User talk:Martin Hogbin/Archive 8

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Boy and girl paradox

Have you seen the variant of this problem when we're told that the boy was born on a Tuesday? I've added a subsection. Richard Gill (talk) 07:42, 4 November 2012 (UTC)

Yes I have seen that. I have been trying to find a more general solution to the problem. There seems to be at least two types of variant. Martin Hogbin (talk) 21:14, 4 November 2012 (UTC)

In what way, more general? And what two variants? There are different ways you can "realistically" imagine getting the information you are told. Each different way gives a different answer. You have to model the "data generating mechanism". The paradox is designed to make this point clear. Just like MHP: what counts is not just what you see, but how you came to see it. If the data itself influences the chance you get to see this data, you had better take account if this. So recognising there are variants of this kind, is part of the solution process.

The variant "and the boy was born on a Tuesday" is more for fun. A new twist to the problem. Richard Gill (talk) 08:45, 5 November 2012 (UTC)

Yes, the entire problem is for fun but there are two kinds of information that you might be told. I do know that what counts is not just what you see, but how you came to see it. The MHP is an excellent and initially unintuitive example of this. The host opens a door to reveal a goat, does he do this by chance or must he do it, it makes a difference.

I will try to explain what I mean by two types of information (not doubt there could be more) a bit later. You might find it interesting, or you might not. Martin Hogbin (talk) 09:32, 5 November 2012 (UTC)

Devlin and Devlin-fixed (Rosenthal, Gill,... )

Focus on the odds on your door (door 1) hiding the car. Initially, 2:1 against. Host opens a (different) door revealing a goat (note emphasis on the word a). He does this under either hypothesis with certainty, so likelihood ratio - the ratio of the chances of getting the new information under the two hypotheses - is 1. Next you hear that it was door 3 he opened. The chance of that (door 3 and not door 2) is 50% under either hypothesis, by symmetry. So likelihood ratio is again 1 and final odds remain 2:1 against. Devlin forgot the last step (the information which door was opened is just as irrelevant, as far as the two hypotheses are concerned, as the information that a door was opened.

The two hypotheses in question are of course: the car is behind door 1; the car is not behind door 1 (it's behind 2 or 3). Devlin's missing step is the step from simple solution to conditional solution.

Bayes (odds form) gives us a systematic way to solve all variants of MHP simply, correctly and efficiently. This is not an academic side-show for students of mathematics. It's "numerical literacy" education of ordinary people to see through common fallacies in probabilistic reasoning in society (the media, law, public policy). That's why Rosenthal gives it so much prominence in his book. You should read it. Richard Gill (talk) 09:02, 5 November 2012 (UTC)

As I understand it, pretty well all probabilities in Bayesian probability are, strictly speaking, conditional, however, there is something of a convention in puzzles of this nature to condition only events which are additional to the game rules.

Under the game rules there are three doors, two goats, one car, and the host must always reveal a goat behind an unchosen door and must always offer the swap. When he has a choice, the host can open either legal door.

Some people take the the w/vS statement to tell us that the host did actually open door 3 to reveal a goat, but we know that he could, under the game rules, have opened door 2.

What we should therefore do is to have a table showing all the possible combinations of car position and host door opened and then condition this on the actual data (host opens door 3). We then make the natural Bayesian assumption that the host is equally likely to have opened door 3 as door 2. When we do this we find that it makes no difference which door the host opened but even though some people might find this quite obvious from the start we should go through this process as a matter of "numerical literacy" . Maybe in some other circumstances this might have been a vitally important point, so let us do it the laborious and pedantic way, just to be sure that we have not missed anything, even thought we might think that this step is actually unnecessary because it is obvious to us that will make no difference to the probability of interest which door the host opens.

Although I do not entirely agree with it, that argument makes some kind of sense to me. Have I got it right? Martin Hogbin (talk) 10:00, 5 November 2012 (UTC)

I am not sure that I quite understand your statement, 'the information which door was opened is just as irrelevant, as far as the two hypotheses are concerned, as the information that a door was opened'. Are you just saying that things would have been exactly the same had the host opened door?

No. I'm saying exactly what I said. You've chosen door 1; suppose the host opens another door revealing a goat, which door as yet unspecified, and asks if you want to switch. Would you be interested to know which door was opened? Obviously not. Why not. Because whether or not the car is behind your door, the chance that either other door is opened is 50-50 .. in the one case by your ignorance of the actual location of the car, in the other case by your ignorance of the host's opening-doors-strategy. Richard Gill (talk) 16:10, 5 November 2012 (UTC)

Are you extending the requirement to condition on the door actually opened by the host to the case where the doors were not numbered, because the host did actually did open just one of them but had a choice of two? Martin Hogbin (talk) 10:13, 5 November 2012 (UTC)

I'm taking the Bayesian line of updating prior odds by likelihood ratios to get posterior odds. For pedagogical purposes I'm doing it rather carefully, in a sequence of little steps.

Initial situation: the player has chosen door 1. 2:1 against that the car is behind it. Now the host is going to open a particular door and reveal a particular goat. I think of that information arriving in separate steps. Imagine that to begin with, the player keeps his eyes shut.

First step: he hears a goat revealed behind one of the other doors, doesn't know which of the two. If asked now whether he would like to switch to the other closed door, he would answer: "Yes please". The host was going to open a door a reveal a goat, whether or not the car was behind door 1, so the chance of his data - hearing the sound of a goat - is 1, whether or not the car is behind door 1. The likelihood ratio is 1:1 and the posterior odds are 2:1 in favour of switching.

Second step: the player of course opens his eyes before making his decision and sees that it was door 3 which was actually opened. He's got hold of another bit of information. Does it change his odds on the car being behind door 1? No: because if the car is there, the chance is 50% that the host opens door 3 (by ignorance of how the host opens a door when he has choice), while if the car is not there, the chance is also 50% that the host opens door 3 (by ignorance of where the car actually is). So the likelihood ratio is 0.50 : 0.50 or 1:1 and the odds on the car being behind door 1 still don't change.

Third step: he takes a close look at the goat and sees that it is a girl goat, or whatever, ... this won't change his odds any more, since yet again, the chance of it being that particular goat is the same for him, whether or not the car is behind door 1.

It's nothing to do with the numbers of doors. Call them left, right and middle if you prefer.

Pedagogical purpose of this exercise: the difference between simple and conditional solution lies merely in the question, do we explicitly explain why the number of the door opened is irrelevant, or do we take it for granted?

Subsidiary purpose: rescue Devlin's combined doors solution (i.e. patch his argument). Richard Gill (talk) 16:05, 5 November 2012 (UTC)

Just for a moment, Richard, you had me thinking that I had missed some subtle aspect of the problem but it turns out that it is the same old thing, does the door number (or door identity) opened by the host change the probability that the originally chosen door hides the car? Whether you state this question simply, as I have just done, or spin it out into a long laborious sequence, as you have done above, makes no difference, it is the same old question.

We have made some progress though, it least you now seem to accept that the identity of the goat revealed and the identity of the door opened are equally (un)important.

On the other hand, you have still not seen through Morgan's sleight of hand. If you really must make a meal of the host door identity that you should do the same with the player's original choice of door. We can, if you like, go through a long and contrived sequence of events. Firstly the player picks a door, without noticing anything about it. He takes the probability that it hides that car as 1/3, as there are three doors with nothing to choose between them. Then he notices that the door has a number '1' on it. He thinks for a while, does a quick Bayesian calculation, and finally concludes that the probability that the car hides the car is still 1/3. Then he notices that it is coloured green. Another calculation confirms that the probability is still 1/3. Why are you not demanding, in the interests of "numerical literacy", that we add this missing step to Devlin's solution as well. Martin Hogbin (talk) 19:46, 5 November 2012 (UTC)

I don't want to make a meal of the door identity question. I just want any arguments which are presented as logical arguments to be correct. Devlin pretends to present a logical argument but gives no justification for his last step. Some people thought it was obvious, some people thought it wasn't obvious. When Devlin was informed of this he suddenly began to doubt, himself.

In order to justify Devlin's last step you need to explain why the question of which particular door he opens (door 2 or 3) is irrelevant to the question of whether or not the car is behind door 1. You can say it is obvious, or you can write out the argument if you think it is not obvious. If you say something is obvious, but you are challenged (as Devlin was) you are obliged to fill in the details. This is not pedantry. After all, by opening door 2 or door 3 he does dramatically change the chance the car is behind door 2 or behind door 3! Some readers are uneasy here. Devlin was uneasy. Nijdam screams murder.

The more you put in the more you get out. I don't think it's a big deal. But let's at least be clear as far as the logic is concerned.

If you want to argue that switching gives the car with probability 2/3, then you have a very easy job. The simple solution. Switching gives the car if and only if your initial choice is a goat, probability 2/3. Combined doors: it's as if you are offered the option of exchanging door 1 for doors 2+3. 1/3 versus 2/3. Very intuitive, completely true.

If you want to argue that switching gives the car with probability 2/3, given you chose door 1 and the host opened door 3, you have a little more work to do. You'll need somewhere to use your (natural) assumption that the host is equally likely to open either door if he has a choice. For instance, you could do the simple solution first and then say that by symmetry the conditional probabilities are the same.

I think the whole idea that some solutions are correct and others incorrect is a waste of time. Every solution is correct in its own terms. Let's just be clear about this.

Secondly, like it or not, almost everyone who is trained in elementary probability, and certainly any beginner in this field, is going to sit down and calculate the conditional probability of finding a car behind door 2 given the player chose door 1 and the host opened door 3 from first principles, like Sevlin did in his second paper back in 1975. Like it or not, the article is going to contain a heap of material on conditional solutions. On wikipedia we are not supposed to make original syntheses, but my job as a mathematician outside of wikipedia is to do this. I like to show people the synergy between simple and conditional, the easy bridges, the smart ways to avoid doing blind computations. I would also like to counter the Morgan et al. propaganda! Richard Gill (talk) 15:06, 6 November 2012 (UTC)

Clear words to that propaganda are long overdue, addressing false starting point, saying "you could if you had, although you haven't..." Gerhardvalentin (talk)

I have no real objection to being pedantic/rigorous/mathematically sound so long as it does not detract from what was intended to be a simple puzzle but, if you do this, you should do it consistently. Just above, you show how you deal with the information you get from the identity of the goat revealed. It is exactly the same way as you deal with the identity of the door opened; you use some maths to show that it does not matter. If you are going to give a reason why the door opened makes no difference, you should also give a reason why the goat revealed makes no difference.

You say, 'If you want to argue that switching gives the car with probability 2/3, given you chose door 1 and the host opened door 3, you have a little more work to do'. Sure, I have no problem with that. You need to show that the fact that the specific door opened by the host (namely door 3) does not affect the probability that the originally chosen door hides the car but you also need to show that the fact the player has originally chosen a specific door (namely door 1) does not affect that probability. Of course it is obvious but so is the irrelevance of the hosts door. Pedantic is fine but pedantic and incomplete just makes one look silly.

Might it be that what is obvious and what needs argument depends on who you are? It's a fact that some people are unhappy with Devlin's argument, including Devlin himself, and Devlin himself did not know how to explain what he was doing.

Let me also point out a mathematical fact. Consider statement (A): "the chance is 2/3 that switching will give you the car, whatever door you first chose and whatever door is opened by the host". Note: statement (A) specifies 6 different conditional probabilities: Prob(car at a|player chose b, host opened c)=2/3 for all the 6 permutations (a,b,c) of (1,2,3). Consider also statement (B) "the initial location of the car is completely random" and statement (C): "the choice of the host is completely random (if he has a choice)". It's an inescapable mathematical fact that (A) is true if and only if (B) and (C) are both true. Somebody who claims (A) has explicitly or implicitly assumed and used both (B) and (C). Someone who claims (A) and uses (B) to support his claim but not (C) has not given a correct argument. This is in essence what Devlin did. Devlin is a mathematician and a figurehead, a leading public figure, in the mathematical community. By his own criteria and the criteria of the community he was writing for, his argument was wrong.

I don't ask you to find all this important. I just hope that intelligent lay-persons could at least appreciate the way mathematicians think. Society uses mathematics all over, and mathematics contributes vastly to society. Our value to science and society is precisely that we care about these issues. MHP on wikipedia is a topic which is interesting both to lay-persons and to people learning mathematics. Since it is a fixed item in every introductory probability and statistics course, this means that a huge proportion of the people consulting the wikipedia page on MHP will actually be students of probability or statistics. I wouldn't even be surprised if it's a majority. Like it or not, laypersons and mathematicians have got to work together on this article. There always have been and always will be roughly equal numbers of sources, editors, and readers of the two categories. Richard Gill (talk) 08:25, 7 November 2012 (UTC)

Unfair to Devlin? (follows on from above)

Can I make it clear that I am not fighting for mathematical sloppiness, I am fighting for something that most mathematicians hold dear - consistency.

The Morgan hypnosis seems to have even affected the way that mathematicians have addressed this problem. I do not think that the problem with Devlin's solution is necessarily where you think. This is what he says (I have bolded a bit):

"By opening his door, Monty is saying to the contestant 'There are two doors you did not choose, and the probability that the prize is behind one of them is 2/3. I'll help you by using my knowledge of where the prize is to open one of those two doors to show you that it does not hide the prize. You can now take advantage of this additional information. Your choice of door A has a chance of 1 in 3 of being the winner. I have not changed that. But by eliminating door C, I have shown you that the probability that door B hides the prize is 2 in 3.'"

How does Devlin know that 'the probability that the prize is behind one of them is 2/3' or that the probability that the car is behind the original door is 1/3? His real mistake is not to explain this. Your assumption is that he uses the fact that the car is initially placed uniformly at random but it equally could be that the player chooses uniformly at random at the start. Now your assumption (C) is not required and the rest of the argument is fine. Martin Hogbin (talk) 10:33, 7 November 2012 (UTC)

Morgan

I do blame Morgan for inventing this arbitrary distinction (that is their conjuring trick). They do not discuss the door originally chosen, or how the car is placed but they do make a big fuss about the host's choice of door. For some reason this has stuck in the minds of people who want to solve the problem 'properly'. If you want to do this you should consider all the possibilities allowed under the game rules and then explicitly condition them if you want the result for specific doors. It is no good waving your arms and saying it is obvious that the player's original choice makes no difference. You really should include the goat too. We are told that there are two goats and that the host reveals only one of them. Martin Hogbin (talk) 19:00, 6 November 2012 (UTC)

Selvin (1975b) gave the first conditional solution, using the natural (symmetric) distributions corresponding to ignorance of how the car is placed and ignorance of how the host chooses a door. Subjective probability! Morgan et al. however took a frequentist approach and showed that if you knew that the car is hidden completely at random, but you don't know how the host chooses a door to open, conditional probability still tells you to switch whatever door is chosen and whatever door is opened. To my mind, this is an academic diversion (useful as a pedagogical example in an introductory probability course). Of course they sold their paper by saying that Vos Savant had been wrong. And they were soundly criticised for this.

An in-depth (professional, academic-in-positive-sense) study of MHP sooner or later has to address the issue: what kind of probability do we want to talk about? Different choices will lead you to different analyses of MHP.

And mathematicians are always interested in generalizing problems. In particular they always ask: can we weaken the assumptions and still get the same conclusion? Morgan et al. made a novel mathematical contribution to that question. Richard Gill (talk) 08:38, 7 November 2012 (UTC)

Yes, Morgan did generalise the problem, but they did it badly. They were inconsistent in their approach in that they generalised one aspect of the problem (the host's door choice) but not another (the player's door choice). It is not clear from their paper whether they did this intentionally or not but I suspect that it was intentional because one generalisation leads to an interesting result and the other does not. Their is no reason why they should not do this but their legacy, unfortunately, is to leave many people thinking that the 'correct' or 'complete' solution to the MHP requires consideration of and conditioning on the specific door opened by the host but does not require consideration of the door originally chosen by the player. There is no logical reason for this. We should either notice at the start the symmetry with respect to door numbers and use this as a reason to ignore both the player's and the hosts choices of door or we should consider both door choices important and do our calculations accordingly. To pull one choice out of a hat and not the other is a conjuring trick. If Morgan had made clear in their paper that they were investigating a specific chosen generalisation of the problem then it would have been seen for what its is, in investigation into a slight problem variant rather than a recipe for how the problem must be approached by everyone.

In their approach, even if you would make the player's initial choice random, you are going to condition on it. A waste of time to look at such a generalization. They do the generalization which leads to new, mathematically interesting results. Richard Gill (talk) 12:32, 7 November 2012 (UTC)

Exactly the same generalising argument that Morgan used with the doors applies to the goats too.Martin Hogbin (talk) 09:45, 7 November 2012 (UTC)

From their frequentist point of view there is simply no point in modelling the goats in more detail. And we already agreed (I thought) that from the subjective probability point of view it is a pointless exercise. I think that if you are a frequentist statistician then their generalization is natural and interesting. So I think you are wrong to call this a conjuring trick. They are writing for a particular audience within a particular culture, and within that context, what they do is sensible and interesting and natural. Only their criticism of Vos Savant is uncalled for.

Please understand me: I'm *not* saying that this an important contribution to the academic literature on MHP! And certainly not saying that it's of any interest to the popular reader. It's one of many footnotes. Richard Gill (talk) 12:35, 7 November 2012 (UTC)

Forget the goats for the moment. What Morgan do inconsistently is to treat the three unknown distributions differently. The producer's door choice for the car is treated as uniform at random but the host's door is not (it is not clear how they treat the player's original door choice).

The culture they are writing for is irrelevant. Mathematicians like to be consistent and Morgan are not. That inconsistency has been absorbed into the way that everybody, including mathematicians, approach the problem. Martin Hogbin (talk) 14:04, 7 November 2012 (UTC)

Yes, Morgan et al. are somewhat inconsistent. I don't think it is a good paper. But they have not influenced how mathematicians like to solve the problem. Mathematicians like to solve the problem with conditional probability (and with K&W uniformity assumptions). See Sevlin 1975b. Clearly, Selvin had a lot of correspondents who didn't like his simple solution, and he switched to conditional! Richard Gill (talk) 12:30, 8 November 2012 (UTC)

I do think that Morgan have influenced how mathematicians choose to solve the problem. I have nothing against conditional probability but, under the rules of the game, the player can choose any door and the host can open any legal door. If we specify both doors then we have two conditions, the door originally chosen and the door opened by the host. If we are going to be pedantic we must condition on both. I know that conditioning on the door initially chosen is, with the correct assumptions, trivially easy, due to an obvious symmetry with respect to doors but that is no reason not to mention it. If we we are going to use this symmetry to dispense with one condition we can use the same symmetry to dispense with the other. That is what I am complaining about. If you are going to fuss about mathematical integrity at least do it properly. Martin Hogbin (talk) 12:52, 8 November 2012 (UTC)

Selvin (1975b) did it Morgan's way (but with the natural, equal door-opening probabilities) 15 years before Vos Savant and Morgan, and he probably did that in response to mathematicians' complaints about his first solution, Selvin (1975a). So you think Selvin 1975b shows lack of mathematical integrity? And you think Morgan influenced later generations of mathematicians? Come off it. Morgan's paper is not influential at all among mathematicians. It's a pedantic paper, badly written, badly argued, makes heavy weather out of mathematical trivialities. Mathematicians solve this problem themselves, without consulting the literature, and they'll almost invariably do it in the Selvin 1975b way. Richard Gill (talk) 13:09, 9 November 2012 (UTC)

I don't see any asymmetric handling of initial door choice and host's door choice in the conditional solutions. Both are conditioned on: it is shown that the probability that switching wins is 2/3 when the player has picked door 1 and the host has opened door 3. I have never seen a solution where one conditions only on host's door (i.e. door 3). That would be bizarre. Many of the "unconditional" solutions condition (only) on the initial pick (i.e. door 1), though (e.g. vos Savant). Typical conditional solutions also do not use symmetry to dispense with conditioning, but instead compute the conditional probability directly. -- Coffee2theorems (talk) 15:23, 9 November 2012 (UTC)

And there is still the matter of the goats. At the very least we must specify that the host does not distinguish between the goats. Martin Hogbin (talk) 13:10, 8 November 2012 (UTC)

I disagree. I thought we were being Bayesian, using, as is traditional (or axiomatic, even) equal probabilities to represent complete ignorance, complete indifference, complete symmetry. We don't know whether or not the host distinguishes between goats. That's why the chance the host opens either goat door when he has a choice is 50-50. Because of our lack of information about this. That's also why the identity of the goat which we get to see gives us no information about the car, either.

If we want to be careful, then we write out in full that we are giving, a priori, the three locations of the car equal probability 1/3, and we are giving the two choices of the host, were we coincidentally to initially choose the car door, equal probability 1/2, whichever door that might be. Because of these uniform initial probabilities it doesn't matter how we choose a door. Probably we choose door 1 because "1" is our favourite number. So we choose door 1 and condition on that choice. In my opinion we don't waste time talking about goats. Sure, goats are distinguishable, but we have no information about them, so it's a complete waste of time to add a sentence saying "we see that the goat has a pink ribbon around it's neck but we have no way of using this information". I might just as well say "we see Monty is wearing a blue necktie but we have no way of using this information."

With the probabilities specified in this way, we can now simulate the game millions of times, which converts the probabilities into long-run relative frequencies. The player always picks door 1. The car is hidden uniformly at random. We'll see the host opening door 3 on half of the occasions. The car will be behind door 2 on 2/3 of those occasions. You want me to simulate the identity of the goat and the way the host chooses a door, given the goats in question? Whatever for? On the other hand, how could I simulate MHP without simulating the host's choice?Richard Gill (talk) 13:03, 9 November 2012 (UTC)

Hope for the future

PS I'm really glad we had a conclusion to the RfC and the article is moving again. There is some "new blood" active, which is really healthy, and a lot of smart ideas and new resources came up in the discussion. I plan to keep away from the page for the time being. Richard Gill (talk) 15:11, 6 November 2012 (UTC)

Yes, hopefully we can now work together. I am not convinced that every newcomer is a new as they seem. Martin Hogbin (talk) 19:00, 6 November 2012 (UTC)

I think TotalClearance is probably the same as the recent anonymous editor 213.102.96.77 in which case he's from Stockholm, and therefore (as far as I can tell) a genuine newcomer. And clearly needs to learn basic wikipedia rules. Richard Gill (talk) 09:39, 7 November 2012 (UTC)

Nijdam's statement

Following your suggestion, I continue some of the discussion with you here. It looks as if you still belief that what I called situation 3, i.e. the formulation with known door numbers, but solved with the simple switching argument, is correct and a good way to present the MHP to the readers. I hope I'm wrong in this, and that you in the mean time have come to some sense and understand the error in it. It has been spelled out maybe too many times. So, if the RfC - for what it's worth - comes to present 3 as an important way of looking at the MHP, Wikipedia will be misleading its readers. Nijdam (talk) 16:15, 8 November 2012 (UTC)

Firstly, it is open to you to correct any misleading of our readers that you consider might be caused by the simple solutions by explaining the exact problem in the appropriate section of the article. That was the idea of my compromise proposal for which there was a consensus.

Secondly, if you are going to present more detailed explanations (in which steps that many consider too obvious to mention are explicitly stated) then you should make sure that you are consistent in your approach. In other words you must not just cover the possibility that the host might have opened door 2 but also the possibility that the player might have chosen door 2 or door 3. For completeness you should also state that we take it that the host does not distinguish between goats. Martin Hogbin (talk) 17:36, 8 November 2012 (UTC)

No. We the player can distinguish between goats but we don't know anything about the goats in the game and we don't know anything about the host's behavour in this matter. I think, Martin, that you are confusing frequentist and subjective probability. Frequentist probability based on actual host behaviour versus subjective probability based on our knowledge of the host behaviour (zero). Maybe the host prefers to open door 2 on rainy days, and door 3 on sunny days. We are not going to explicitly take account of everything that we know nothing about. Richard Gill (talk) 13:26, 9 November 2012 (UTC)

I guess you dream of goats. Nijdam (talk) 18:06, 8 November 2012 (UTC)

I guess you dream of doors. Martin Hogbin (talk) 00:35, 9 November 2012 (UTC)

In fact I do occasionally, but not in connection with the MHP. Nijdam (talk) 13:30, 9 November 2012 (UTC)

A question for Nijdam (and Richard)

Rather than continuing this somewhat confrontational discussion, I would like your help to clarify something. Here is my question. Suppose the problem statement had been:

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

With the usual game rules, strictly speaking should we use conditional probability to solve this problem? Martin Hogbin (talk) 00:35, 9 November 2012 (UTC)

I would say: Let's number the doors 1, 2, and 3, from left to right as seen from the audience. I'll be using these numbers in my solution, later.

You want me to solve the problem in a routine way using probability theory, and I suppose you are interested in a Bayesian solution i.e. based on subjective probability determined by Laplace's rule of indifference. A priori, the car is equally likely behind any of the three doors, and a priori, if I happen to pick the car door, the host is equally likely to open either door to reveal a goat. How I pick my door, doesn't make any difference.

Suppose for the sake of argument, that I picked door 1 and the host opened door 3. In that case you would want me to tell you the probability that the car is behind door 2, given I picked door 1 and the host opened door 3. Keep my initial choice fixed throughout as door 1, we are therefore interested in Prob( car is behind door 2 and host opened door 3 )/P(host opened door 3) = Prob( car is behind door 2 and host opened door 3 )/ (P(car is behind door 2 and host opened door 3)+P(car is behind door 1 and host opened door 3)) = 1/3 / ( 1/3 + 1/3 * 1/2) = 1/3 / (1/3 +1/6) = 1/3 / 1/2 = 2/3 Richard Gill (talk) 13:19, 9 November 2012 (UTC)

You won't be surprised, I totally agree with Richard. The mere fact you do not mention the door numbers we may take as an example, only allows us to choose them our own. Nijdam (talk) 13:25, 9 November 2012 (UTC)

Nijdam, obviously you are standing next to your shoes again: We do not need to choose them our own. Door numbers are irrelevant, don't you see, do you? What solely matters is to distinguish between the door originally selected by the player, and the group of the two unselected "host's" doors. This applies to any permutation of doors likewise. We don't need to be allowed to choose our own door numbers, because we do not "need" any. We just got to understand what the tricky story exactly is about. Door "numbers" are of no relevance. Gerhardvalentin (talk) 16:21, 9 November 2012 (UTC)

Of course, I also have other derivations of this same conditional probability under these same assumptions, using symmetry in various ways to short-cut the calculations. The version I gave is the version Selvin gave in 1975b and I think that every teacher of an introductory probabiilty course will *automatically* and *independently* do it this way first - not because that is how some people did it in the past, but simply because it's straightforward, routine, no thinking, just calculate.

PS by "the usual game rules" I suppose you mean the minimal disambiguation: we knew in advance that Monty was certainly going to open a goat door different from the door we chose first. Notice that I *derived* the K&W assumptions from Laplace's principle. Writing down these initial (uniform resp. conditionally uniform) probabilities was part of the solution, not part of the assumptions/rules/givens. Richard Gill (talk) 13:29, 9 November 2012 (UTC)

Yes that is fine. We assume the actual rules of the game and, from our Bayesian perspective take all distributions about which we have no knowledge as uniform. Martin Hogbin (talk) 13:40, 9 November 2012 (UTC)

We put uniform distributions on *sets of alternative facts* about which we are (or were) ignorant. If we chose door 1, and if the car happened to be there, the *facts* of whether the host opens door 2 and 3 were, a priori, equally likely. Richard Gill (talk) 15:36, 9 November 2012 (UTC)

You have to be careful about the principle of indifference, or you will get inconsistency. Saying "we take all distributions about which we have no knowledge as uniform", if taken literally, is not a consistent rule. It results in us initially believing in the car being uniformly distributed, the host's choice being uniformly distributed, and the car being uniformly distributed after a door is opened. This set of assumptions is the usual incorrect answer, and it is not merely incorrect, it is actually inconsistent (logically contradictory).

You can assume the final car distribution to be uniform, but not while making the other two assumptions. Of these possible applications of the principle of indifference, people feel the first two assumptions are the natural ones, so they get used. It is fairly obvious that they result in a complete and consistent probability model (these assumptions essentially constitute a minimal specification for a simulation using dice, and that basically ensures consistency). Then you either reason within the model in the standard way (conditional probability), or throw in extra assumptions to simplify things. In the latter case, you need to show that no inconsistency arises from adding the extra assumptions. -- Coffee2theorems (talk) 17:19, 9 November 2012 (UTC)

Of course you have to be careful. It's not knowing nothing which is important; it's symmetry of your knowledge (or lack of knowledge) which is important. Richard Gill (talk) 18:51, 9 November 2012 (UTC) PS cf. Two envelopes problem, boy or girl problem - two other famous probability paradoxes where there are "conflicting" superficial symmetries. You have to think carefully to figure out what are the real symmetries. In MHP there is no symmetry between the door the player first chose and the door the host left closed. These two doors were determined by an unfolding history of events in which the location of the car has an influence on one of the choices but not on the other.

I was going to say something similar. In case of the initial distributions, if we decide to stick strictly to the information given in the question, then we know absolutely nothing about these distributions and, if we are Bayesians, we must take them to be uniform.

After the host has opened a door (assuming the standard rules) we do have information about the distribution in the two unopened doors so we can no longer take this as uniform. Failure to make this distinction gives the Bayesian approach an unfairly bad press in some cases. Martin Hogbin (talk) 13:34, 10 November 2012 (UTC)

Thank you both. I hope you will bear with be as I make some changes to the setup. I am trying to establish exactly why we do certain things.

Suppose that the doors are not arranged in a way that suggests an obvious way of numbering them. Would you both be happy to call them I(initially chosen) H(opened by the host) and R(remaining)?

Would it matter if door R was hidden from view but we knew that it was there? Martin Hogbin (talk) 13:40, 9 November 2012 (UTC)

The player needs to be able to choose one (name, point, pick..) from three objects and the host needs to distinguish between them, and sometimes choose as well. No need to see any doors. We can use envelopes, boxes, names, letters. No need to *see* anything. Eg the player chooses a letter, one of A, B or C. We know that these three letters correspond, in order, to "car", "goat", "goat", or to "goat", "car", "goat", or to "goat", "goat", "car". And we know the host knows too. E.g. the player picks B, and knows that the host will either say "would you like to switch to A? I can tell you that C gives a goat" or vice versa. Richard Gill (talk) 15:30, 9 November 2012 (UTC)

You can also solve the problem by saying, right up front, "by symmetry, the identity (the numbers on, names of...) of the doors is irrelevant, What's important are their roles". Several respected academic writers do it that way. You can then thereafter refer to the doors by roles. Door with car CA; door chosen CH, door left closed CL, door opened O. CA = CH or CL. Probabilities 1/3 and 2/3. That's all Richard Gill (talk) 15:48, 9 November 2012 (UTC)

I have no problem with your last suggestion, that we declare the door numbers irrelevant by symmetry.

What I am trying to establish is exactly when and why we must use conditional probability. In my example above, in which the doors are named I,H, and R, and in which door R is hidden from view, must we use conditional probability to be strictly correct? Martin Hogbin (talk) 18:57, 9 November 2012 (UTC)

I don't understand the question. I just said: all the doors can be hidden from view, but the player has to choose one, and the host has to know what is behind each one. The player also has to be able to distinguish between the door which he first chose and the other door which the host left open. The doors have to be distinguishable, and therefore we can name or number them as we like, and whether or not we can see them. What do you want to know? What are you assuming? I don't care how you name the doors and which doors can be seen and which doors can't be seen. BTW: if you ask the question how *must* I solve MHP then Nijdam will give you an answer, and probably Coffee2Theorems will give you an answer, but I will say that you're asking a stupid question (read my paper). Richard Gill (talk) 19:51, 9 November 2012 (UTC) PS I can tell you several ways you *can* solve MHP. Do you want to know whether or not you should switch? Do you want to know the chance that the car is behind door 2 given you chose door 1 and the host opened door 3? Do you want to have my advice? My advice is choose your initial door completely at random and then switch, whatever, and not bother your head about the probability of winning half way through any particular run of the game. A complete waste of time. Garbage in (subjective probabilities) garbage out (subjective probabilities). Richard Gill (talk) 20:01, 9 November 2012 (UTC)

I am asking whether you would consider the simple solutions to solve this problem completely, with no missing steps from a strict mathematical perspective. Martin Hogbin (talk) 20:56, 9 November 2012 (UTC)

Martin: I am not dogmatic about how MHP has to be solved. If you want a dogmatic answer ask a dogmatic person.

I think the mathematician should offer a spectrum of different (mathematically correct) solutions. If you assume A1, then B1. If you assume A2, then B2. Etc.... The consumer has to decide whether they want to assume A1 or A2 or A3 ... *and* they have to decide whether they're happy with B1 or B2 or B3 ... . The point is, an informal verbal question about an imaginary quiz show does not define a formal mathematical problem. It's a matter of discussion, taste, background, tradition, ... how to translate Vos Savant's question into a mathematical problem. Creativity and imagination is required here. Experience. It's an Art, not a Science. Different translations give different problems and different problems have different solutions. Read my paper!!!! Richard Gill (talk) 11:01, 10 November 2012 (UTC)

Thank for your patience Richard but you are missing my point completely. Let be start again below. Martin Hogbin (talk) 13:36, 10 November 2012 (UTC)

Another attempt to explain my points to Richard and Nijdam

Let me start with this question. Which clearly prompts some kind of conditional response.

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

Now let us consider Devlin's solution. Richard, you split problem into conceptual steps. below is you description, slightly edited by me.

Step 0: Initial situation: the player has chosen door 1. 2:1 against that the car is behind it.

Step 1: Now the host is going to open a particular door and reveal a particular goat. I think of that information arriving in separate steps. Imagine that to begin with, the player keeps his eyes shut. He hears a goat revealed behind one of the other doors, doesn't know which of the two.

If asked now whether he would like to switch to the other closed door, he would answer: "Yes please". The host was going to open a door a reveal a goat, whether or not the car was behind door 1, so the chance of his data - hearing the sound of a goat - is 1, whether or not the car is behind door 1. The likelihood ratio is 1:1 and the posterior odds are 2:1 in favour of switching.

Step 2: The player of course opens his eyes before making his decision and sees that it was door 3 which was actually opened. He's got hold of another bit of information. Does it change his odds on the car being behind door 1? No: because if the car is there, the chance is 50% that the host opens door 3 (by ignorance of how the host opens a door when he has choice), while if the car is not there, the chance is also 50% that the host opens door 3 (by ignorance of where the car actually is). So the likelihood ratio is 0.50 : 0.50 or 1:1 and the odds on the car being behind door 1 still don't change.

Step 3: he takes a close look at the goat and sees that it is a girl goat, or whatever, ... this won't change his odds any more, since yet again, the chance of it being that particular goat is the same for him, whether or not the car is behind door 1.

The point that is made by some mathematicians is that Devlin does not explicitly mention step 2 - the information gained by knowing which specific door was opened. Even if there is no information revealed by that step, or the information does not change the original odds, we should still mention it (because, one might argue, in different circumstances that step might reveal some information that does change the original odds).

Have I understood you correctly?

Yes, Devlin skipped Step 2. And when challenged he did not have an answer. Your parenthetical remark "because, one might argue..." is irrelevant. Every step of an argument has to be justifiable. Richard Gill (talk) 19:14, 10 November 2012 (UTC)

I am not asking if you support this solution or whether this is the best way of solving the problem. All I want to get clear is what exactly is the problem that some people find with Devlin's solution. Martin Hogbin (talk) 14:16, 10 November 2012 (UTC)

The reason Step 2 needs explanation is because you obliged me to take this step, and every correct step has a justification. You asked me to give a probability given the host opened door 3. So I took Step 2, and I explained why it is correct. Didn't you see the explanation? You will notice that it refers to our ignorance of the host's door opening procedure. Till then we had only made use of our ignorance of the car's location. Till then we had only obtained a simple solution: (A): "chance of winning by switching given we chose door 1 = 2/3". Step 2 strengthens this to a conditional solution: (B): "chance of winning given we chose door 1 and the host opened door 3 = 2/3". Statements A and B are different. They mean something different. It costs a tiny bit more work to explain why B is true, than to do this for A. Don't you realise that? Richard Gill (talk) 18:53, 10 November 2012 (UTC)

(A) is true because the car is equally likely behind every door. (B) is true because the car is equally likely behind every door and because the host is equally likely to open any legal door, given location of car and door initially chosen by host. Richard Gill (talk) 19:23, 10 November 2012 (UTC)

Yes I did know that. I would say though that the reason we need to explain step 2 is that in different circumstances, for example if we knew the host had a door preference, missing out that step could make us give the wrong answer. We are making our solution more general.

I also think you have not started at the beginning. You should have had:

Step -3: Initial situation: the player has to choose a door.

Step -2: Now the player is going to choose a particular door. Imagine that to begin with, the player chooses a door without looking at its number or position. If asked now for the probability that his chosen door hides the car, the player would say 1/3. His choice was random.

Step -1: The player now looks at the door number and sees that it was door 1 which he has chosen. He's got hold of another bit of information. Does it change his odds on the car being behind door 1? No: the number gives him no information that allows him to change the probability that the car is behind door 1.

Contrived, yes, but so were your steps, they are conceptual not real steps. In slightly different circumstances, for example if we knew the car had not been placed uniformly, then ignoring step -1 would give us the wrong answer. We have generalised the problem a bit more. Martin Hogbin (talk) 23:08, 10 November 2012 (UTC)

Taking care of step 2 carefully does not make the solution more general. It is the necessary bridge from (A) to (B). If you want to draw conclusion (B) you have to argue for it and some way or other you will use the fact that the two doors are equally likely to be opened because we don't know anything about how the host chooses.

We can go back to step -3 if you like. Player chooses a door. He knows nothing and the car is equally likely behind any of the three doors. Because he knows nothing about how the car is hidden he chooses the door closest to him. He's at the left hand side of the stage as seen from the audience. Call the doors 1, 2 and 3 (left, right, middle). After his choice, left hand door (which we call door 1), he still knows nothing about where the car is, so the car is still equally likely behind any of the three doors (left, middle, right, or for short, 1, 2,3). This is not contrived. It's quite simply: boring.

By the way, I did not say that his choice was random. I don't care how the player chooses. He names or points or whatever... He makes a choice.

On the other hand, look at the argument for Step 2. It was not entirely trivial. It brought in a completely new ingredient into play (our ignorance of how the host chooses a door when he has a choice).

A logical derivation is like wikipedia: you don't have to source everything but everything has to be sourcable. You have to source everything likely to be challenged. Devlin was challenged and didn't know what to say.

Now it's my turn. Tell me how you would logically derive (B), going right back to the start. Richard Gill (talk) 07:10, 11 November 2012 (UTC)

As we are told nothing about the way the player chooses a door we must take it as random. So going back to step -3, after the player has chosen a door we know the probability that it hides the car is 1/3, because the player has chosen at random. Now we are told that the player has, in fact, chosen door 1. We need to prove that the probability that this door hides the car is still 1/3. Your statements A and B above now become A - the probability that a player who chooses randomly will pick the car is 1/3, B the probability that the chosen door hides the car of a player who chooses randomly but has picked door 1 is 1/3. They are not the same thing at all. We need to do some extra work to show B, in fact we need to use the fact that the car was initially placed randomly. Martin Hogbin (talk) 09:35, 11 November 2012 (UTC)

You are answering the problem in which the player always chooses door 1. This is not the same as the problem in which the player chooses randomly but we consider only the condition that the player has in fact chosen door 1. In the case of the standard problem they happen to have the same answer but the two questions are conceptually different. Martin Hogbin (talk) 10:36, 11 November 2012 (UTC)

No. I am solving the problem from the player's point of view, as per your request, and given that he chose Door 1, as per your request, and given the host opened door 3, as per your request. You asked me to derive (B). I'm not assuming that the player always chooses door 1, any more than that the host always opens door 3. I use subjective probability, the whole thing is a once off affair. Richard Gill (talk) 11:47, 11 November 2012 (UTC)

I think we must still take the players initial door choice as uniform. The player can, under the usual rules of the game, pick any door. If we do not take the players initial choice as uniform what distribution should we take.

Let me put it this way. A car is placed behind door 2 of three doors. I pick a door but do not notice what number it is. What is the probability that I have chosen the car? Martin Hogbin (talk) 12:10, 11 November 2012 (UTC)

The player has picked a door. It doesn't matter how he did it. You asked the question. In response, I derived for you, player's probability that the car is behind the middle door, given that he first chose the left hand door, and the host opened the right hand door. I carefully explained each step. Do you retract your own question now? Richard Gill (talk) 12:50, 11 November 2012 (UTC)

The doors are numbered, the question I am referring to is at the top of this section. The player picks a door, if we (or the player) do not know the number of the door then we can say that the probability that the door hides the car is 1/3, regardless of how the car was placed. Once we know the door number (or other ID) we now need to know how the car was placed to get a probability that we have chosen it. Martin Hogbin (talk) 13:00, 11 November 2012 (UTC)

Then I misunderstood your question. I solved the problem from the point of view of the player, with the doors identified by him, however he likes. Moreover, the probabilities I introduced corresponded to the player's ignorance of how things are done. I do not assume the car is hidden at random. I assume the player knows nothing about how the car is hidden. I didn't assume anything about how the player chose his door since it is unncessary - I watch developments unfold, from the point of view of the player, from the moment he has chosen his door.

Meantime, you still haven't said whether or not you agreed that (A) and (B) are different things, and you took no notice at all of my motivation for my Step 2. I might as well be talking to a brick wall. Richard Gill (talk) 13:10, 11 November 2012 (UTC)

Of course A and B are different things!

I agree with you we do not assume anything but, because we are given no information as to how the player makes their initial choice, the car is placed, or the host makes his choice, we must take these to be uniform.

Do you agree that these are also different things?

C The chance that the chosen door hides the car given that the player has chosen a door.

D The chance that the chosen door hides the car given that the player has chosen a door and observed that the door chosen is number 1.

They both have the same value but only because the car is placed uniformly Martin Hogbin (talk) 15:35, 11 November 2012 (UTC)

(1) The car is not placed uniformly! I don't know how the car is placed. C and D are different things but both have the value 1/3 because we don't know anything except that there are three doors....

(2) Whose chance are you talking about? My chance, Monty's chance, or the player's chance?

(3) The player has to choose a door. And Monty has to know (at least) which door hides a car, which doors hide goats. From the player's point of view, the doors can be distinguished, but he doesn't know what's behind them, the host does, and the player knows the host knows. I will assume that the doors are identified by position: left - middle - right as seen by the audience. I'll number (=name, label) them accordingly 1 -2 - 3.

(4) I am not interested in variants of the problem in which after choosing his door, say the left hand door, the player notices that there were some numbers written on each of the three doors, for instance "1", "-3", and "7". 17:44, 11 November 2012 (UTC)

Breakthrough: you agreed A and B are different things! Then I hope you agree that a careful and complete and correct derivation of B might differ from a careful complete and correct derivation of A. Then you will also agree that the real issue is, (I) should we be intetested in A or in B? Secondarily, (II) do we want to be careful and complete and correct, or not? I think that (I) and (II) are the issues which face the wikipedia editor. Hopefully all editors agree that A and B are different things. They disagree how important is the difference, and how important it is to write out every little detail in full.

Might we agree to that? I want to take a MHP-Wiki-break. — Preceding unsigned comment added by Gill110951 (talk • contribs) 17:53, 11 November 2012

If you want to stop now that is up to you but you have still missed my point completely. You seem to have got it stuck in your head that I am arguing that your A and B are the same or that the difference does not matter or that I do not care. That is not the case

My argument is that my C and D are also different things, my argument for this is almost identical to yours:

C The chance that the chosen door hides the car given that the player has chosen a door.

D The chance that the chosen door hides the car given that the player has chosen a door and observed that the door chosen is number 1.

The chance is on the basis of the information given in the question (normal game rules assumed). We have also agree to take a Bayesian perspective so because we are given no information as to how the player makes their initial choice, the car is placed, or the host makes his choice, we must take these to be uniform. Martin Hogbin (talk) 18:06, 11 November 2012 (UTC)

Fine. So you are assuming that there are distinguishing marks on the doors, which the player only becomes aware of after he has chosen a door? This is a new variation on MHP which introduces new complications. Alongside of C and D we also need introduce E, when the player has also seen the host open a door, and F, when he has taken notice of all kinds of distinguishing characteristics of door and goat. The step from D to E will be the interesting one, analoguous to the earlier step from A to B. We need to go all the way to the moment he makes his choice.

It is not really a variation, it is a conceptual distinction. The question only says '...you're given the choice of three doors...You pick a door'. It does not say how the player picks the door. Just like the door being opened to reveal a goat, we can imagine the initial door being chosen in two stages. First the player chooses a door, then he see which door it is.

Anyway, you agreed A and B are different things, so I hope you agree that giving a justification of A is a different thing from giving a justification of B. Finally I want to know if you understand and agreed with my justification of B (so-called Step 2). You'll never get to E and F if you don't take the step from A to B.Richard Gill (talk) 12:00, 12 November 2012 (UTC)

Yes I do agree with you and it looks as though you agree with me now. My overall point was the same one made by you in your email. Not only is there 'only a hair-breadth's difference between the "simple" and the "conditional" solutions' but the 'conditional' solutions given in the article have similar problems to those of the 'simple' solutions in that they do not treat the player's initial door choice or the identity of the goat revealed properly. Thus the benefit of these solutions is less than a hair's breadth. Add to that the fact that it was not the questioner's (vos Savant's) intention that we should distinguish between possible doors opened by the host and the 'conditional' solutions become even less beneficial.

I don't agree with you. The difference between simple and conditional is small, but it's a real differencel. It exists. It's easy to bridge. But the issue of the player's choice is not an issue at all. The issue of the goat identity is a non issue. Richard Gill (talk) 22:50, 12 November 2012 (UTC)

Have you got my point at all? There are three equal problems with the simple solutions.

The simple solutions do not take account of the door identity originally chosen by the player.

The simple solutions do not take account of the door identity opened by the host.

The simple solutions do not take account of the goat identity revealed by the host.

The current 'conditional' solutions in the article only account for the second of these problems. Martin Hogbin (talk) 15:38, 13 November 2012 (UTC)

I did get your point but you did not notice that I had addressed it. The conditional solutions don't take account of the door identity originally chosen by the player since this is a non-issue. We are solving the problem from the point of view of the player and he's already chosen a door. From this point on, it is still true that it is equally likely for him that each door hides the car and moreover that if his door happened to hide the car, for him, the host is equally likely to open either other door. The conditional solutions do not take account of the goat identity revealed by the host since this is a non-issue, too. We don't take account of the history of the universe up to the moment the player chose his door. We don't take account of everything else the player might or might not notice about the doors, goats, whatever. The conditional solutions do take account of which door is opened by the host because it is essential to do so, in order to justify their conclusion. Which door was opened certainly changes the chance the car is behind the door which got opened!!!! It certainly changes the chance of the door that didn't get opened! It happens not to change the chance of the initially chosen door, and that needs justification. Richard Gill (talk) 16:13, 15 November 2012 (UTC)

The idea of the conditional solutions is to carefully discuss the evolution of the player's justified beliefs from the moment after he chose his first door up to the moment at which he must decide whether or not to switch. Of course if you want to be careless and jump to conclusions you are welcome to do so, but in analysing solutions of an infamous brain teaser it is better to be careful. The result is a correct derivation of the conclusion "given the player chose door 1 and the host opened door 3, the chance is 2/3 that the car is behind door 2". A correct and complete derivation cannot fail to make explicit use of the assumption that if the player's initial door happens to hide the car, then as far as the player is concerned, the host is equally likely to open either other door!

If you only want to derive the conclusion "given the player chose door 1, the chance is 2/3 that the car is behind the door left closed by the host" then you won't use that assumption.

I'm not saying that which conclusion you have to work towards. I'm saying that the arguments to get to either conclusion are different. And the argument to get to the more detailed conclusion is slightly more involved, uses more inputs. Richard Gill (talk) 16:22, 15 November 2012 (UTC)

On the other hand my proposal was to give the 'conditional' solutions equal prominence in the article and I do not to renege on that. All I wanted to show was that this offer was a genuine concession and compromise. Martin Hogbin (talk) 16:21, 12 November 2012 (UTC)

This was never in doubt! Richard Gill (talk) 22:22, 12 November 2012 (UTC)

How the problem starts

Here is the original statement

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, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

It is often interpreted to be asking a conditional question which might be more clearly worded like this:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door and the host, who knows what's behind the doors, opens another door which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Given that the player has originally chosen door 1 and the host has opened door 3, what is the probability that the player will win by switching

Does anyone have any argument with this restatement of question. It is this question that I want to talk about here. We assume the standard game rules. Martin Hogbin (talk) 13:48, 14 November 2012 (UTC)

The conditionalists take the original statement, and then state that what we have to do is to compute the conditional probability that the car is behind door 2 given the player initially chose door 1 and the host then opened door 3. I don't see why you rewrite the original statement.

This seems to me to be a completely sensible move. The player has to decide, knowing which door he initially chose and which door the host opened to reveal a goat.

Having made this translation from ordinary language into the language of probability theory, the required conditional probability is easily computed to be 2/3 if we furthermore assume that all doors are equally likely to hide the car and that the host is equally likely to open either door if he has a choice. Just routinely use the definition P(A|B)=P(A&B)/P(B), with the initial choice of the player, door 1, fixed, throughout. In other words: "P" for probability stands here for the probabilities which are valid for the player after the player has chosen door 1 but before anything else has happened.

Given that the player initially picked door 1, P(B)=P(host opens door 3) = 1/2 by symmetry (the host has to open door 2 or door 3), and P(A&B)=P(host opens door 3 & car is behind door 2)=P(car is behind door 2)=1/3 by symmetry (the car has to be behind door 1, door 2 or door 3). Note that the first use of symmetry involves the choice of door opened by the host. Given the player initially chose door 1, the host is equally likely to open door 2 or 3. Ignorance of host choice method is implicitly being used here, alongside of ignorance of car location.

Of course, I also silently used the assumption that the host certainly opens an unchosen door to reveal a goat. I only explained the not quite so obvious bits of the argument.

Note we refer to the doors by number, merely for convenience. Standing on the stage, one would point at them, or choose a door by walking up to it (pointing with your whole body!). Talking about what is going on one might refer to them as left, middle and right. The numbers are not important. The numbers are just names, labels, so that we can talk about the three doors in a convenient (not verbose) way. Both player and host must be able to distinguish between the doors, by common sense understanding of the (imaginary) real world situation we are talking about. Richard Gill (talk) 16:34, 15 November 2012 (UTC)

For some reason you answer more that I ask. I was just making sure that we were both agreed on the question we were intending to answer. But since you have continued...

Why have you started the answer half way through the game, that is to say after the player has chosen a door?

Let us start at the beginning, as the question suggests when it says, '...you're given the choice of three doors... You pick a door'. We can ignore the evolution of humanity and everything before the player chooses his door because it is not mentioned in the question.

We can try to answer the question from the perspective of the player but that is not really possible, we are not told what is in the players mind or how the player picks a door. We can only use the information on the subject given in the question and that is nothing. The game rules make it clear that the player has a free choice of any door. We are therefore obliged, as good Bayesians, to take the door to be picked door uniformly at random.

Do you agree so far? If you do not mind could you just let me know whether you agree with me so far. Martin Hogbin (talk) 19:12, 15 November 2012 (UTC)

Disagree. The player has chosen Door 1. No need to model how he did that. Totally irrelevant that for me his choices were equally likely. We have to help him make his future decision. Richard Gill (talk) 12:36, 16 November 2012 (UTC)

Do you agree that under the rules of the game the player can choose any door? Martin Hogbin (talk) 15:15, 16 November 2012 (UTC)

Yes. Richard Gill (talk) 17:46, 16 November 2012 (UTC)

Would you agree that if the question had not mentioned any door numbers at all, or any way of identifying the door originally chosen, we would have to take the original door to be chosen uniformly at random? Martin Hogbin (talk) 22:12, 16 November 2012 (UTC)

No. We are answering the question whether the player should switch in a certain situation. The location of the car is independent of the player's initial choice, so the player's choice can be taken as fixed. He made his choice. It's done. From this point onwards, the car is equally likely behind each of the three doors. I am not going to make assumptions which I am not going to use. The assumption that I do make, and use, is that the player's choice is independent of the location of the car, which is for him completely unknown, so that for him, the car is equally likely behind any of the three doors.

For the player and the host the doors are identifiable. Whether we use numbers or locations or something else as an aid in solving the problem is irrelevant. Whether or not the question mentions door names, colours, numbers or locations is irrelevant.

Bulleted list item

Independence

If you want to use the argument of independence then you must say so. It is certainly not clear how the conditional solutions presented by the article make use of this independence.

It is clear. They do condition on the player's choice. The probabilities of where the car are, remain 1/3, 1/3, 1/3, after the player chose door 1. Definition of independence: conditional probability equals unconditional. Richard Gill (talk) 16:01, 17 November 2012 (UTC)

PS use of independence is mentioned explicitly in the text accompanying the Bayes formula solution (the section with the long formulas which no one will read) Richard Gill (talk) 08:58, 18 November 2012 (UTC)

In the other hand you might consider it too obvious to mention. In that case, why not go the whole way and say that, from the player's perspective, whether the car is behind the originally chosen door is independent of the specific door opened by the host, so we can take that as fixed as well? The simple solutions are thus justified. Martin Hogbin (talk) 13:29, 17 November 2012 (UTC)

Yes. You can say that. I have been saying it for years, didn't you notice? In my opinion it's a legitimate argument and I have been trying to persuade the conditionalists of that myself. Finally I was forced to write it up in an own publication. I consider it implicit in several earlier publications. But whenever I use words like "symmetry" or "independence" a heap of people from both camps complain. (Boris Tsirelson understands me, but no-one else). Richard Gill (talk) 16:01, 17 November 2012 (UTC)

PS: I wouldn't say "fixed" in this context - we can't fix the door the host opened. We can calculate the chance, given the door the host opened. Different concepts. (If we talk about fixing the choice of the player either meaning makes sense and comes down to the same thing). Conditioning in probability simply means selecting, a posterior. Fixing usually means fixing something in advance. It's the difference between an active intervention a priori and a passive selection a posteriori. The whole topic of determining causality with statistics turns around the question when these two ways of "fixing" are the same. Richard Gill (talk) 09:02, 18 November 2012 (UTC)

By the way lots of sources do take the player's initial choice as completely random; to start with: Selvin 1975a. It adds further symmetry to the problem making the door numbers completely independent of the relationships between the roles of the doors. It makes some solutions easier, others more elaborate (eg Grinstead and Snell have to give a gigantic table). Richard Gill (talk) 16:16, 17 November 2012 (UTC)

Surely you see my point now.

I see your point but I don't agree with it. Maybe I will come back and discuss your table below, later. Right now I'm interested in whether you find my attempts to make the conditional solutions section of the article more accessible, succesful or not. (1) I modified the conditional probability table so that it now contains six equally likely situations and hence enables simple counting to derive probabilities and conditional probabilities, no longer any need to do arithmetic with fractions. (2) I made similar extended explanation of the decision tree, both in the text and the caption of the diagram. (3) I wrote a derivation of the conditional solution via Bayes rule, odds form, so that formulas are no longer needed and instead we have a very simple calculation. (4) I also wrote a few words on "Devlin fixed" in the conditional solution section.

The formalists won't like this because it is not formal enough. But the question is whether in this way one can at least make the conditional solution comprehensible to people like you.

(I prefer to build conditional solutions on simple by use of symmetry, but it seems everyone else wants to keep the two kinds of solutions as different as possible; while I want to show how close they are). Richard Gill (talk) 13:08, 18 November 2012 (UTC)

PS and anyone who wants to keep to the status quo will comlain about "own research". I notice again that the conditionalists tend to use frequentist language while the simplists tend to use Bayesian. For the Bayesians, simple and conditional solutions are even closer since for them, probability comes from symmetry. It's symmetry which we have first. You can correctly deduce from symmetry from the outset that you'll disregard door numbers. The frequentists on the other hand think of the probability distributions as being given from outside, knowledge of use of true random processes to generate choices for instance. So for them symmetry is something which can be seen a posteriori, not as a driving force from the outset. Richard Gill (talk) 13:12, 18 November 2012 (UTC)

Consistent solutions

General reader solution

The reader assumes, probably unwittingly, a symmetry with respect to door number and ignores both the door number originally chosen and the door number opened by the host in calculating the posterior probability that the car is behind the originally chosen door. Essentially the simple solutions.

Simple solution fixed

We state at the start that there is a clear symmetry with respect to door number so we can ignore the number of the door opened by the host and the number of the door originally chosen in calculating the posterior probability that the car is behind the originally chosen door.

Alternatively we state that whether the car is behind the originally chosen door is independent of the specific door originally chosen by the player and the specific door opened by the host. (We might note that whether the car is behind a specific unchosen door is obviously not independent of the door number opened by the host).

We are now free to use the simple solutions.

Conditional probability solution

We start by considering all the possibilities allowable under the game rules. We have to decide how to assign initial probabilities to elements of our sample space.

If we are Bayesians or decide to apply the principle of indifference then we take the distributions of the original car placement, the door originally chosen, and the door opened by the host to be uniform at random. (Alternatively we take all these distributions to be completely unknown, in which case the problem is insoluble and we cannot decide whether it is better to switch or not.)

Bayesians apply the principle of indifference. Frequentists don't use it. So it is not "or" but "and hence". Richard Gill (talk) 13:14, 18 November 2012 (UTC)

We then condition this sample space according to the door number originally chosen and the door number opened by the host.

The Bayesian can notice from the start that choosing the door is performed by the player at the outset, and the probabilities we need in the problem come from he, the player, applying the principle of indifference to what is still unknown (location of the car, how the host will choose if he has a choice). Anyway, the Bayesian by definition is interested in *his* probability of getting the car by switching, *at the last moment that* he has to make his choice, ie *conditional on everything that he knows by that stage*. So he is going to condition on his door choice anyway. He does not need to model his uncertainty before choosing, as to what he was going to choose. Richard Gill (talk) 13:17, 18 November 2012 (UTC)

Inconsistent solutions

Simple solution badly criticised

We give a simple solution

We are now criticise this solution for not taking account of the door number opened by the host but ignore the fact that we have also not taken account of the door number originally chosen.

Simple solution badly fixed

We give a simple solution. We fix this solution by pointing out at the start that there is a symmetry with respect to the door opened by the host but fail to point out there is also a symmetry with respect to the door number originally chosen.

Morgan's conditional probability solution

We start by considering all the possibilities allowable under the game rules. We have to decide how to assign initial probabilities to elements of our sample space.

We tacitly apply the principle of indifference to the distributions of the original car placement and the door originally chosen and take them to be uniform at random. We take the distribution of the host's door choice to be completely unknown and parameterise it.

We then condition this sample space according to the door number originally chosen and the door number opened by the host and come up with an 'elegant solution'.

Partial conditional probability solutions

We start with the tacit conditioning of our sample space according to the door originally chosen then explicitly condition it according to the door opened by the host.

Discussion

You keep talking about "the conditional probability". You've fallen into the trap so many judges, journalists, newspaper-readers, politicians... fall into. You can't talk about *The* conditional probability. All probability is conditional either on some conceptual collection of "many repetitions" (frequentist) or on some baseline set of knowledge, which can depend on *whose* knowledge, and at *which* time.

First problem is to decide which/when/whose probability is relevant/interesting... For the Bayesian taking the point of view of the player, the answer is *axiomatic*: the player has to decide to switch after, say, choosing door 1 and after, say, door 3 was opened by the host. Therefore *by definition* (or axiom) the player has to figure out the probability the car is behind door 2 given he chose door 1 and the host opened door 3.

Yes, I know all that, I am not sure what you think I do not understand about it. I have also said many times that all probabilities are, strictly speaking, conditional. We have lots of conditions, that the player obeys the generally accepted rules, that the host's name is Monty, that the player initially chooses door 1.

The systematic way to attack the problem is then to go back to the first (in time) given, and then work step by step forward. No point at all in going back further in time. We go back just as far as we have to, in order to be able to correctly reason our way from then, forwards to the decision moment.

Yes we do have to decide where to start. But I think it was you who suggested a reasonable convention was start with the basic rules of the game and then consider any additional constraints as conditions. That seems an eminently sensible suggestion to me.

Almost every source that mentions conditional probability, starting with Morgan, clearly indicate that they also treat the problem that way. The distinction that is always made in the literature is between the general case, in which a player plays the game, no door numbers are mentioned and the player decides on their strategy before the game starts, and what we might call for convenience the conditional case in which the player initially chooses door 1, and the host opens door 3 then the player has to decide what to do.

Even if you decide to start after the player has initially chosen a door there is still some explaining to do, which is currently missed out in the article. If we are only told that the player has picked a door (which we can only take as uniformly at random since we are given no information to indicate otherwise) the the answer is 2/3, regardless of the initial car placement and the hosts door policy (as we have both commented many times). Once we are told that the player has, in fact, chosen door 1 we need to consider both these distributions to solve the puzzle. The players initial door choice is not quite such a simple matter as it might at first seem. Martin Hogbin (talk) 19:56, 18 November 2012 (UTC)

I agree that there is some explaning to do. I think it should be done this way: We assume the player knows nothing (except the rules) and we take his point of view. How we make our initial choice is therefore irrelevant, and once done, it's done. After we have made our initial choice, for us the car is equally likely behind each door, and the host is equally likely to make either choice in the case the car happens to be behind our initially chosen door 1.

Unfortunately, as far as I know, I am the only "authority" who explicitly discusses the frequentist/subjectivist contrast in how a mathematical model should be set up to solve MHP. So I have a COI. Moreover since the literature is confused and confusing, and my work is recent and not famous, one can argue that Wikipedia has to be confused and confusing, too. My work was even inspired by the confusion on Wikipedia!

Thinking about the problem in the Bayesian way forces one to compute the conditional probability of ... from first principles. Having found it is 2/3 and hence equal to the obvious simple solution unconditional probability 2/3, one might start thinking, and ponder about whether or not there was a shorter way of getting to the same answer (for the conditional probability). And of course, there are many ways to do this.

Agreed. I am not saying the currently shown 'conditional' solutions are wrong they just inconsistent and incomplete. We should make them better. Martin Hogbin (talk) 19:56, 18 November 2012 (UTC)

This is why the simple solutions and the conditional solutions need both to be taken seriously, and it is the bridges between them which are important. Like it or not, readers come from both popular and academic backgroun. Editors too. Sources too. Richard Gill (talk) 13:24, 18 November 2012 (UTC)

Agreed again, but let us treat the unknown distributions consistently or explain why we do not do so. Martin Hogbin (talk) 19:56, 18 November 2012 (UTC)

I've argued that for the Bayesian who is taking the player's point of view in order to advise the player, we don't even have to talk about the distribution of the player's choice. "Consistency" does not apply here. Richard Gill (talk) 09:16, 19 November 2012 (UTC)

To be clearer what I am getting at, we cannot treat the player's initial choice as trivial. Depending on exactly what question is asked, what our assumptions are, and what exactly we mean by 'the probability' it may or may not matter. As the questions can be taken to state two constraints that are additional to the game rules it is, in my opinion, most logical to take them both as conditions, to be applied to the general case based only on the game rules. Regardless of how we choose to treat the problem though is always incumbent upon us to explain exactly what we are doing and not to miss out any steps, rationales, or justifications if we want to claim we have a complete answer. The current 'conditional' solutions fail to do this and are therefore little better than the much-criticised 'simple' solutions. Martin Hogbin (talk) 08:59, 19 November 2012 (UTC)

I agree with some of what you say, not all. I agree that the section on conditional solutions does not explain where its assumptions are coming from, what rationale is there for this approach. Exactly like most of the sources themselves. AFAIK RD Gill is the *only* source who explicitly and neutrally (he claims) goes into this matter. I have been trying to improve this section of the article in these respects. Please help make it better still. Tell me if I succeeded. Richard Gill (talk) 09:32, 19 November 2012 (UTC)

Perhaps we can just agree that the 'conditional' solutions in the article and in most sources are not quite as good as they are cracked up to be by some people.

Well, that has been my own opinion too, for years. I think the article is much much better now. Balanced, intelligent. Richard Gill (talk) 17:41, 19 November 2012 (UTC)

My proposal was intended to help 'simplists' and 'conditionalists' to work together in harmony. Unfortunately Rick and most of the other 'conditionalists' seem to have left the article, which is a great pity. Maybe they will return or we will get some new editors although I do sense that we may be being handed a long length of rope to play with. As I am basically a 'simplist', I will leave it to the 'conditionalist' to write the section that they are so keen to have. I am always willing to give my opinion if asked though. Martin Hogbin (talk) 13:58, 19 November 2012 (UTC)

Don't worry, they'll be back (cf Terminator). In the meantime, time to have some fun elsewhere. Did you see the "sleeping beauty" problem? Richard Gill (talk) 17:43, 19 November 2012 (UTC)