Hogbin (talk) 15:24, 26 June 2014 (UTC)

Words, words, words. Give me a formal proof.Nijdam (talk) 06:54, 27 June 2014 (UTC)

I might need your help with that. However, I will try to set the appropriate sample space and demonstrate my solution with that if that is acceptable. Martin Hogbin (talk) 08:54, 27 June 2014 (UTC)

That's okay; I'll be of help as much as I can. Nijdam (talk) 19:09, 27 June 2014 (UTC)

Distinguishable goats

For the sake of simplicity, I deleted the parts above and give a summary of the essential parts.

The sample space is described by the variables

C the door number of the car

X the door chosen by the player

H the door opened by the host

G the door hiding goat 1, this is dependent on C.

All have possible values of 1,2,3 and C and X are independent. Because of the symmetry we only consider the case {X=1}, and condition on this event.

We explicitly assume that the goats and cars are placed uniformly at random so

P(C=1) = P(C=2) = P(C=3) = 1/3

and

P(G=g|C=c) = 0 for g = c, and 1/2 g ≠ c.

Possible outcomes permitted under the game rules, with probabilities are:

P(C=1,G=2,H=2) = P(H=2|C=1,G=2) P(G=2|C=1)P(C=1) = 1/6 * P(H=2|C=1,G=2) = 0 (case "3") or 1/6 (case "G")

analogously

P(C=1,G=2,H=3) = 1/6 * P(H=3|C=1,G=2) = 1/6 (case "3") or 0 (case "G")

P(C=1,G=3,H=2) = 1/6 * P(H=2|C=1,G=3) = 0 (case "3") or 0 (case "G")

P(C=1,G=3,H=3) = 1/6 * P(H=3|C=1,G=3) = 1/6 (case "3") or 1/6 (case "G")

P(C=2,G=1,H=1) = 1/6 * P(H=1|C=2,G=1) = 0

P(C=2,G=1,H=3) = 1/6 * P(H=3|C=2,G=1) = 1/6

P(C=2,G=3,H=1) = 1/6 * P(H=1|C=2,G=3) = 0

P(C=2,G=3,H=3) = 1/6 * P(H=3|C=2,G=3) = 1/6

P(C=3,G=1,H=1) = 1/6 * P(H=1|C=3,G=1) = 0

P(C=3,G=1,H=2) = 1/6 * P(H=2|C=3,G=1) = 1/6

P(C=3,G=2,H=1) = 1/6 * P(H=1|C=3,G=2) = 0

P(C=3,G=2,H=2) = 1/6 * P(H=2|C=3,G=2) = 1/6

The strategy of the host is reflected in the still unknown conditional probabilities P(H=h|C=c.G=g). The host only has a choice when C=1, i.e. the car is behind the first chosen door. If he prefers door 3 ("3"), he'll open it. If he prefers goat 1 ("G"), he'll open the door with this goat. I introduced the relevant probabilities in the equations above. Nijdam (talk) 18:21, 17 July 2014 (UTC)

I understand that the strategy of the host is reflected in the still unknown conditional probabilities P(H=h|C=c.G=g) but not this is not Morgan's method they make no mention of G=g. P(H=h|C=c.G=g) is not the same as P(H=h|C=c).

Before proceeding, can we confirm how we have labelled the doors and goats. We have used the single observed instance of the game to label them. We label he door originally chosen by the player 1, the door opened by the host 3, and the goat revealed 1.

In the example above, let us consider the case where the host cares nothing about doors but has a goat preference, in particular, we can consider the case where the host always reveals goat 2 where possible, irrespective of door number. Goat 1 is the goat that, in this instance, was revealed. Now we have to condition on G=H because we know that in this instance the host did reveal goat 1 so the door opened must be the door hiding goat 1. Do you agree.Martin Hogbin (talk) 17:25, 18 July 2014 (UTC)

Discussion

By tradition, we label the door originally chosen 1, and the door opened by the host 3. We have to label the doors and the goats ourselves. We label the goat that was revealed 1 and the other 2. The door that was hiding goat 1 we also label G.

Can we do this? Martin Hogbin (talk) 12:58, 17 July 2014 (UTC)

Concerning the chosen door and the opened one, I see no problem, but on the other hand why shouldn't we just number the doors and have the player choose door 1 and the host open door 3. The revealed goat is a different story. Up till now I do not know which goat is revealed. Nijdam (talk) 18:03, 17 July 2014 (UTC)

This brings us to an interesting point. We do indeed not know which goat has been revealed but the same applies to the doors. Up until the host opens a door to reveal a goat, we do not know which door will be opened.

In the question as stated by me, we are not given identifying labels for either the doors or the goats. We can choose to give the doors labels but how do we do that? I numbered them after their eventual functions within the puzzle. Door 1 is the door that will be chosen by the payer and door 3 is the door that the host will open. Can we really do this? If 'yes' the we can do the same with the goats. Goat 1 is the goat that will be revealed and goat 2 is the other one.

If we decide that it is not correct to label the doors after their eventual functions then we have to decide how we can label them at all. Doors (and goats) are, in principle, distinguishable so we could argue that there must be a leftmost and rightmost door from the player's perspective and we could label them on that basis; 1 is leftmost etc. I would argue that the goats must also have some distinguishing characteristics, one would be lighter in colour that the other, one would be heavier, so we could label the heavier goat 1.

The only way now that we can distinguish between goat labelling and door labelling is by making the assumption that the player can see the doors at the start of the game. Even if we accept this assumption, though, it is, in my opinion, a weak argument. It is what what we know rather than what we literally see that is important in probability calculations. There is no real justification for giving the sense of sight special significance.

So my position is that we can, and should, either label both doors and goats or label neither and treat the problem as essentially symmetrical with respect to doors and goats. Martin Hogbin (talk) 09:29, 18 July 2014 (UTC)

I think it is commonly accepted the player is not blind. Distinguishing is no big issue. All ways of labelling are quivalent. For instance just label one of the doors as number 1, choose another to be door 2 and the remaining one as door 3. If you like to label the chosen door (in this instance of the play!!) as number 1, okay. But in repeating the game, in the next instance of the game the chosen door may have another number. Labelling the goats is also easy, just paint a big number 1 on one of them, and we're done.

Note that we can only label the door according to their function in this instance of the game. It is not correct to always call the chosen door door number one.Nijdam (talk) 16:15, 18 July 2014 (UTC)

Yes, I agree with that. The door originally chosen by the player in this instance is number 1 and you agree that we can call the goat revealed in this instance also number 1?

No problem. Nijdam (talk) 22:03, 18 July 2014 (UTC)

Comment by SPACKlick

Sorry to jump into this discussion but Nijdam, would it not be worth including |X=1 in those 12 options above, seeing as how it is already mentioned? It makes some of the unknown terms 1 or 0.

P(C=1,G=2,H=2,X=1) = 1/6 * P(H=2|C=1,G=2,X=1)

P(C=1,G=2,H=3,X=1) = 1/6 * P(H=3|C=1,G=2,X=1)

P(C=1,G=3,H=2,X=1) = 1/6 * P(H=2|C=1,G=3,X=1)

P(C=1,G=3,H=3,X=1) = 1/6 * P(H=3|C=1,G=3,X=1)

P(C=2,G=1,H=1,X=1) = 0

P(C=2,G=1,H=3,X=1) = 1/6 * P(H=3|C=2,G=1,X=1)=1/6

P(C=2,G=3,H=1,X=1) = 0

P(C=2,G=3,H=3,X=1) = 1/6 * P(H=3|C=2,G=3,X=1)=1/6

P(C=3,G=1,H=1,X=1) = 0

P(C=3,G=1,H=2,X=1) = 1/6 * P(H=2|C=3,G=1,X=1)=1/6

P(C=3,G=2,H=1,X=1) = 0

P(C=3,G=2,H=2,X=1) = 1/6 * P(H=2|C=3,G=2,X=1)=1/6

Highlighted in bold are those cases where H=3, italics are those cases where G=3. I'm not sure Martin's additional limitation of only looking at the subset of cases where the host reveals a specific goat from a specific door adequately reflects the original problem. SPACKlick (talk) 13:15, 17 July 2014 (UTC)

@SPACKlick. No need, it just complicates the notation. The P I use is the conditional probability given {X=1}. Of course you're free to introduce the condition {X=1} in the notation, with P the unconditional probability.Nijdam (talk) 18:03, 17 July 2014 (UTC)

On the other hand the mentioning of the condition {X=1} immediately shows, as you did, the impossible outcomes. Nijdam (talk) 18:06, 17 July 2014 (UTC)

SPACKlick, I hope you and Nijdam do not mind but I have given your comments a new heading so that I can explain what this discussion is all about here. I can move it to my talk page if Nijdam prefers.

I do not like the Morgan solution (generally called the 'conditional' solution round here), mainly because, in my opinion, it is neither fish now fowl. It is not the simple door-symmetric solution obviously intended by the writers, which is the solution given by all respondents up to the publication of the Morgan paper (I discount the original Selvin problem and following comments some ten years earlier). Neither, though, is the Morgan solution particularly thorough. It makes some some arbitrary (and generally unstated) assumptions, which are that the car is initially randomly placed, the player chooses randomly, but the host does not act randomly. If we decide that, even if initial distributions are all uniform, we should take a conditional (Morgan-like) approach then I think that we must also take the player's initial door choice into account.

There has, over several years, been much discussion as to what exactly should be taken into account in setting up this problem mathematically. It has been argued that, if we are to take the (completely unknown) host door preference into account, the we might as well take other completely unknown factors into account, such as what colour the host's tie was.

I think there is a reasonable set of rules to decide what is important, and that is to only consider possible choices that are mentioned in the problem statement. We are clearly told that there are three doors, one car, and two goats. I have argued for some time that, on that basis, we should also take the identity of the goat revealed into account in producing a complete solution. Whether we should do that and, if so how, is what this discussion is about. Martin Hogbin (talk) 09:59, 18 July 2014 (UTC)

No problem with the move and I have been following the discussion for a while so I get what you're discussing. I'm going to ramble a bit here and will edit another day when I can think straight. The point I was trying to make about labelling both doors and goats is. When we label the door the player picked, the remaining two doors and the two goats are still symmetrical therefore we can label the remaining doors by the hosts action and we can label the goats by the hosts action, because these are trivially independent of the door label chosen for the players door. However the Action that labels the goat revealed and the door it was behind links those two. So you remove consideration for a set of outcomes. The outcomes possible are; (where contestant always picks the door shown first, Capital is the door the host opens g is goat 1 and t is goat 2)

cgT cGt gcT gTc ctG cTg tcG tGc

We can label the doors 123 or 132 without making a difference. Any if we select that the host always opens door 3 (and we list the doors in order 123) sample space becomes a duplicate.

cgT ctG gcT gcT ctG cgT tcG tcG

If we then also specify that the host always reveals G and not T half of the sample space disappears. I think it's because the doors have their distinction tied to other parts of the puzzle PRIOR to the host opening a door. Even though they are not revealed, if you've picked the door with Goat 1 behind it, you have goat 1, so if you treat them as distinguishable, they are distinguishable from then on. Any later labels you apply are additional to that original label. SPACKlick (talk) 13:07, 18 July 2014 (UTC)

I do not quite follow what you are saying but do you agree that the goats are as distinguishable as the doors so, if we choose to distinguish between possible doors opened by the host then we must also distinguish between the goats. I have not seen a published solution which does this. Martin Hogbin (talk) 15:18, 18 July 2014 (UTC)

Added after the below discussion. Yes I agree the Goats are distinguishable just like the doors are, however I disagree with labeling them a posteriori.

As you will know, Martin, you may not like the 'Morgan' solution, it is however the correct solution, Another correct solution uses symmetry, but also to calculate the same conditional probability. What you call the 'simple door-symmetric solution' , can only be the simple solution, which is not correct. My advise, better not like a not correct solution. Concerning the tie the host weares, there is no mentioning of it in the problem statement. The goats are mentioned however, but nowhere is made a difference between the two. You will experience in the above analysis that distinguishing between the two goats will make no difference. Nijdam (talk) 16:28, 18 July 2014 (UTC)

I agree about the tie, that is what I said above, it is not mentioned in the problem statement so there is a logic to ignoring it.

Do you agree though that, if the host were always to reveal goat number 2 (not the one revealed in this instance) wherever possible that, in this instance, the probability of winning by switching would be 1? Martin Hogbin (talk) 17:08, 18 July 2014 (UTC)

Correct. Nijdam (talk) 22:42, 18 July 2014 (UTC)

And do you agree in the above case, that P(H=3|C=1)=1/2 (assuming the goats are placed uniformly).

Okay. Nijdam (talk) 13:39, 19 July 2014 (UTC)

Using the Morgan formula, this value of conditional probability gives us the answer 2/3 which is not correct. So what am I missing?

Please, show me.Nijdam (talk) 13:39, 19 July 2014 (UTC)

I am not sure what the problem is. Morgan make the value of P(H=3)|C=1) a parameter q.

You have just agreed above that P(H=3|C=1) = q =1/2. Morgan then show that the probability of winning by switching is given by

1/(1+ q) = 1/(1 + 1/2) =2/3.

You have also agreed above that, if the host were always to reveal goat number 2 (not the one revealed in this instance) wherever possible that, in this instance, the probability of winning by switching would be 1. I am sure that you would also agree that, if the host were always to reveal goat number 1 (the one that was revealed in this instance) wherever possible, the probability of winning by switching would be 1/2.

Thus Morgan give the answer of 2/3 whereas the correct answer is anywhere from 1/2 to 1 depending on the host's goat policy.

In general P(C=1|X=1) ≠ P(C=1|X=1,H=h) ≠ P(C=1|X=1,H=h,G=g) although applying the obvious symmetries inherent in the question

P(C=1|X=1) = P(C=1|X=1,H=h) = P(C=1|X=1,H=h,G=g) = 1/3. Martin Hogbin (talk) 08:46, 20 July 2014 (UTC)

Note: P(C=1|X=1)=1/3; P(C=1|X=1,H=2) =P(C=1|X=1,H=3) =1/3;

See under Sample space.Nijdam (talk) 10:51, 21 July 2014 (UTC)

I will look under Sample space and reply there but the above seems quite clear cut to me. For a host with a preference for goat 2 and who is indifferent to door number, the Morgan calculation gives a probability of winning by switching of 2/3 whereas the actual probability is 1, as shown by the above calculation or very simple logic. If the host shows the dispreferred goat the player must have originally chosen the preferred goat and thus must always win by switching. Where is the error in this argument? Martin Hogbin (talk) 12:59, 21 July 2014 (UTC)

Here, I think, you get into trouble as a consequence of your idea of labelling the goats.Nijdam (talk) 15:07, 21 July 2014 (UTC)

I see what you are doing now. You have given some probabilities in our sample space values that depend on the goat positions. That seems and odd way of doing it as you have a sample space probability that changes with every instance of the game. Martin Hogbin (talk) 09:43, 19 July 2014 (UTC)

I've not, so what are you referring at? Nijdam (talk) 13:39, 19 July 2014 (UTC)

In that case I do not understand what you mean by:

P(C=1,G=2,H=3) = 1/6 * P(H=3|C=1,G=2) = 1/6 (case "3") or 0 (case "G") Martin Hogbin (talk) 14:15, 19 July 2014 (UTC)

A little further I explained that (case "3") refers to a host with preference for door 3, and (case "G") to a host preferring goat 1. In fact I've written down two probabilities, one for case "3" and one for case "G".Nijdam (talk) 15:47, 19 July 2014 (UTC)

OK, I understand. If we are going to consider a host who might choose by door or goat then we need to add a parameter here. The host may choose by door half the time and by goat the other half. Martin Hogbin (talk) 09:49, 20 July 2014 (UTC)

Right, here is, as I now understand, where your variable M comes into sight. The point is, however, that we do not know, which value M takes. And when we specify the occurred value of M, we just may use one of the probabilities I gave.Nijdam (talk) 21:36, 20 July 2014 (UTC)

And, if the host were always to reveal door number 2 (not the one revealed in this instance) wherever possible that, in this instance, the probability of winning by switching would be 1? This clearly is addressing the "Morgan variant" outside the paradox, but it's clearly not addressing the intended paradox. The intended paradox is characterized in that the location of the prize remains unknown. This circumstance is due to the fact that the host of the intended paradox is obliged to secrecy regarding the car-hiding door (Norbert Henze). Literature unison: It is assumed that the host selects symmetrical when he should have the choice between two goats (white goat and black goat :). Better: As to the intended paradox, WE ARE TO ASSUME that the host selects symmetrical if he has a choice. So we HAVE to give him this chance! Otherwise it's never the intended paradox, but just a aberrant variant. (Nijdam, please excuse my intruding here). Gerhardvalentin (talk) 22:16, 18 July 2014 (UTC)

Gerhard, we are indeed talking about an aberrant variant, one of my invention in which the host has a goat preference. Do not be alarmed though, it is just a point of discussion. Martin Hogbin (talk) 09:06, 19 July 2014 (UTC)

Sample space

XCGH - cond. prob.{X=1} with pref. goat 1

1122 - 1/6

1123 - 0

1132 - 0

1133 - 1/6

1213 - 1/6

1233 - 1/6

1312 - 1/6

1322 - 1/6

2….

3….

What you call 'Morgan' formula: the average conditional probability of getting the car when switching, given X=1 and H=3. Averaged with respect to the two goats

P(C=2|H=3)=P(H=3|C=2)/[P(H=3|C=1)+P(H=3|C=2)]=1/[1/2+1]=2/3

The conditional probability of getting the car when switching, given X=1, H=3 and the preferred goat is shown.

P(C=2|H=3,G=3)=P(H=3|C=2,G=3)P(C=2|G=3)/P(H=3|G=3)=1/2

The conditional probability of getting the car when switching, given X=1, H=3 and the preferred goat is not shown.

P(C=2|H=3,G=1)=P(H=3|C=2,G=1)P(C=2|G=1)/P(H=3|G=1)=1

"Averaging" (Law of total probability)

P(C=2|H=3)=P(C=2|H=3,G=1)P(G=1|H=3)+P(C=2|H=3,G=2)P(G=2|H=3)+P(C=2|H=3,G=3)P(G=3|H=3)

=1x1/3+0x0+1/2x2/3=2/3

Where is your problem?Nijdam (talk) 10:53, 21 July 2014 (UTC)

The problem is that we have defined goat 1 to be the goat that was revealed, just like we defined door 3 to be the door that was opened and 2/3 ≠ 1/2. Morgan calculate the probability after the door has been opened to reveal one of the goats but before the player has noticed which goat it is. Martin Hogbin (talk) 16:04, 21 July 2014 (UTC)

No this cannot be a problem. As I mentioned before, instead of choosing yourself which of the goats will have number 1, you let the progress of the play decide. But once decided, the goat with number will keep this number. That means in a following instance of the play, the revealed goat may well be number 2.

Of course, just as the host might open door 2 next time but we are being asked for the probability for this instance. That is the whole point of the Morgan paper. They insist that we must calculate P(C=1|H=3) not just P(C=1). I say we must calculate P(C=1|X=1, H=3, G=3).

I agree, but we can't define G as g=h otherwise we miss possibilities that affect those probabilites. if G represents "The revealed goat" then P(G=3|H=3)=1, if we define G as the darker goat P(G=3|H=3) is on average 1/2 but it depends on the strategy of the host. We must define G and H independently for P(C=2|X=1, H=3, G=3) to accurately represent the problem. This is what I mean by labeling the goats Prior to the action of the host. We could label them after, if and only if we labelled the doors prior. In actuality it's more rigorous to label them all before.

Doors 1 2 3

GoatA GoatB Car (ABC)

The problem then for a complete morganesque solution is that the question doesn't even hint at which goat was revealed so you end up with the question being

P(C=2|X=1, H=3, A=3) + P(C=2|X=1, H=3, B=3)

Since P(A=3 U B=3|X=1,H=3)=1 [With given X=1 being redundant] The above line reduces to

P(C=2|X=1,H=3) which is Morgan's solution. SPACKlick (talk) 12:38, 22 July 2014 (UTC)

Sorry SPACKlick, I have some difficulties in understanding your comment.

What do you mean by "we can't define G as g=h"?

If G is the revealed goat, then G=1 ("Mary") or G=2 ("Kathy"), so never G=3.Nijdam (talk) 12:58, 22 July 2014 (UTC)

Earlier you defined G as the door containing Goat 1 and H as the door the host opens. If we label the goats 1 and 2 based on which one is revealed then G===H which is the linking I'm objecting to. If G is now the number of the revealed goat (1 and 2) then everything I've been saying is irrelevant. To make this clear. If we choose to label the goat Goat 1 because it is revealed then the probability of the host revealing Goat 2 is 0, and the host can have any goat preference he wants it doesn't affect things. SPACKlick (talk) 14:11, 23 July 2014 (UTC)

Well, I'll try to understand what you mean. Firstly: be precise! If G is the revealed goat, it is not a door number. In my definition, G is the number of the door hiding goat 1. Which of the goats has number 1? We have to choose one. We may number the (in this instance of the game) revealed goat as 1, and paint this number on its back. And as I explained above, in subsequent instances goat number 1 may not be revealed, but number 2. So be sure the probability of either goat to be revealed is 1/2 if the host has no preference. If he reveales goat 1 if possible, it is 2/3. Nijdam (talk) 10:07, 24 July 2014 (UTC)

I am being precise. 1) G means the door number behind which we find Goat 1. 2) We label the revealed goat Goat 1 after it is revealed 3) from 1)and 2) G===H. Unless we label the Goats independent of the Doors we run into this conflict, the labels must be arbitrary to maintain the original problem. SPACKlick (talk) 11:02, 24 July 2014 (UTC)

Sorry, but you have to read better. The only thing you can say is that this time the event {H=G=3} has occurred. Next time this may be different. Nijdam (talk) 12:27, 24 July 2014 (UTC)

I read fine. Tell me which of my premises are false?

• The door the host opens is labeled a posteriori 3
• The Goat revealed behind that door is labelled a posteriori "Goat 1"
• The hosts preferences wrt doors have been specified with respect to door number
• The hosts preferences wrt to goats have been specified with respect to "Goat 1" and "Goat 2"

The labeling ceases to be arbitrary when the two independent classes of Door and Goat are labeled with respect to the same function or action. The host opening a door serves to label both the Goats and 2 of the doors. I don't know how to make this clearer. By the method by which you are labeling your items P(H=3)=1 [which is fine by symmetry] P(G=H) = 1 [which on its own would be fine but...] together P(G=H=3)=1. This limits your sample space and reduces the effect of any preferences or strategies of the host. To say "It might be goat 2 next time" is to miss the fact that it could never be goat 2 the first time, irrespective of host preferences.

In essence functional labeling afterwards makes host preferences or strategies meaningless. You can get away with it by appeal to symmetry where only 1 class of objects is a posteriori labelled or where each class is independently labelled. Get it now?SPACKlick (talk) 14:45, 24 July 2014 (UTC)

Maybe your reading is fine, but you do misunderstand what I wrote. You're right that some authors try to call the opened door in each repetition of the game door number 3, which then is a stochastic label, placed on possible different doors in different instances of the game. This of course causes trouble for the right interpretation. I explicitly mentioned I do not act this way. I only let this instance of the game determine on which door I have to paint the number 3, and which goat I will call goat 1. Thes nimbers then are fixed. It's s a stupid way of doing, but Martin asked me if this was possible. I could as well, and completely equivalent, have numbered doors and goats in advance. P(H=3)=1/3, P(G=3)=1/3 and P(G=3|H=3) is depending on strategy of the host. If the host has no preference for one of the goats this probability will be 1/2. Nijdam (talk) 15:10, 24 July 2014 (UTC)

You say "I could as well, and completely equivalent, have numbered doors and goats in advance" but this is not completely equivalent. Numbering in advance makes the labels of the items independent, therefore any preference for door 3(p) and any preference for goat 1 (m) combine to form a strategy. You can only have such a strategy with pre-labeling. I'm starting to get the sense that the way your doing it is for the doors and goats to be pre-labeled something and "Door 1" and "Goat 1" are additional labels being hung on them a posteriori. If that is the case then you are right, it has no effect. SPACKlick (talk) 15:38, 24 July 2014 (UTC)

[Edit conflict]We are addressing a version of the question given above in which no door numbers are given. We had a discussion on whether you can, and should, label door numbers in that case. I think that we agreed that you could number them but the only meaningful way to do so was based on their roles in this particular instance of the game. Therefore, we might as well follow tradition and call the door chosen by the player 1, the door opened by the host 3, and the remaining door being 2. Exactly the same argument can be applied to the goats; the goat that was revealed is goat 1 and the other is goat 2.

I do not think that this causes any logical problems (but if it does they would apply equally to doors and goats). We recognise that in other instances of the game different doors may have been chosen and a different goat revealed but the fixed labels used refer to the roles of the objects in the described instance.

We also assume, as Morgan did, that the the question requires a conditional probability given the specific doors chosen and the specific goat revealed in this instance (if not all we need is a simple solution).

To make the problem clear now, I consider a particular case in which we know that the host has no preference for doors but always reveals goat 2 where possible. In this case, because the goats are randomly placed, P(H=3|C=1) = 1/2 which, using the Morgan formula, gives the probability of winning by switching as 2/3. In fact, because we have seen goat 1 revealed behind door 3 we know that switching wins with certainty. Martin Hogbin (talk) 15:02, 22 July 2014 (UTC)

Except you above defined Goat1 as the revealed goat. You chose to apply your labels that way: You've labelled The doors

1(The one the contestant picked)

2(The one nobody did anything with

3(The one the host opened)

And you've labelled the prizes

Car

Goat 1 (The Goat the host revealed)

Goat 2 (The Goat that remains hidden)

Because this labelling wasn't independet The probability Goat 1 is behind door 3 is 1. In the above notation P(G=3,H=3) = 1

Whereas if you label the goats before they are revealed

Car

Goat 1 (The darker goat)

Goat 2 (The lighter goat)

The probability of Goat 1 being the goat that is revealed is (on average) 1/2. All I'm saying is that while arbitrarily labeling distinguishable objects (Goats, Doors, Hosts, Prizes, Contestants & Cars) and then conditioning on them is fine and an accurate representation of the problem. If your labels are not independent then you can't condition on them, they're no longer arbitrary and the labels themselves start to affect the solution. SPACKlick (talk) 14:11, 23 July 2014 (UTC)

That is exactly what vos Savant did with the doors in the original problem. She says 'the host, who knows what's behind the doors, opens another door, say No. 3'. That is the first time that we hear the door number 3, it is not in any way defined in advance. If yours is a valid argument against my solution then it is also a valid argument against the Morgan solution.

I do not think it is a valid argument though, as I have argued in the second paragraph above. Martin Hogbin (talk) 19:02, 23 July 2014 (UTC)

It's not exactly the same if you do it with both doors and goats. The labeling is arbitrary if you just do it with doors, the labeling is arbitrary if you just do it with Goats, the labeling is not arbitrary, but dependent, if you do it with both doors and goats. This may turn out not to matter by symmetry but if you want to consider strategies where the host has door preferences, Goat preferences and combinations thereof you make part of the sample space impossible with dependent labeling. As I also mentioned above, neither form of functional labeling is particularly rigorous, the doors and goats should be distinguished prior to the first event, and then paired down by symmetry where possible.SPACKlick (talk) 09:42, 24 July 2014 (UTC)

There is no problem with labelling the doors and the goats any way that we like. What matters then is how the host chooses what to do next. He can have a goat preference and always reveal a preferred goat where possible or he can he can have a door preference and always open a preferred door where possible; he cannot do both. I have already mentioned that fact and said that we need to have a parameter (like Morgan's 'q') which indicates the probability that the host will choose what to do by goat rather than by door.

To simplify things in the discussion with Nijdam, I was considering the particular case where the host has a preference for one of the goats but cares nothing about the doors. Having solved that problem we could then go on to consider the more general case. Martin Hogbin (talk) 15:26, 24 July 2014 (UTC)

You have calculated the two conditional probabilities and averaged them (assuming a probability of 1/2 for each case) to arrive at the correct numerical answer. Morgan did not do this thus their solution is incomplete. To quote (nearly) , 'their answer is correct. Their method of proof, however, is not'.

Not assuming the probabilities to be 1/2, but calculating them. But you're right I first did another calculation, but changed it, without changing some numbers. I give the full calculation now.

Morgan did not meet this situation. Nijdam (talk) 18:16, 21 July 2014 (UTC)

Exactly, that is all that I am saying. Your mathematics is fine but Morgan took a short cut.

This is exactly analogous to the way the simple solutions are not complete but do arrive at the correct numerical answer in the case that the host chooses evenly between legal doors. Martin Hogbin (talk) 16:17, 21 July 2014 (UTC)

Far from that. Nijdam (talk) 18:04, 21 July 2014 (UTC)

I cannot see the difference. If you make the natural and simplifying assumption that the specific goat revealed in the instance mentioned in the problem is not important then you get the Morgan solution. If you further make the natural and simplifying assumption that the specific door opened in the instance mentioned in the problem is not important then you get the simple solution. Martin Hogbin (talk) 09:20, 22 July 2014 (UTC)

I'll have a deeper look into it, but for the moment I'd say, in the 'standard' MHP somre of the doors are specifically mentioned, and hence need to be labelled. To make thing easier, just use a car and two empty doors. The criticism on the simple solution, by Morgan's argument, keeps. Nijdam (talk) 12:58, 22 July 2014 (UTC)

You can fall back on the argument that the doors were numbered in the published problem but the goats were not but I find it hard to justify the distinction between arbitrary labels assigned by ourselves and arbitrary labels assigned by someone else (vos Savant). The numbers in the published question mean nothing in themselves the only meaning they have, either to vS or us, is to identify the role of the door in the instance of the problem described. We could all agree to use cumbersome titles like 'the door originally chosen by the player this time' but it is much easier to give it a shorter label, like door 1.

It could be argued that the numbering of doors by vos Savant is an instruction to the problem solver to consider door identity important and the lack of goat label is an instruction to the solver to consider the goat identity unimportant. In principle I could see some merit in that argument but, in this case, we know that vS did not want us to concern ourselves with door numbers. She intended to ask the question to which the simple solution is the correct one. Martin Hogbin (talk) 15:13, 22 July 2014 (UTC)

I think the fact that the doors are numbered is an indication, from VS, that the doors have to be treated as distinguishable (by host and contestant) but the goats do not. (I couple it with the "say, door 1" indicating that any of the distinct doors could fulfill the same function) The problem would be the same whether it was empty doors, indistinguishable particles, "identical" twin goats, differing coloured goats or one goat and one empty door. The defining feature of the two is "not car". That said, it's worth having considered the two goats as distinct and distinguishable because it aids proving that it has no effect on the problem. SPACKlick (talk) 14:53, 24 July 2014 (UTC)

Vos Savant's intention

Goats and doors, not being bosons, are always distinguishable, the question is whether we chose to distinguish between them.

It is possible to take the numbering of doors as an instruction from vos Savant to treat the door identities and important and to distinguish between the two doors that the host might open. However, we know from what she has since written that it was not her intention to do so, so that assumption is, in my opinion, a little perverse.

I do not understand your last comment. If we distinguish between the goats then it clearly does affect the answer. If the host has revealed his dispreferred goat then because the goats are randomly placed the probability that the host will open door 3 is 1/2 and using the Morgan calculation gives a probability of winning by switching of 2/3. We, however, know that under those circumstances the player will always win by switching because he must have originally chosen the host's preferred goat. Martin Hogbin (talk) 15:15, 24 July 2014 (UTC)

That all presumes the probability is calculated with the information of what the hosts preferred goat is, or the level of his preference (samewise for doors). There is no indication that any information about host strategy should be included in the calculation other than to simulate it as a universal distribution. on the Arguments page, I found an interesting quirk of accounting for a universal distribution of goat and door preference. SPACKlick (talk) 15:30, 24 July 2014 (UTC)

I have responded there. Martin Hogbin (talk) 15:47, 24 July 2014 (UTC)

Analysis

To show with a Morgan type argument that the simple explanation is not a correct solution: consider a simplified MHP with a car and two empty doors. The simple explanation always give 2/3 as the chance to get the car, also when the host has a preference for door 3. The correct calculation of the conditional probability shows the dependence.

In the standard MHP we have a car and 2 goats. Even if the Morgan calculation should turn out to be flawed, the above reasoning stays valid. Hence the simple explanation is not a sound solution!

But is the Morgan calculation flawed? No! I give the conditional probabilities given X=1, with a probabilty p as preference for goat 1.

XCGH - cond. prob.{X=1} with probability p as pref. goat 1

1122 - 1/6 p

1123 - 1/6 (1-p)

1132 - 1/6 (1-p)

1133 - 1/6 p

1213 - 1/6

1233 - 1/6

1312 - 1/6

1322 - 1/6

Direct calculation gives: P(C=2|H=3) = (1/6+1/6)/(1/6+1/6+1/6(1-p)+1/6p)=2/3.

Or by considering the positions of the goats:

P(C=2|H=3,G=1) = 1 and P(G=1|H=3) = (1/6)/(1/2)=1/3

P(C=2|H=3,G=2) = 0 and P(G=2|H=3) = (1/6(1-p))/(1/2)=(1-p)/3

P(C=2|H=3,G=3) = (1/6)/(1/6p+1/6)=1/(1+p) and P(G=3|H=3) = (1/6(1+p))/(1/2)=(1+p)/3

Hence

P(C=2|H=3) = 1x1/3 + 0x(1-p)/3 + 1/(1+p) x (1+p)/3 = 2/3

as it should be (or else I would have made an error in my calculations)

Nijdam (talk) 21:58, 23 July 2014 (UTC)

How is your reasoning? Formulate it in terms of probabilities.Nijdam (talk) 09:56, 24 July 2014 (UTC)

Draft Monty Hall Re-write

There is a draft of a significant re-write of Monty Hall here. I'd appreciate your comments especially as regards WP:NPOV and with an eye to the history of the page.SPACKlick (talk) 14:25, 9 July 2014 (UTC)

I may look into it, sectionwise, if you present th changes in such a way that the differences with the lemma are clear. Nijdam (talk) 16:05, 21 July 2014 (UTC)

There is an edit in the history labelled "Revert to Article", this provides a diff with the article. SPACKlick (talk) 07:55, 22 July 2014 (UTC)

Null hypothesis reversion with query "What's the intention?"

The intent of my attempted edit was to consolidate a number of citations of editions of Fisher's Design of Experiments into one thorough citation (currently reference 9) attached to the quotations section.172.249.8.109 (talk) 03:51, 24 July 2014 (UTC)

Well, a lot of text in the ref box seems to have disappeared, so where did it go? Nijdam (talk) 10:16, 24 July 2014 (UTC)

It did? A copy of the article dated June 11 shows about half the number of references. Was there a lot of hidden content? From my perspective the article has approximately doubled in size in the last 6 weeks. We are obviously looking at very different things. I am very confused.172.249.8.109 (talk) 04:20, 25 July 2014 (UTC)

Motivation for my edits: talk section "Marked article for much needed editing" from Thelema418 in December of 2013. Her comments have been neither challenged nor addressed. She complained that "some of the citations do not come from academic sources", so I felt free to change references. There are now almost three times as many inline citations as previously.172.249.8.109 (talk) 04:17, 26 July 2014 (UTC)

A question for you.

Nijdam, I would value your opinion on the correct sample space to use in this problem.

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Confidence Interval

Nijdam, I've just updated (actually rewritten) the new section on Misunderstandings. I know from your previous contributions that you are knowledgeable about this subject, so I would appreciate it if you could review it and let me know if you disagree with anything. Thanks. Dezaxa (talk) 06:51, 27 October 2014 (UTC).

Clarification motion

A case (Monty Hall problem) in which you were involved has been modified by motion which changed the wording of the discretionary sanctions section to clarify that the scope applies to pages, not just articles. For the arbitration committee --S Philbrick(Talk) 21:48, 27 October 2014 (UTC)

Cumulative frequency analysis

You PRODded this back in May, and it was deleted. Undeletion has been requested at WP:REFUND, so per WP:DEL#Proposed deletion I have restored it, and now notify you in case you wish to consider AfD. Regards, JohnCD (talk) 19:16, 15 January 2015 (UTC)

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