Talk:Monty Hall problem/Archive 34

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Names for simple solutions.

I have given the various simple solutions names based on the sources that give them but I think it would be better to have descriptive names (like 'Combining doors') for them. What do others think? Martin Hogbin (talk) 14:07, 5 November 2012 (UTC)

Criticism of the simple solutions

I have moved the 'Criticism of the simple solutions' section to the start of the 'Solutions considering the specific doors the player chooses and host opens' section as it forms a natural introduction to the reasons that some people consider the 'conditional' necessary or preferable. I would have thought that this position was preferable for the 'conditionalists' as it mentions the issues with the simple solutions earlier in the article. Martin Hogbin (talk) 11:05, 5 November 2012 (UTC)

Not preferable, as it conflates conditional solutions with the criticism - and in POV fashion is clearly suggesting use of conditional probability to solve the MHP is in some way controversial. Perhaps the section could use some introductory text neutrally explaining the difference, like perhaps:

Most sources in the field of probability, including many introductory probability textbooks, solve the MHP by showing the conditional probabilities the car is behind door 1 and door 2 are 1/3 and 2/3 (not 1/2 and 1/2) given the contestant initially picks door 1 and the host opens door 3 (Morgan et al. 1991 Chun 1991 Gillman 1992 Grinstead and Snell 2006:137-138 Lucas et al. 2009). In contrast, the simple solutions above show that a player with a strategy of switching wins the car with probability 2/3 (Grinstead and Snell 2006:137-138 Carlton 2005). For example, this means out of a sample of 300 shows where the player initially chooses door 1, a strategy of switching should win about 200 times - 100 times when the car is behind door 2 and another 100 times when the car is behind door 3. The solutions in this section focus on the specific case where the player has initially picked door 1 and the host has then opened door 3. In the same example of 300 shows, this should happen about 150 times (the other 150 times the host opens door 2). The following solutions show that even in this case, where the player can see two closed doors and an open door 3 showing a goat, the probability of winning by switching is still 2/3.

The point here is that use of conditional probability is the standard approach in the field of probability and it shows something slightly different than what the simple solutions show. The criticism should come after the conditional solutions are presented (per WP:WEIGHT). -- Rick Block (talk) 17:23, 5 November 2012 (UTC)

I think you are still thinking in terms of the old argument, which solutions are 'correct'. Morgan started the whole 'conditional' thing, and they did it by criticising the 'simple' solutions. They did that because it made a nice intro into their conditional argument. Should we not do the same?

Side remark: Selvin (1975b), in the second paper ever written on MHP, presents the conditional solution. In the first ever paper on MHP, Selvin (1975a), he had already presented the simple solution. He did not even comment on the difference between his two solutions. His first paper assumed that the player's choice was random and the location of the car was random: nine equally likely possibilities. His second paper contains the standard and elementary conditional probability calculation, from first principles, of the probability the car is behind door 2 given the player chose door 1 and the host opened door 3 ("by first principles" means: by using the definition P(A|B)=P(A and B)/P(B)). Now of course he assumes the host's two choices are equally likely, when there are two choices, as well as that the location of the car is random. Now he lets the the initial choice of the player be fixed.

I still think you are missing something Richard, as discussed on my talk page. Martin Hogbin (talk) 18:22, 6 November 2012 (UTC)

The point I want to make: the whole "conditional thing" is just the routine (i.e. without thinking, just calculate) way of solving this problem for people accustomed to elementary probability theory. Morgan et al. first introduced a host whose choice of door to open might be biased and used this to criticize Vos Savant's simple solution, 15 years later. Richard Gill (talk) 14:14, 6 November 2012 (UTC)

On the other hand, I am not going to fight over it. The advantage of the current structure is that those that like the 'conditional' solutions can decide what format presents them in the best way. If you think 'get them in quickly' makes them sound more authoritative by all means do that. Martin Hogbin (talk) 20:01, 5 November 2012 (UTC)

Time to clean up?

There are two sections on Bayes theorem in the article: Monty_Hall_problem#Formal_solution and Monty_Hall_problem#Bayes.27_theorem. Perhaps the second one could be deleted? Richard Gill (talk) 16:27, 3 November 2012 (UTC)

Perhaps both can be deleted? The article on Bayes Theorem contains MHP as an example. A heap of formula manipulation is not useful for anyone here. We need ideas, simple logical arguments, simple tables perhaps (TotalClearance's six line table is splendid), but not heaps of formulas. If you really want to solve MHP by Bayes Theorem then apply Bayes Theorem in the form "posterior odds equals prior odds times likelihood ratio". Richard Gill (talk) 14:54, 4 November 2012 (UTC)

Deleted the second one. I don't think both should be deleted, as Bayes' theorem is a common way to solve this problem, and is simple enough. If having the usual Bayes' theorem solution and odds solution both is too much, I'd suggest deleting the odds form solution. Many people are not familiar with odds or likelihood ratios. The odds version isn't any simpler, either. Partly that is because of the choice of language, though; the odds version would be clearer written this way:

Bayes' theorem gives

${\begin{aligned}P(C=c|H=3,S=1)&\propto P(H=3|C=c,S=1)\overbrace {P(C=c|S=1)} ^{1/3}\\&\propto P(H=3|C=c,S=1),\end{aligned}}$

where $\propto$ stands for "proportional to", i.e. equal up to a multiplicative constant. The host's behavior is described by

$P(H=3|C=1,S=1)={\tfrac {1}{2}},$

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

$P(H=3|C=3,S=1)=0,$

leading to

$P(C=2|H=3,S=1)={\frac {1}{{\tfrac {1}{2}}+1+0}}={\tfrac {2}{3}}.$

While this is how I would do it, it's only a hair's breadth simpler, and not simpler to those who are not familiar with ${\text{posterior}}\propto {\text{prior}}\times {\text{likelihood}}$. -- Coffee2theorems (talk) 17:34, 4 November 2012 (UTC)

Jeff Rosenthal promotes Bayes in odds form in book and article about MHP. Then it becomes "Devlin fixed" (combining doors, fixed). Focus on the odds on your door hiding the car. Initially, 2:1 against. Host opens a 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.

Why I like this solution: people need to be taught about Bayes in odds form - that's how to make conditional probability, and updating probabilities when you get new information, intuitive! It's the smart way to do Bayes in practice, all over science. It builds on the simple solution and shows the conditional solution is only a hair's breadth away! It replaces arithmetic with intuition. It fixes the Devlin combined doors solution...

I would like to inform ordinary people about how you can do correct probabilistic reasoning in your head, using a few intuitive ideas. Ordinary people aren't going to look at formulas. The conditional solution needs to be delivered in words, pictures, or a simple table (like TotalClearance's) otherwise it is indeed an "academic side-show". I've succesfully used this solution in talks to lawyers, doctors. It works: both for gaining understanding about MHP and for gaining understanding about probability. Richard Gill (talk) 08:12, 5 November 2012 (UTC)

I'm not saying that the formulas should replace a table. There is already a table and a corresponding decision tree in the article. I think the table in my earlier proposal and the Lucas-type solution are even simpler. The formulas are an addition to that. There are many articles on Wikipedia with formulas, including something as basic as 0.999... (featured article!), and I don't see why this article in particular should avoid them like the plague. There are all kinds of readers out there, and some prefer the Bayes' theorem. If you can read the formulas, then they are a clear and short way of putting it. That's why the notation exists.

You usually say that we're not trying to teach people mathematics here, so it's pretty odd that you now want to use the MHP article as a platform for teaching people the odds form of Bayes' formula. What I wrote above is surely simpler. If you prefer words to formulas, then you can put it this way: Since the prior is uniform, the posterior is simply the normalized likelihood. The likelihood is 1/2 : 1 : 0, which normalizes to 1/3 : 2/3 : 0 by multiplying it with 2/3. These are the probabilities of the car being behind each door. This is basically the Lucas solution, there isn't any more arithmetic here than in your solution, and it is much more intuitive as it doesn't have odds and likelihood ratios. If you want to teach people one generally useful thing, then surely it is the fact that the posterior is proportional to prior times likelihood, which is the form used basically everywhere in Bayesian statistics. -- Coffee2theorems (talk) 16:32, 5 November 2012 (UTC)

(1) I'm interested in intuitive verbal solutions to MHP which everyone can understand. (2) The English speaking world knows what odds are: it's a basic, non-technical concept. (3) The point of the Selvin-corrected solution is that it is intuitively obvious that posterior odds should equal prior odds when the new information is equally likely under the two alternatives. So you don't need to introduce technical concepts like "likelihood ratio". (4) Conclusion: there is no arithmetic at all in that solution, no technical terms, no unfamiliar concepts. Just plain common sense intuitive - and correct - verbal reasoning. Finally (pedagogical extra): it is a nice intro to Bayesian reasoning (posterior is proportional to prior times likelihood). AFTER you have given the solution, you can tell the interested reader that this was an example of Bayes in action. Richard Gill (talk) 20:38, 7 November 2012 (UTC)

Picture

The picture next to Devlin's solution is highly misleading and should be removed and never used again. Nijdam (talk) 10:50, 6 November 2012 (UTC)

It is not misleading but there is an argument that it requires additional logical steps to be correct. We have a section called 'Criticism of the simple solutions'. This would be the best place to explain what exactly is wrong with Devlin's solution as it is shown in this article and how any deficiencies in it could be remedied. Martin Hogbin (talk) 12:35, 6 November 2012 (UTC)

There is a missing assumption and hidden step in Devlin's argument. It's because the host is equally likely to open either door if he has a choice, that the chance the car is behind door 1 is not altered by revealing a goat behind door 3. Devlin did not mention this. When he was criticised on this point, he got confused, and he switched to a careful but (in my opinion) unenlightening routine conditional probability calculation.

How to fix his argument: the prior odds on car behind door 1 are 2:1 against. The probability the host will open door 3 if the car is behind door 1 is 0.5 (we have no knowledge of the host's strategy so the two possibilities are equally likely). The probability the host will open door 3 if the car is not behind door 1 is 0.5 (doors 2 and 3 equally likely, given not door 1, since we have no knowledge of how the car is hidden). So whether or not the car is behind door 1, the chance the host will open door 3 to reveal his goat is 0.5. The fact is therefore irrelevant to the question whether or not the car is behind door 1. This solution is given by yours truly but was previously expounded in a popular book as well as several articles by Jeff Rosenthal.

So I don't see anything misleading in the present text and picture, and they give a nice opportunity for "conditionalists" to "fix" the solution later ( = plug the gap), it's a beautiful vehicle (in my opinion) for illustrating Bayes theorem, odds form. I have succesfully explained MHP and Bayes to lawyers and to doctors in this way. Richard Gill (talk) 14:35, 6 November 2012 (UTC)

Then, please Richard, tell me what the picture shows in understanding the solution. Nijdam (talk) 09:49, 8 November 2012 (UTC)

No. I don't find any of the pictures of much use. I can't tell you, therefore, what use they have to people who do find them useful. Different people think different ways. I don't understand you, and you don't understand me. Richard Gill (talk) 16:01, 9 November 2012 (UTC)

Then why comment here at all? Nijdam (talk) 18:21, 9 November 2012 (UTC)

Everything is missing in the picture, and I'm not interested in reading for the so many'th time how the MHP may be solved. The picture doesn't and does not contribute to understanding. No, on the contrary, people understand this as follows: The not chosen doors have together probability 2/3 on the car (first picture), and when door 3 is opened (2nd picture) the opened door does not show the car, HENCE the remaining door must have the 2/3 probability. Is this misleading, or what? Maybe Martin does not understand this, but you, Richard, of all people, should know. Nijdam (talk) 21:30, 6 November 2012 (UTC)

Mathematicians are rather more careful with logic than non-mathematicians. Hence a mathematician could indeed find this misleading. Yet we are talking about a popular solution, widely cited in the literature, especially the popular literature. And Keith Devlin is even a top professional mathematician. So: criticise argument and picture, later in the article, giving sources for your criticism. Preferably in a constructive way. Richard Gill (talk) 09:27, 7 November 2012 (UTC)

Relevant sources should be considered, and shouldn't be "not even ignored" by editors here. After years of preaching, yet being still unaware of the fact that – after the guest did first select door #1 – the odds on door #2 : door #3 to be opened by the host, is exactly  1:1  in MvS's imaginary one time game show, is a miserable testimony. See Krauss and Wang, for example.
It's of no avail for the article that some editors obviously still are unaware of the relevant sources that say so, moreover saying "why exactly" this is so. Gerhardvalentin (talk) 22:43, 6 November 2012 (UTC)

Nijdam is saying that there is a "gap" in the visual representation of the argument. It corresponds to the already known "gap" in the verbal (written) argument. In my opinion, the gap is tiny and easily filled (symmetry). This has been done by various reliable authors (e.g., Rosenthal). It seems hard if not impossible to explain to a non-mathematician that there is a gap at all. So far, for instance, neither Martin nor Gerhard were prepared to agree that from a pure logic point of view, this argument includes a missing step. While the author (Devlin) himself, did see that his argument was incomplete, and he withdrew it in embarrasment!

Thus the picture is a not-misleading illustration of a flawed argument: they both have the same flaw. It seems to me that Nijdam ought to compose a little section criticising the argument, for later in the article, where we will report criticism of simple solutions. Maybe he could show at the same time how to patch the argument, and while he's at it, he could try to create a not-misleading visual representation of the patched argument. Citing reliable sources for the material which he employs. Richard Gill (talk) 09:14, 7 November 2012 (UTC)

I agree that is a good way forward. Rather than giving our readers less information by removing something let us give those who are interested more by explaining the exact problem and how it might be solved. Martin Hogbin (talk) 10:15, 7 November 2012 (UTC)

• "Missing step": Morgan et al., in focusing on quite another issue than MvS, have managed to leave the inaccurate, yes incorrect impression that in MvS's story, there could be any difference of chance that the host opens #2 or #3 of his doors, in case the guest should have chosen door #1, whether or not the car is behind door #1. MvS's story does exclude, just from the outset, that there could be any difference. Basing on MvS's story, such difference can never occur in her one-time event. So yes, as a result of Morgan et al.'s heretical story, you have to "do that missing step" and to just accentuate and and to proof that in MvS's story, there never can be any difference regarding the host's opening of one of his two doors, whichever it may be. Thank God, meanwhile there are reliable sources to prove this given fact. One just has to cite those sources. Gerhardvalentin (talk) 23:50, 7 November 2012 (UTC)

Statement

I like to make the following statement, to make sure what this whole discussion is about.

1. Some sources consider the MHP to be the problem as originally formulated with door 1 chosen and door 3 opened, showing a goat. This formulation is not solved by the simple argument that switching gives the player the car with 2/3 chance, but by calculating conditional probabilities.
2. Other sources for some reason like to consider the MHP to be the problem in which the player - or actually the audience - is asked in advance, even before choosing a door to decide whether to switch or not after the host will have opened a door showing a goat. In this formulation the simple switching argument is sufficient.
3. There even are sources (reliable?) who are confused, and present the first formulation, but use the simple switching argument as a solution.

It's my opinion (Wikipedia policy) we have to make this distinction clear to our readers. If somehow we mainly emphasize situation 3, we fail. Nijdam (talk) 10:58, 8 November 2012 (UTC)

Your opinion is very clear, Nijdam. You say "This formulation is not solved by the simple argument that switching gives the player the car with 2/3 chance, but by calculating conditional probabilities". That's your opinion and the opinion of some sources. And of some editors. But not all sources, not all editors. There's no point in discussing this again. The discussion has gone on for five years or so and as far as I know, nobody involved ever changed their minds.

If you do want to discuss this more, I suggest my talk page. Martin Hogbin (talk) 12:41, 8 November 2012 (UTC)

According to the conclusion of the RfC (simple solutions first, without health warning, conditional solutions and criticism of simple solutions later), the literature on this issue will be surveyed in a separate and later section of the article. Go ahead and work on that part of the article. The better that part of the article is written, the more chance you have of getting the point across, which you find so crucial! It's in your own hands! Richard Gill (talk) 12:20, 8 November 2012 (UTC)

Richard, you know, and have admitted so, that 1 and 2 are correct and 3 is erroneous. Aren't you worried that Wikipedia might direct the readers mainly to 3 as a correct way of considering the MHP? — Preceding unsigned comment added by Nijdam (talk • contribs) 16:06, 8 November 2012

Richard may not worry because thinks that you will explain exactly why 3 is erroneous in the appropriate section. If you will not do that then he or I or someone else will. Martin Hogbin (talk) 17:54, 8 November 2012 (UTC)

(A) I do not believe that 1 and 2 are correct and 3 is erroneous!

(B) I hope any possible defects of (3) will be explained in such a good way later in the article, that anyone who is smart enough to understand the issues will easily be able to learn about them, and come to their own conclusions.

(C) I think, Nijdam, that you still haven't appreciated any of the things I was trying to say in my article in Statistica Neerlandica about model building and problem solving. I think you also don't understand that popular readers only need to know why sticking with your initial choice is not so smart. They are not interested, and don't need to be interested, in technical issues about conditional and unconditional probabilties.

(D) One can also approach MHP with ideas from decision theory - and many people do, and many sources do! There is not One and Only One Truly Correct Solution to MHP (in my opinion). Richard Gill (talk) 13:47, 9 November 2012 (UTC)

I was just about to say much the same. The consensus arising from the RfC was first to have a section giving your type 3 solutions. Later on we have a section currently entitled 'Solutions considering the specific doors the player chooses and host opens' (I think this needs changing) in which there is also a section called 'Further analysis of the simple solutions'. In these sections you can explain exactly what is wrong with the type 3 solutions, what steps are missing and what alternative solutions are available. I agree that the distinction that you mention must be made clear to our readers; we have agreed a way to do this. I, and no doubt Richard and others, are happy to work with you to achieve this aim. Martin Hogbin (talk) 12:41, 8 November 2012 (UTC)

Frequentists vs. Bayesians

Frequentists vs. Bayesians... --Guy Macon (talk) 23:50, 9 November 2012 (UTC)

Cool! We need more humour at MHP. Richard Gill (talk) 11:13, 10 November 2012 (UTC)

Section removed from lead

These two rather unrelated sentences have been removed from the lead by TotalClearance. What is the claimed problem with them. Martin Hogbin (talk) 23:20, 6 November 2012 (UTC)

Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy.

Decision scientist Andrew Vazsonyi described how Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until Vazsonyi showed him a computer simulation confirming the predicted result (Vazsonyi 1999).

TotalClearance, is your complaint that this text does not refer to something in the body of the article. If so, I suggest that the best solution would be to add something to the body. Martin Hogbin (talk) 23:40, 6 November 2012 (UTC)

It is obvious and well known that the first impression regarding the chance to win by staying versus the chance to win by switching, at the end of MvS's world famous tricky story, is mostly false and incorrect. This fact is vividly portrayed and demonstrated by reporting on the first reaction of one of the world's most famous mathematician Paul Erdős. TotalClearance deleted that representative reference without giving any sound argument. Now he deleted it for the 4th time meanwhile, without giving any additional sound argument. I am going to re-insert that reference, because it is of great importance to evaluate and to realize the hidden problem. Should TotalClearance remove it once more without additional sound argument, I will be going to report about his presumed vandalism. Gerhardvalentin (talk) 00:06, 7 November 2012 (UTC)

I am not sure what TotalClearance's objection is but we are waiting to hear from him. Martin Hogbin (talk) 10:17, 7 November 2012 (UTC)

When I removed those two sentences on November 3. 8:15 h, I started a new section "The lead" on this talk page at the same time November 3. 8:16 h giving a short reason. But neither Martin Hogbin nor Gerhardvalentin nor anybody else took the opportunity to discuss this modification. Instead of giving arguments Martin Hogbin reverted two times (November 6. 00:25 hand November 6. 23:38 h) and Nijdam one time (November 6. 21:34 h). Four days after my fruitless trial of discussion Gerhardvalentin is talking about "vandalism" and "missing arguments", but he had time enough in advance to give his arguments referring to the two sentences. What's the matter with him? --TotalClearance (talk) 10:34, 9 November 2012 (UTC)

Sorry TotalClearance, I missed your comments on the talk page. The fact remains that nearly all sources state that the probability of winning by switching is 2/3 under the normal assumptions. The sentence about Paul Erdos was also well sourced. Martin Hogbin (talk) 19:59, 9 November 2012 (UTC)

I think, TotalClearance, that no-one understood your reasons. You were not bothered by the anecdote, but by the suggestion that the answer was "2/3". Your criticism was that the answer is only 2/3 under certain assumptions, and they hadn't been stated yet. I disagree with you. The answer is 2/3 if you take a subjective probability approach and use uniform distributions to represent total ignorance (Laplace). We don't need to make any new assumtions to get to 2/3. We just follow standard principles of reasoning in the face of uncertainty. Richard Gill (talk) 13:54, 9 November 2012 (UTC)

I don't want to discuss this nonsense anymore. Only one additional remark: Vos Savant herself responded to those who criticized the ambiguity in her narrative as follows:"Pure probability is the paradigm, and we published no significant reason to view the host as anything more than an agent of chance who always opens a losing door and offers the contestant the opportunity to switch" (1991, italics added, cited in Mueser/Granberg 1999). Her assumptions lead us to the following solution table, with door 1 chosen and the host opening a losing door randomly:

Prob.	behind door 1	behind door 2	behind door 3	opened door	result if staying at door #1	result if switching to the door offered
1/6	Car	Goat	Goat	2	Car	Goat
1/6	Car	Goat	Goat	3	Car	Goat
1/6	Goat	Car	Goat	1	Goat	Car with P=1/2
1/6	Goat	Car	Goat	3	Goat	Car
1/6	Goat	Goat	Car	1	Goat	Car with P=1/2
1/6	Goat	Goat	Car	2	Goat	Car

Therefore the probability of winning by switching is 1/2, independent of the opened door, and not 2/3. So, switching is better than staying, but the odds are the same for both the unchosen doors. But in case that the host opened an unchosen door (2 or 3), switching is not better than staying. It seems that vos Savant actually didn't understand the consequences of her assumptions. --TotalClearance (talk) 23:50, 9 November 2012 (UTC)

You seem to have overlooked the fact that Vos Savant also insisted that her intention was, that the reader would understand that the host always opens a *different* door to the door chosen by the player, in order to reveal a goat. She also said that almost all her letter-writers, including those who disagreed with her answer, understood her intention correctly. Richard Gill (talk) 11:06, 10 November 2012 (UTC)

We don't know this, and in the letters, she had published, the words referring to those necessary rules are rare. Moreover she had reasons enough to find excuses for her dubious solution, because in the end she could lose her reputation. --TotalClearance (talk) 16:01, 10 November 2012 (UTC)

We do know this, the original question said (my bold), '...the host, who knows what's behind the doors, opens another door'. It is quite clear from this that the rules should exclude the host's opening the door originally chosen by the player. Martin Hogbin (talk) 16:23, 10 November 2012 (UTC)

Moreover we know her solution (2/3); and we know the problem goes back to Selvin 1975a, Sevlin 1975b, Nalebuff 1987; who also all make the same rules clear. Richard Gill (talk) 18:40, 10 November 2012 (UTC)

Martin Hogbin said:"It is quite clear from this that the rules should exclude the host's opening the door originally chosen by the player." Why that? In this one-time-game the host opens another door, but the possible outcomes may include the host's opening the door originally chosen by the player. There is no reason to believe that the host must open an unchosen goat door. From the fact "the host opened an unchosen goat door" cannot be deduced the assumption "the host had to open an unchosen goat door".

Richard Gill said:"...and we know the problem goes back to Selvin 1975a, Selvin 1975b, Nalebuff 1987; who also all make the same rules clear." Most of the readers of the Parade magazine don't know that. Vos Savant's question should stand for itself without any relation to former similar problem statements. Or did she refer to Selvin and Nalebuff explicitly in her column of the Parade magazine? --TotalClearance (talk) 11:08, 11 November 2012 (UTC)

Our job on Wikipedia is not to give a completely new solution of the question Vos Savant posed in Parade magazine (because we believe that everyone till now understood it in a different way) WP:NOR, but to survey the published literature WP:V on Monty Hall problem. It's huge. There is great diversity of solutions. We have to do this in mind in a neutral way WP:NPOV. About the only thing that's agreed is that the host has to open an unchosen door and reveal a goat. (Thanks TotalClearance for spotting my typos - I've corrected them). Richard Gill (talk) 14:50, 11 November 2012 (UTC)

As Richard says above, almost the only thing agreed on by all sources that cover this subject is that the host cannot open the door already chosen by the player. The article does not make this quite clear but it should.

If you think anyone might be interested you could add a bit in 'Variants' section in which the host has the option of opening the door that the player has already chosen, if you can find a source for this. Martin Hogbin (talk) 15:28, 11 November 2012 (UTC)

Please answer my questions in the subsection "Anecdotes in the lead?" below. --TotalClearance (talk) 16:17, 11 November 2012 (UTC)

Anecdotes in the lead?

We should avoid "anecdotical evidence". We know that the 2/3-solution is correct only under certain circumstances. For a NPOV I removed a part of the lead. --TotalClearance (talk) 08:16, 3 November 2012 (UTC)

Well, it is not considered "anecdotical evidence", but merely an anecdote. That it is possible to make such assumptions that the odds will be equal, has no connection with Erdos disbelief. Nijdam (talk) 13:37, 9 November 2012 (UTC)

If vos Savant's solution, which is stated previously, would be correct, why does it need an anecdotical support in the lead? And why is not reported in the lead: that vos Savant's problem statement "is not well-formed" (Gardner, one of the world's most famous authors of brainteasers), that her solution was proved not to be corresponding to her problem statement (Monty Hall, one of the world's most famous game show hosts) and that vos Savant herself acknowledged the ambiguity of her original statement? The miserable state of the article, which fails the NPOV conditions, may already be observed in the lead. --TotalClearance (talk) 18:22, 9 November 2012 (UTC)

Vos Savant: The lead doesn't say (and shouldn't say) that Vos Savant's solution is correct. Wikipedia can't report that some solutions are correct, some incorrect. It can only report the opinions on this matter which are published in the literature. Read WP:Truth. Richard Gill (talk) 17:19, 11 November 2012 (UTC)

The article says:"Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy." What about this? --TotalClearance (talk) 17:55, 11 November 2012 (UTC)

What about this, then? The sources are agreed that switching is the best strategy. Wikipedia reports what the literature says, it reports what is commonly believed. Maybe the literature is wrong. Then wikipedia will be wrong. You need to study WP:V and WP:Truth. If you want to change people's ideas about MHP then write an article about it and get it published in a suitable journal. Wait 10 years till it becomes a main-stream point of view. Then come back to wikipedia. Richard Gill (talk) 13:01, 12 November 2012 (UTC)

Monty Hall: Monty himself corresponded with Selvin (see Selvin, 1975b), agreed with Selvin's 1975a "simple solution", and himself gave another "simple solution" (which Selvin reports with approval) quite reminiscent of the so-called "combined doors" problem! Monty did point out that in reality the show did not go that way. Richard Gill (talk) 17:19, 11 November 2012 (UTC)

You said:"Monty did point out that in reality the show did not go that way." This sentence should be inserted in the lead. --TotalClearance (talk) 17:55, 11 November 2012 (UTC)

Why? the article is about MHP as it is understood by the sources - a popular brain-teaser - not about the real Monty Hall show (there are other articles on Wikipedia about the man and about the show). Richard Gill (talk) 13:03, 12 November 2012 (UTC)

Martin Gardner: Gardner's only issue was that Vos Savant hadn't stated explicitly that we are to understand that Monty certainly opens a different door revealing a goat. But luckily, most readers understood her problem this way, and every single other source of which I am aware, writing on MHP, makes this explicit. Read Jason Rosenhouse's book or (the MHP chapter in) Jeff Rosenthal's book. Richard Gill (talk) 17:19, 11 November 2012 (UTC)

You said:"Gardner's only issue was that Vos Savant hadn't stated explicitly that we are to understand that Monty certainly opens a different door revealing a goat." Gardner's only issue? This is the crucial point! --TotalClearance (talk) 17:55, 11 November 2012 (UTC)

TotalClearance, there are references given in the article in the 'Problem' section stating how the problem is generally understood. There are no sources that I am aware of that consider that the host might open the door originally chosen by the player. Martin Hogbin (talk) 18:16, 11 November 2012 (UTC)

The Problem: Added 'unchosen' to clarify standard problem

Martin Hogbin has added 'unchosen' to clarify the standard problem. But did he verify that the mentioned author Barbeau2000:87 really used the term 'unchosen' in his paper? --TotalClearance (talk) 18:10, 11 November 2012 (UTC)

I did not check Barbeau for those exact words, perhaps someone else will, but we could have put Krauss and Wang (which I have checked) or any of several other references. Martin Hogbin (talk) 18:28, 11 November 2012 (UTC)

Marilyn vos Savant herself says "opens another door", and she says

• "Then the host purposly [...] a loosing [...] from the two unchosen."

So what is the problem of TotalClearance, is he talking of quite another scenario, and not of the given clear one-time show scenario that asks for the decision "switch or stay"? The decision must be made for exactly that given scenario. Gerhardvalentin (talk) 09:38, 13 November 2012 (UTC)

Vos Savant said:"...and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat." From this statement cannot be deduced that the possible outcomes must exclude the host's opening the door originally chosen by the player, just because it is a one-time show scenario. Several times vos Savant failed to explain explicitly the host's restraint to open an unchosen goat door:

• "My original answer is correct. But first, let me explain why your answer is wrong. The winning odds of 1/3 on the first choice can’t go up to 1/2 just because the host opens a losing door."
• "So let’s look at it again, remembering that the original answer defines certain conditions, the most significant of which is that the host always opens a losing door on purpose."

In fact she used the rule "the host always opens an unchosen door which has a goat" implicitly from the beginning of her reasoning. But this rule cannot be deduced from her problem statement as a compelling one.

If Barbeau is mentioned as the source for the standard analysis it should be assured that he really used the correct and complete problem statement in his paper. --TotalClearance (talk) 18:23, 13 November 2012 (UTC)

The Barbeau reference is to a small book. With a very small chapter on MHP. I have a copy and will check later. I seem to recall that Barbeau gave a simple solution. He certainly makes the standard disambiguation. Whether or not he specifically uses the word "unchosen" seems to me quite irrelevant. Nowadays, Jason Rosenhouse's book is a better reference. Richard Gill (talk) 10:36, 14 November 2012 (UTC)

TotalClearance, Wikipedia is based on what is written in reliable sources. We have sources that say that the Whitaker/vos Savant statement is ambiguous, so we say that in the article. It is ambiguous in many respects, not just the one that you are concerned about. We also have sources saying how most people understand the question, so we put that in the article. We also have sources referring to way that most people see the problem as the 'standard assumptions', so we may use that term in the article and explain exactly what it means. There are no sources that I am aware of that solve the problem in which the host may open the door originally chosen by the player or that state the most people see the problem this way. You may have understood the problem that way but that is irrelevant. Martin Hogbin (talk) 13:31, 14 November 2012 (UTC)

To anyone who is interested in the Barbeau (2000) reference: I can scan it and put it in a dropbox folder and share it with anyone who is interested. However it is a rather disappointing item. Edward J. Barbeau's book is a compendium of many brain teasers which were discussed in the author's own column in the journal College Mathematics Journal. Between 1993 and 1998, in response to the controversy unleashed by Vos Savant's Parade article, MHP was discussed there 6 times (unfortunately back numbers of the journal were not available on-line, last time I looked). In his book, Barbeau gives more than 40 references to MHP and closely related problems but almost no analysis. This is what he says:

"The standard analysis of MHP is based on the assumption that after the contestant makes the first choice, the host will always open an unselected door and reveal a goat (choosing the door at random if both conceal goats) and then always offer the contestant the opportunity to switch. The contestant initially selects a door concealing a goat with probability 2/3. With a policy of always switching, she will win a car with this probability."

Note: simple solution, yet mention of host unbiasedness! I get the impression Barbeau does not find the problem very interesting. Richard Gill (talk) 16:01, 15 November 2012 (UTC)

Conclusion: Barbeau used the word "unselected". Seems to me that "unchosen" will do just fine. Richard Gill (talk) 14:58, 19 November 2012 (UTC)

Vos Savant and the media furore

We currently have a paragraph on this subject in the lead but nothing about it in the body of the article. What should we have and where should it go?

Do we need a section on this? The lead says what needs to be said, I think. And right at the end of the article there is more history. Richard Gill (talk) 12:32, 16 November 2012 (UTC)

It is what the problem is most well known for. The lead should be a summary of the article. At present we have information in the lead that is not in the body. Martin Hogbin (talk) 09:50, 17 November 2012 (UTC)

Copy and paste to the "History" section then. Richard Gill (talk) 09:55, 19 November 2012 (UTC)

BTW the history refers to "more than 40 published (academic) papers" on MHP. That was in ca. 2000. Ten plus years on it must be more than 50 now. Richard Gill (talk) 09:55, 19 November 2012 (UTC)

Just did a recount using Rosenhouse book for 2000-2009 and a guess for 2010 onwards (see references in article), brings me to 75. Richard Gill (talk) 12:49, 19 November 2012 (UTC)

Added "table of 6" solution

I boldly added the table of 6 (TotalClearance's proposal!) between the conditional probability table and the decision tree solutions. We now have a lot of citations for this solution in some of the major works on MHP. I think (and I hope others agree) that it makes the conditional probability solution a whole lot more accessible. No more arithmetic with fractions. Just simple counting. Richard Gill (talk) 15:37, 19 November 2012 (UTC)

I also separated the statement of the standard problem at the beginning of the article into two parts: first the "deterministic" constraints on the host, second the usual probability assumptions. Since it can be argued that part of the problem is to deduce (or at least, give reasons for) any probability assumptions which are made. Richard Gill (talk) 17:55, 19 November 2012 (UTC)

Odd section on Odds

The section entitled "odds" has become totally superfluous. It says now that you can compute odds from probabilities, and probabilities have already been computed. This section should contain a derivation of conditional 2/3 using the odds form of Bayes theorem, following Rosenthal and other authors. For some (esp. Anglo-Saxon) readers, it makes the calculations a whole lot more simple, intuitive and illuminating. This way it can include the "Devlin fixed" solution making a bridge to the simple solutions.

The section contains an orphaned bit of calculations left from earlier solutions. I'll delete that, first, and boldly compose a new odds section, later. Richard Gill (talk) 12:08, 16 November 2012 (UTC)

Good idea. Martin Hogbin (talk) 21:59, 16 November 2012 (UTC)

Done. Richard Gill (talk) 06:23, 20 November 2012 (UTC)

Why isn't Economist solution not among the simple solutions

The simple solution of "The Economist" in which the location of the car is fixed but unknown and the player chooses completely randomly is missing from among the simple solutions. Such a decision oriented solution is not uncommon in the literature and connects to the academic decision theoretic stuff later in the article. Can it be added? Richard Gill (talk) 08:35, 18 November 2012 (UTC)

I would support its addition, with a simple explanation of what has been done, along the lines you give above. As the player has chosen at random, the initial car placement and the host's door policy become unimportant so this solution makes the 'conditional' ones redundant. They have obviously interpreted vos Savant's 'say door 3' in the way that she intended, in other words not to specify the condition that the host has actually opened door 3.

I will add this solution with some words which might need references, to try to explain the 'Economist' approach. Martin Hogbin (talk) 13:43, 19 November 2012 (UTC)

Yes please! I rearranged the "strategic dominance" subsection to tie in with (and refer to) this. Richard Gill (talk) 06:24, 20 November 2012 (UTC)

New intro to conditional part

I added a simple calculation of conditional 2/3 (by imagining 600 repetitions) at the beginning of the section on conditional solutions. This way the reader knows what we are talking about before launching into all kinds technical discussions and literature surveys and alternative derivations.

I put it in a separate (initial) subsection.

Then I added a sentence or two before each of the subsequent subsections introducing the topic of that subsection so that the reader can see more quickly what is coming and why, and thereby can more easily decide whether to read in detail or skip to the next subsection. Richard Gill (talk) 18:21, 20 November 2012 (UTC)

The conditional solution table

To model the host's behaviour opening door 2 or door 3, if he has the choice, completely we consider The solution shows the six possible equally likely arrangements of one car and two (distinguishable) goats behind three doors. and The table shows the result of staying or switching after initially picking door 1 in each case. In the first case the host opened door 2, in the fourth case door 3, according to the rules of the game show:

behind door 1	behind door 2	behind door 3	opened door	result if staying at door #1	result if switching to the door offered
Car	Goat	Goat	2	Car	Goat
Goat	Car	Goat	3	Goat	Car
Goat	Goat	Car	2	Goat	Car
Car	Goat	Goat	3	Car	Goat
Goat	Car	Goat	3	Goat	Car
Goat	Goat	Car	2	Goat	Car

Now we have to pay attention to With respect to the final situation in the game show we consider the cases with door 3 opened, corresponding to the rows 2, 4 and 5. according to the final situation in the game A player who stays with the initial choice wins in only one out of three of these equally likely possibilities, while a player who switches wins in two out of three. The probability of winning by staying with the initial choice is therefore 1/3, while the probability of winning by switching is 2/3. We can see that the cases with door 2 opened are giving the same result and we are able to observe a symmetry between the non-chosen doors 2 and 3. --TotalClearance (talk) 09:03, 3 November 2012 (UTC)

So the goats are called "Goat 1" and "Goat 2" and the difference between the bottom and top halves of the table is the order of Goat 1 and Goat 2 behind the doors. And the host's rule is that he opens the door of "Goat 1" if he has a choice: the left hand goat in the first three rows, the right hand goat in the second three rows. The host can distinguish between the goats but the player can't (or rather: doesn't, probably because in advance he doesn't know the two goats personally, nor the rule being used by the host). Then indeed there are six equally likely possibilities, with Door 2 and Door 3 being opened by the host equally often. Given that Door 3 is opened, there are three equally likely possibilities. Switching wins the car in 2 our of 3 cases, both unconditionally and conditionally.

Note that TotalClearance says, dogmatically, we have to pay attention to the cases with door 3 opened according to the final situation in the game. Not everyone agrees with this (neither editors, nor readers, nor sources). The conclusion of the RfC is indeed that in the first part of the article we are not going to pay attention to this issue.

I have not seen this solution before. It's a cute and I think original "conditional solution". Richard Gill (talk) 09:20, 3 November 2012 (UTC)

Why do you need all that? This is all you need. We know the door numbers are unimportant so everything else is certainty.

behind door 1	result if staying at door #1	result if switching to the door offered
Goat	Goat	Car
Car	Car	Goat
Goat	Goat	Car

Martin Hogbin (talk) 10:08, 3 November 2012 (UTC)

TotalClearance is proposing what he considers to be a "simple" conditional solution. It gives you a stronger conclusion than your table, Martin, but you have to do more work to get it. You could also add to your table, Martin, the comment that the door numbers are irrelevant by symmetry, and then you would also get the conclusion "given the host has opened Door 3, switching gives the car with probability 2/3". The more you put in, the more you get out. No free lunches.

But in view of the conclusion of the RfC, TotalClearance is going to have to try to get his preferred solution somewhere later in the article. And he's going to have to source it. Richard Gill (talk) 10:40, 3 November 2012 (UTC)

What exactly is stronger about the conclusion you reach from TotalClearance's table. We know door numbers and goat ID are irrelevant so we can conclude from my table that "given the player has originally chosen door 1 and the host has opened Door 3, switching gives the car with probability 2/3". The door numbers are discarded at the start.

One can *argue* that door numbers are irrelevant and then one can *conclude* what you just wrote. But you did neither. TotalClearance put in more assumptions and got out a stronger conclusion. You put in less assumptions and got a weaker conclusion. I think it's a matter of opinion which solution is better. TotalClearance on the other hand seems to think that you have to do it his way. His solution seems to me to be quite original. Richard Gill (talk) 15:31, 3 November 2012 (UTC)

TotalClearance's table chooses to distinguish between the two goats but does not include all the possibilities. For example, there is no cases where Goat 2 is revealed behind door 2. If we are going to distinguish between goats then we must have a complete table showing all the possibilities. THis will be quite complicated, bearing in mind the next problem with the table.

Yes there is. It's the third of the six possibilities (six permutations of three distinguishable objects: Car, Goat 1, Goat 2). If Goat 1 is behind door 1 and Goat 2 is behind door 2 then the host will be forced to open door 2 and reveal Goat 2, even though he prefers to reveal Goat 1. Richard Gill (talk) 15:31, 3 November 2012 (UTC)

Sorry, I should have said that there is no case where the player has originally chosen the car and Goat 2 is revealed behind door 2. Richard's version of the table can be found as the first tablehere for reference.

Yes. Of course. TotalClearance assumes (I think) that the host prefers to reveal Goat 1. If the car is behind door 1, the order of items behind the three doors is Car, Goat 1, Goat 2 in the first half of table (row 1) and Car, Goat 2, Goat 1 in the second half of the table (row 4). In the first case the host opens door 2, in the second case he opens door 3. In neither case does he reveal Goat 2. If the player knew about this, and knew the two goats in advance (goats are distinguishable, right!?) then the host's behaviour would give further information to the player, changing his probabilities that the other door hides the car, yet again. The player would not only condition on which door was opened but which goat he saw there, and get probability 1/2 in one case, 1 in the other case. Averaging out - when correctly weighted - to 2/3. Richard Gill (talk) 07:53, 4 November 2012 (UTC)

TotalClearance has also fallen for the Morgan conjuring trick. There is nothing special about the door number opened by the host. If we decide that door numbers are important (and goats are not) than we must show all the possible doors that the player might have initially chosen. That would result in this table.

You can argue there is nothing special about the door number opened by the host but you didn't - you took it for granted. This is nothing to do with some conjuring trick. TotalClearance wants to know what is the chance that the car is behind door 2 when you chose door 1 and the host opened door 3. (So he's not interested in two-thirds of the big table which you put below here: the player didn't choose door 2 or door 3). People like TotalClearance think that that is how the question has to be answered. Some editors think that (I don't, by the way), some writers think that, some readers will think that. The result of the RfC is that people like TotalClearance are going to have to be patient, and do their thing later in the article. Richard Gill (talk) 15:39, 3 November 2012 (UTC)

You still have not spotted Morgan's sleight of hand. The number of the door that the host opens is no more important than the door number that the player originally chooses. We know that neither makes any difference but the pedantic way of doing things is to show all the possible doors that the player might have chosen and all the doors that the host might have opened and then condition on the doors mentioned in the question. The intuitive way is to accept that door numbers are irrelevant, thus arriving at my table. Martin Hogbin (talk) 01:02, 4 November 2012 (UTC)

Player initially chooses	behind door 1	behind door 2	behind door 3	opened door	result if staying	result if switching to the door offered
Door 1	Car	Goat	Goat	2	Car	Goat
Door 1	Goat	Car	Goat	3	Goat	Car
Door 1	Goat	Goat	Car	2	Goat	Car
Door 1	Car	Goat	Goat	3	Car	Goat
Door 1	Goat	Car	Goat	3	Goat	Car
Door 1	Goat	Goat	Car	2	Goat	Car
Door 2	Car	Goat	Goat	3	Goat	Car
Door 2	Goat	Car	Goat	3	Car	Goat
Door 2	Goat	Goat	Car	1	Goat	Car
Door 2	Car	Goat	Goat	3	Goat	Car
Door 2	Goat	Car	Goat	1	Car	Goat
Door 2	Goat	Goat	Car	1	Goat	Car
Door 3	Car	Goat	Goat	2	Goat	Car
Door 3	Goat	Car	Goat	1	Goat	Car
Door 3	Goat	Goat	Car	2	Car	Goat
Door 3	Car	Goat	Goat	2	Goat	Car
Door 3	Goat	Car	Goat	1	Goat	Car
Door 3	Goat	Goat	Car	1	Car	Goat

Just like the other table, this reduces to mine. If we want to distinguish between goats then the table gets more complicated still. How does this help our readers? Martin Hogbin (talk) 14:11, 3 November 2012 (UTC)

In case of vos Savant's solution, row 3 with the car behind door 3 doesn't fit to the standard problem conditions and in my experience many people argue consequently, that the odds must be the same for door 1 and 2 referring to the rows 1 and 2 being equally likely and row 3 being impossible. But we know that this reasoning is wrong...

The conditional solution table can be considered as a "mental model representation" of the MHP as described in table 1, page 5, in Krauss' and Wang's paper. They say:"Most representations found in the literature consist of more than three single arrangements and specify Monty Hall’s behavior in each arrangement." Such an illustration often is used also if you have several arrangements which are not equally likely. In that case it is very helpful to expand the respective table to all arrangements being equally likely. So you are able to solve a problem by counting off the matching cases. This is an easy and intelligible way for ordinary people to understand conditional probability and meets didactic purposes.

IMO this solution table should be given to the reader directly after the formulation of the standard problem because it fits perfectly. Since the result of the RfC is not a law, with the help of a few editors here we can insert this table early in the article. --TotalClearance (talk) 22:12, 3 November 2012 (UTC)

The result of the RfC represents a consensus after years of arguing here. Content is determined by consensus. Martin Hogbin (talk) 01:06, 4 November 2012 (UTC)

IMO the result of the RfC is not a consensus but a decision of a small majority, which could be changed in certain cases. --TotalClearance (talk) 08:24, 4 November 2012 (UTC)

That may be your opinion but, as I have said, we have a conclusion to years of argument. Please do not start it again.

Anyway: TotalClearance's solution is completely original, as far as I know! Congratulations. Do you agree with my additions to your table? I put it also on my talk page at [1]. TotalClearance also says, tellingly, "This is an easy and intelligible way for ordinary people to understand conditional probability and meets didactic purposes". I think that TotalClearance is interested in using MHP in a probability class, in order to explain conditional probability. But the wikipedia page on MHP is (I think) mainly about explaining MHP to a very wide audience, not about explaining conditional probability to students in a probability class. What the trained probabilists tend to forget, is that there are other ways to explain why staying with your initial choice is an unwise strategy if you want to get the car. Richard Gill (talk) 07:53, 4 November 2012 (UTC)

I am interested in the phenomenon that people are arguing wrong if they apply vos Savant's solution to the standard problem, with door 1 chosen and door 3 opened. Expanding her table for modelling the host's behaviour completely helps people to understand what happens in the game show. Furthermore this conditional solution table is the next logical step following vos Savant's solution, and it doesn't need any elaborated arguments like strategy, symmetry and "combined doors" to solve the problem. --TotalClearance (talk) 08:55, 4 November 2012 (UTC)

OK, now i understand your solution. It's the same as the decision tree in the article, but in which you split the two lower branches (each of probability 1/3) into two identical branches, each of probability 1/6. As if the host tosses a coin to aid his decision which door to open whether he has a choice or not. The top branch of each pair is heads, the bottom branch is tails. If he has a choice, then heads means open door 2 and tails means open door 3. Instead of having four distinct outcomes, with probabilities 1/6, 1/6, 1/3, 1/3, you have 6, all with probability 1/6. This is neat and a useful pedagogical device. Much better than (what seems like) pages of formulas, in the two Bayes sections (which in my opinion should both be deleted). Richard Gill (talk) 14:48, 4 November 2012 (UTC)

I do not see how that table is helpful. Lines 2 and 5, for example, are identical (as you are not distinguishing between goats). So this represents the host tossing a coin to decide which door to open. If it comes up heads he opens door 3 and if it comes up tails he opens door 3. Two distinct outcomes? Crazy! Maybe if I toss a coin and it comes up heads I become a different person. Martin Hogbin (talk) 21:31, 4 November 2012 (UTC)

Decision tree: 1/3 = 2/6, isn't it? --TotalClearance (talk) 22:14, 4 November 2012 (UTC)

Yes 1/3 = 2/6 = 7/14 = 243/486 So what? Where do the two distinct outcomes come from? Looks like double Dutch to me. Martin Hogbin (talk) 22:21, 4 November 2012 (UTC)

What is the problem? The aim is to show that it makes no difference at all which of those three door has been first selected by the guest, nor which one of "his two doors" has been opened by the host. Sources say that there is "no difference". The scenario is described and known: The guest first selected his door, and thereafter the host is to open one of his two unselected doors with a goat behind and to offer switching. Now it is a fact that there are many various weighty reasons to switch, nobody "must" only base on maths. The second goat and the car (or vice versa) are now behind the two still closed doors: the guest's door and the still closed host's door. As the modus operandi of the host, in case he has got two goats, forever will be completely unknown in this one time problem, it is completely useless to jump to conclusions based on "which one" of his two doors he opened. Doing this anyway, that may serve for purposes of maths only, but never to support the decision asked for. Using conditional probability in maths classes is out of position to ever be able to decide whether – in the actual problem – the probability to win by switching "could be" between [1/2 to 2/3], or "could be" between [2/3 to 1]. Any of such assumption is completely worthless for the actual problem and for the actual decision asked for. Sources say therefore that the host's opening of his nearby door or of his distant door is "all the same". For the reader, it should clearly be shown that this makes "no difference". Use decision trees or various different tables, all of this will be helpful for the reader to "see" that it makes no difference "which" door the host has opened. And use the "combined door" visualization. Redundancy is most welcome to achieve this goal. Gerhardvalentin (talk) 00:05, 5 November 2012 (UTC)

The aim is to show that under K&W conditions the player should switch because the probability of winning by switching is 2/3. It is but a step from vos Savant's illustration, stated as the first simple solution, to the solution table which is solving "the full problem":

Prob.	behind door 1	behind door 2	behind door 3	result if staying at door #1	result if switching to the door offered
1/3	Car	Goat	Goat	Car	Goat
1/3	Goat	Car	Goat	Goat	Car
1/3	Goat	Goat	Car	Goat	Car

The first simple step is to expand the row 1 according to the host's behaviour, if he has a choice, similar to the decision tree:

Prob.	behind door 1	behind door 2	behind door 3	opened door	result if staying at door #1	result if switching to the door offered
1/6	Car	Goat	Goat	2	Car	Goat
1/6	Car	Goat	Goat	3	Car	Goat
1/3	Goat	Car	Goat	3	Goat	Car
1/3	Goat	Goat	Car	2	Goat	Car

The next simple step is to expand the remaining rows, too. Because of 1/3 = 1/6 + 1/6:

Prob.	behind door 1	behind door 2	behind door 3	opened door	result if staying at door #1	result if switching to the door offered
1/6	Car	Goat	Goat	2	Car	Goat
1/6	Car	Goat	Goat	3	Car	Goat
1/6	Goat	Car	Goat	3	Goat	Car
1/6	Goat	Car	Goat	3	Goat	Car
1/6	Goat	Goat	Car	2	Goat	Car
1/6	Goat	Goat	Car	2	Goat	Car

Therewith you have the correct solution table for the situation: door 1 chosen. Analoguous the tables for the other situations have the same appearance, with other doors chosen and other doors opened. But we don't need them, because of the car is initially equally likely to be behind each door, the player's initial choice doesn't change anything on principle. --TotalClearance (talk) 17:10, 6 November 2012 (UTC)

Neither does the host's door choice change anything, obviously.

What event exactly does the probability of 1/6 in line 3 refer to? Martin Hogbin (talk) 23:29, 6 November 2012 (UTC)

It doesn't refer to any event. As mentioned before, you can make this work by letting the host always flip a coin, and then add "coin flip result" column in the table.

Both of you are saying "doesn't change anything" without specifying what that means. That's not useful. TotalClearance's argument can be made to work by saying something like "The proofs that switching wins with probability 2/3 given any other initial choice of door and choice of host's door are similar." at the end, or "Assume without loss of generality that the player initially picked door 1 and the host opened door 3." at the beginning. Either way, the meaning is that a proof template has been provided, and that template works for any door choices with trivial changes. I do not see any such concrete interpretation for your words. TotalClearance's table shows that the probability is 2/3 for one particular case, but your first table does not show that it is 2/3 for any of these cases. If you throw in mumbo-jumbo about the law of total probability, you do end up showing that the average of the probabilities for all possible combinations of doors chosen is 2/3, but that still doesn't show that it is 2/3 for any of them, only that their average is 2/3; to finally show that they are all 2/3 requires even more mumbo-jumbo. -- Coffee2theorems (talk) 19:06, 9 November 2012 (UTC)

Distinguishable goats?

If you really want to distinguish the goats then vos Savant's solution is incomplete respectively you have to determine which of the possible arrangements is occuring in her solution. For example:

behind door 1	behind door 2	behind door 3	result if staying at door #1	result if switching to the door offered
Car	Goat 1	Goat 2	Car	Goat 1 or Goat 2
Goat 1	Car	Goat 2	Goat 1	Car
Goat 1	Goat 2	Car	Goat 1	Car

If the player knew about this arrangement, and knew the two goats in advance, in case of the host showing Goat 2 he should switch, in case of the host showing Goat 1 the player should stay. Then his odds of winning the car are more than 2/3. --TotalClearance (talk) 09:53, 4 November 2012 (UTC)

Distinguishing between goats certainly complicates things. Martin Hogbin (talk) 10:46, 5 November 2012 (UTC)

Brilliant solution (6 equally likely cases)

I think TotalClearance's idea for making the conditional probability solutions simple and transparent is brilliant. Imagine the host has a preference to open door 2 or door 3, and his two preferences are equally likely. Then (given the player chose door 1) we have 6 equally likely cases:

• car is behind door 1, host prefers to open door 2, opens 2
• car is behind door 1, host prefers to open door 3, opens 3
• car is behind door 2, host prefers to open door 2, opens 3 (forced)
• car is behind door 2, host prefers to open door 3, opens 3
• car is behind door 3, host prefers to open door 2, opens 2
• car is behind door 3, host prefers to open door 3, opens 2 (forced)

As far as the player is concerned, nothing has changed. What the player gets to see, and the probabilities thereof are the same. In three of the six equally likely cases, the host actually opens door 3 (once forced, twice by choice). Of those three occasions, two times the car was behind door 2, once it was behind door 3. So the probability of winning by switching given the host opened door 3 is 2/3.

A brilliant solution, hopefully already known in the literature, and completely in line with Kraus and Wang's advice to do probability with frequencies, not relative frequencies. Better still, by counting equally likely cases out of admissible cases. Richard Gill (talk) 13:37, 18 November 2012 (UTC)

What preference is it the host has? I do not understand what it means the host has a preference to open door 2 or door 3. Seems to me something like: do you want Pepsi Cola or Coca Cola? Oh I prefer Pepsi or Coca. Nijdam (talk) 21:27, 18 November 2012 (UTC)

If the car happens to be behind door 1, the host will either open door 2 or 3. I call that his "preference". I don't know it, so for me either preference is true with probability 1/2. I imagine him having this preference (to open door 2 or door 3) independently of where the car actually his. He can follow his preference if the car is behind door 1. Otherwise, his choice is forced.

I'm trying to make the conditional solution easier for people who are not mathematicians, who didn't learn probability theory. So it doesn't matter if you Nijdam don't understand this. The question is whether someone like Martin or Gerhard understands it.

You as a mathematician understand that if I have a probability space in which some outcomes (little omega) have probability p and others have probability 2p, I am allowed to split each of the "double" outcomes into two outcomes of probability p. Now I have a bigger probability space (big Omega) on which all the old events and their probabilities are still defined, unaltered. But now I can compute probabilities by counting.

Use your imagination, man! Richard Gill (talk) 09:46, 19 November 2012 (UTC)

The six equally likely cases solution can be found in Jason Rosenhouse's (2009) book, page 54, and is attributed to Steven Krantz (1997), Techniques of Problem Solving, Publisher: American Mathematical Society, Providence. It's also in Kraus and Wang where it is attributed to Johnson-Laird, P. N., Legrenzi, P., Girotto, V., Legrenzi, M. S., & Caverni, J. P. (1999). Naive probability: A mental model theory of extensional reasoning. Psychological Review, 106, 62–88. So mathematicians recommend this little trick as a way of making probability problems more simple, and psychologists recommend this this little trick as a way of making probability problems more simple. Yet neither Martin (a simplist) nor Nijdam (a conditionalist) can make any sense of the solution at all. I can't see how to explain it any better. Sigh... Richard Gill (talk) 17:29, 22 November 2012 (UTC)

Speculation about 'The Economist' solution.

Richared, you gave your understanding of what 'The Economist' were intending in their solution:

We think of the location of the car and the choice of the host (if he has one) as being fixed, unknown. All randomness is in the player's initial choice of door. If he initially picks the door with the car (probability 1/3), then by switching he'll get a goat. If he initially picks a door with a goat (probability 2/3), then by switching he'll get the car.

My own opinion is that they were following an approach that I have proposed, which to ignore the doors completely and concentrate only on the objects behind them: car, goat A, and goat B. There is no need to make any assumptions as to which door was chosen.

As neither of us can tell what was really in the minds of the writers I suggest we both leave out our personal interpretations. Martin Hogbin (talk) 16:07, 21 November 2012 (UTC)

The illustration shows a fixed picture of car, goat, goat. What varies (with equal probabilities 1/3) is the initial choice of door of the player. I have read the original and it tells, as I recall, the same picture. All randomness in the story lies in the choice of door by the player. The location of the car is fixed. The choice of the player is random. (And we don't even talk about the choice of the host). Richard Gill (talk) 18:18, 21 November 2012 (UTC)

That may be the way that you see it but it is not the way that I do. The pictures show the car and the two goats (A and B) being in fixed positions relative to the page but I see this a diagrammatic only. The description makes no mention of doors: it refers only to the objects behind the doors. I see this solution as a vindication of my assertion that the doors are completely irrelevant and that it is only what is behind them that matters. The problem can be perfectly well described without the use of doors or any form of numbering of the objects. The player picks one object, the doors only being present to ensure that the player does not know what object they have picked, the host then removes an object from the remaining two, and the player has to choose between his original choice and the object remaining. No doors, numbers, or positions are necessary. For some reason, the doors seem to have taken on a life of their own in the minds of many solvers of this puzzle. Martin Hogbin (talk) 21:44, 21 November 2012 (UTC)

"For some reason, the doors seem to have taken on a life of their own in the minds of many solvers of this puzzle". For two very good reasons. One, because vos Savant named them. Two, because if you want to solve MHP through principled use of probability theory in which you carefully go step by step through the history leading up to the moment of the player's choice, you are forced to. People will this background will automatically solve the problem in this way. And they'll use it in the courses they teach and the books and articles they write. It is not some kind of mass-delusion, started by evil Morgan. No. It's the careful solution given by Selvin 1975b, in response to the just criticism which he got from numerous correspondents on his rather sloppy first attempt Selvin 1975a. Have you read these papers???? Richard Gill (talk) 15:47, 23 November 2012 (UTC)

I have no objection to being thorough and careful I only object to solutions which claim to be this but are not. I thought you had agreed to this on my talk page. I agree that most solutions in sources use doors but there really is no need to. There are sourced solutions that do not mention doors at all. Martin Hogbin (talk) 10:47, 24 November 2012 (UTC)

Martin, now you are imposing your personal interpretation on the article in the Economist. Read it! Bear in mind that the author comes from economics, and writes for readers from economics. MHP was placed in the mathematical economics literature (which is all about decision theory, game theory) by Nalebuff in 1987. It's a trivial exercise in game theory, which anyone from that background will understand immediately. Already Nalebuff mentioned the game theory approach to MHP (strategies of player, strategies of host). The central result is von Neumann's minimax theorem which solves two-party finite zero sum games if the parties may use randomized strategies. The contestant's optimal strategy if he knows nothing of the game organizers plus host's (and the contestant wants to get the car, the game organizers plus host want to prevent that) is to pick an initial door completely at random and then switch.

You are speculating that the writer of the Economist article thinks like you. A way of thinking which is not even discussed in published literature. Pure speculation and highly biased. I am arguing from knowledge of the context in which the article was written, and knowledge of the preceding literature. Do I have a personal bias? Sure, I have personal favourite approaches, and I have a personal bias to value diversity. Apart from that, I study the literature and use "good faith" (assumptions of intelligence, thoughtfulness) on the part of the various writers out there, in order to obtain a fair (undogmatic) panoramic view of diversity. Richard Gill (talk) 08:43, 22 November 2012 (UTC)

Richard I am not suggesting that we add my interpretation of this solution to the article. Just that we leave our own personal interpretations out of it.

I have to admit that I have based what I have said on what was shown in the article where there is no mention of doors. Do you have a copy of the article that I can read. If it clearly supports your view that the door chosen by the player is to be taken as fixed then we can, of course, say that in the article. Martin Hogbin (talk) 09:17, 22 November 2012 (UTC)

The *picture* which we have in our article shows the location of the car as fixed, and the door chosen by the player as variable. I have always supposed that our picture was derived from the diagram which was originally present in the published Economist article. And this was also the reason why I asked if we could have this proof back in the article: as representative of lots of sources which take the choice of the player as random, the actual location of the car as fixed. We have also frequently seen editors on this talk page spontaneously offer the same solution as their own original idea. It seems to be a common way ordinary people like to solve the problem. Richard Gill (talk) 15:39, 23 November 2012 (UTC)

Hm. Our Wikipedia article links to a web page of The Economist which contains the text of the original article but not the accompanying illustration. So I cannot check this now. I have always imagined that the author of the diagram in our article had copied it from the original article. Richard Gill (talk) 12:37, 22 November 2012 (UTC)

Yes, I have just had a look too. It makes me wonder of we even have the right reference. The words in the Economist section in the article bear little relationship to those in 'The Economist', and there is no diagram there either. Martin Hogbin (talk) 18:50, 22 November 2012 (UTC)

There is a reference to a diagram - a probability tree. So the actual published (paper) version of the article contains much more information than what is on internet. Can you get a photocopy from a UK library? Richard Gill (talk) 12:07, 23 November 2012 (UTC)

I will give that a go. Martin Hogbin (talk) 10:36, 24 November 2012 (UTC)

Conditional solution diagram

Should the diagram of Morgan's conditional solution have some introductory text, similar to that for the decision tree, for example: 'Diagram showing probability of every possible outcome if the player initially picks door 1'. Maybe we should also add, 'Similar diagrams exist for the cases where the player initially chooses door 2 or door 3'.

This is not Morgan's conditional solution! It is Selvin's (1975b) solution. (A few lines of completely elementary, direct calculation; absolutely routine for anyone who knows a tiny bit of probability theory).

Why consider the other cases? Keep it simple. Richard Gill (talk) 18:32, 15 November 2012 (UTC)

The diagram may be confusing to some because it contains no explicit renormalisation of the conditioned sample space. the reader sees two figures 1/6 and 1/3 and has to know how to obtain the required probability from these. This is shown but not explained in the text above. Might there be some advantage in clarity in showing the results of a large number (say 900) trials in this diagram? This would leave the more natural explanation that of the 450 trials in which the host has picked door 3, the player who switched wins 300 times. Maybe it would be better to consider just 6 trials or combinations? Martin Hogbin (talk) 16:36, 14 November 2012 (UTC)

Is there a case for another diagram showing what happens if the host does not choose evenly. Martin Hogbin (talk) 16:38, 14 November 2012 (UTC)

I don't see the problem. The present text has "Assuming the player picks door 1, the car is behind door 2 and the host opens door 3 with probability 1/3. The car is behind door 1 and the host opens door 3 with probability 1/6. These are the only possibilities given the player picks door 1 and the host opens door 3. Therefore, the conditional probability of winning by switching is (1/3)/(1/3 + 1/6), which is 2/3." Do you mean that this needs to be further explained?

Yes that is exactly what I mean. You know about these things and have automatically normalised the probabilities. The diagram is surely not aimed at the expert or even the student of probability. For them, the explanation in proper notation that you have given below is better. The diagram is aimed at the interested general reader, they will see 1/3 and 1/6 and from those figures must calculate a figure of 2/3. Surely it is easier in natural language to say something like, 'in 2 out of the 3 cases that met the conditions specified in the question the player who swaps will win'. I am not going to fight over this, I was rather hoping that those who were so adamant that the conditional solutions were presented in the article would comment. Martin Hogbin (talk) 20:00, 15 November 2012 (UTC)

Here is some such explanation: The calculation is done by using the fundamental rule P(A|B)=P(A&B)/P(B), in words: "probability of A given B is the probability that A and B occur together, divided by the probability that B occurs". In this case, A="car behind door 2" and B="host opens door 3"; and the probabilities (P) are all computed given the player's initial choice (door 1). Therefore, P(B) = P(host opens door 3) = P(host opens door 3 and car is behind door 2) + P(host opens door 3 and car is behind door 1) = 1/3 + 1/6 = 1/2 because "host opens door 3 and car is behind door 2" and "host opens door 3 car is behind door 1" are mutually exclusive events which together make up "host opens door 3".

Filling in yet more details: of course, P(host opens door 3 and car is behind door 2) = P(host opens door 3 | car is behind door 2) x P(car is behind door 2) = 1 x 1/3 = 1/3, while P(host opens door 3 and car is behind door 1) = P(host opens door 3| car is behind door 1) x P(car is behind door 1) = 1/2 x 1/3 = 1/6. Two more instances of the same fundamental rule as before, now in the form: P(A&B) = P(A) x P(B|A). Richard Gill (talk) 18:22, 15 November 2012 (UTC)

PS at the end of the section is the rather odd reference to Behrends. I read this source and expanded this sentence as follows: "Behrends (2008) concludes that 'One must consider the matter with care to see that both analyses are correct'; which is not to say that they are the same. One analysis for one question, another analysis for the the other question".

Behrends does not say that the two analyses are somehow the same. He is at pains to emphasize that they have different aims. Each achieves its own aim in an appropriate (correct) manner (according to Behrends). Richard Gill (talk) 18:37, 15 November 2012 (UTC)

PPS. I still don't see the point of later in the article rewriting this little calculation in huge scary mathematical formulas, after introducing a heap of notation. That just doesn't belong in an article on Monty Hall Problem. In an elementary text book on probability theory for university mathematics courses, after introducing notation, giving Bayes' theorem etc, you might like to show how it all works out in a simple example. The present article is an article about Monty Hall Problem. Not an article about formal probability theory. Just refer to the article on Bayes Theorem, in which MHP is used as a cute example! Richard Gill (talk) 18:49, 15 November 2012 (UTC)

I amplified the text next to the decision tree and the caption to the decision tree. Is that clear enough, now? Richard Gill (talk) 19:27, 15 November 2012 (UTC)

I would not use the word 'renormalise' it is too technical. Martin Hogbin (talk) 09:32, 16 November 2012 (UTC)

Please suggest an alternative expression. To get conditional probabilities, one declares all probability zero, for outcomes not satisfying the condition. The probabilities of the remaining outcomes no longer add to one (they add to the probability of the condition). We divide by this sum to make them add to one again. The technical word for this is renormalization. The text implicitly explains exactly what it is - multiplying everything by the same number, 1/sum, so that they remain in the same proportion, but now adds to one. (This is why odds are convenient - no need to renormalize). Richard Gill (talk) 12:27, 16 November 2012 (UTC)

It is hard to think of anything simple with probabilities, that is why I changed the pictorial diagram to show plays of the game. We can then intuitively normalise by saying that the player wins by switching 2 times out of 3.

TotalClearance had a very clever suggestion, precisely for this purpose. Let me present it in my own words. Imagine that the host always tosses a fair coin. If he needs to choose a door to open, he consults his coin (eg: heads and tails correspond to lower numbered, higher numbered door). Given the player initially chooses door 1 there are 6 = 3 x 2 equally likely cases: 3 locations for the car, times 2 outcomes for the coin toss. Make the table. Or if you like, the decision tree, but with 6, not 4, endponts. In 3 of those 6 cases, the host opens door 3. Of those 3 cases, the cars is behind door 2 twice and behind door 1 once. So 2 out of 3 times that door 3 is opened, switching to door 2 wins the car.

People don't gain any understanding from doing arithmetic with fractions. But they do understand counting. Kraus and Wang emphasize this as a succesful strategy for solving this kind of problem: from fractions to whole numbers. And counting equally likely cases is, I think, more simple and direct and intuitive than imaging 600 repetitions, say, in about 300 of which.... Small numbers are better than bug numbers. But talking about 6 repetions is not helpful. Talking about 6 equally Ikely cases is. Richard Gill (talk) 14:16, 17 November 2012 (UTC)

In the decision tree we would have to say something like considering only the half 1/2 (1/6 + 1/3) of the total cases where the host has opened door 3, of those 2/3 the player wins by switching. Not very good English or very clear. Martin Hogbin (talk) 10:00, 17 November 2012 (UTC)

I have boldly added a counting argument to the discussion of the decision tree, inspired by TotalClearance's table of six equally likely cases, and following Kraus and Wang's advice to calculate chances by plain counting, not by arithmetic with fractions. References also now added (I chose two out of the at least four citations possible). Richard Gill (talk) 15:00, 19 November 2012 (UTC)

I think the diagram is awful now, but I will leave the decisions on the conditional diagrams to those who think such things are necessary in the first place. Martin Hogbin (talk) 08:45, 23 November 2012 (UTC)

Since I put in the "six equally likely cases" explanation as a separate subsection (proof by expansion of Vos Savant's table) we could just as well remove the splitting of one third into two times one sixth from the conditional probability diagram, again. But I agree: those who think this diagram is useful are the ones who should be doing such modifications (or not). I tried to make the diagram more useful by leading it to six equally likely cases. But if this doesn't work, we shouldn't do it.

Whether or not you like conditional solutions, the question is, does the diagram help you to understand them? Does the diagram help people like Martin and Gerhard at least to understand what the conditionalists are talking about? Richard Gill (talk) 12:15, 23 November 2012 (UTC)

I presume that you are talking about Gerhard and myself when we first encountered the problem. The strange thing is that all this conditional stuff is much easier to understand that the problem itself, and far less interesting. The question that you should be asking yourself is, 'Do the conditionalists know what I (and Gerhard) are talking about?'. Speaking for myself, I found that once I had got my head round the basic puzzle, and things like why it matters that the host knows where the car is, the conditional solution was quite simply explained by considering the case where the host does not choose evenly. You can then, very pedantically (and inconsistently), insist on treating standard case in the same way.

To come back to the question of what pictorial diagrams will help explain the conditional solutions best I think you still need to ask who exactly is the audience for them. I think it is vanishingly small. At one end of the scale we have those who have just come to terms with the basic puzzle (or maybe not even done that). They are unlikely to be interested in making it any more complicated at all. At the other end of the scale we will have experts, who surely will not need the pictures. Then we have students of elementary probability. I would have thought that some standard structure with which they might be familiar (like the tree diagram) would be best for them. So, my honest opinion is that they are unnecessary, but I am aware that some people insist on having them, and that this was agreed to, so have them we must. Martin Hogbin (talk) 15:07, 23 November 2012 (UTC)

Yes that's what I meant. And yes, I agree. My guess is that the tree and the table are enough, the diagram is superfluous. And yes, to explain the solution, pedagogically it is smart to start with a biased host. Maybe we can try a rewrite on these lines in a few weeks. Richard Gill (talk) 08:23, 26 November 2012 (UTC)

I think we should wait and see what supporters of the 'conditional' solutions think before we make any major changes. As you know, my personal preference would be to treat the 'conditional' solutions in a slightly more 'academic' way with tables and formulae etc but some may consider this to reduce their prominence in the article, which we agreed not to do. Martin Hogbin (talk) 09:33, 26 November 2012 (UTC)

I should hope that the conditional solution can be made just as accessible as the unconditional solution. It can be done intuitively and without formulas and without tedious arithmetic; and therefore it should be done in that way, prominently. "Formal" probability-theory based solutions are only useful for beginning students of probability theory; and they are useful not for understanding MHP, but for learning how to do probability theory (which is their precise purpose, in the sources concerned). The important thing is that anyone who wants to know what the conditional probability solution is about, can find out. After people understand both simple solutions and conditional solutions, they can form their own opinion as to which kind of solution is most appropriate.

Changes made

I have been bold and made changes to the 'conditional' diagram to explain exactly what I mean. Comments, improvements and reversions welcome, especially from the 'conditionalists'. Martin Hogbin (talk) 10:01, 16 November 2012 (UTC)

I've been bold also, and reverted your changes. The diagram does not show six plays. Nijdam (talk) 11:31, 16 November 2012 (UTC)

By all means revert, but you have only reverted half of my changes. I changed the table to show the average result of 6 plays.

Do you see my point though? In the original, the reader sees two figures at the bottom of the unshaded part of the diagram showing 1/6 and 1/3, and from these they must deduce that the probability of winning by switching is 2/3. No doubt, normalising the conditioned sample space is second nature to people like yourself and Richard but I think the general reader will find it more natural to use whole numbers. Martin Hogbin (talk) 15:02, 16 November 2012 (UTC)

Okay, I'll have a look at it. Nijdam (talk) 20:20, 16 November 2012 (UTC)

How about adding one row in the conditional probability table just after the row with four cells giving the probabilities 1/3, 1/6, 1/6, 1/3.

Let the next row have six cells all with probability 1/6. In other words, split both of the two 1/3's.

Now we have six equally likely cases and we can just count. In 3 of the 6, door 3 is opened. Of those 3 cases, in 2 the car is behind door 2, in 1 the car is behind door 1. Richard Gill (talk) 15:18, 17 November 2012 (UTC)

I boldly made these changes. OR? Or simple arithmetic. In any case, following Kraus and Wang's advice (do probability by counting, not by arithmetic on fractions). Richard Gill (talk) 15:51, 17 November 2012 (UTC)

Not OR! The six equally likely cases solution can be found in Rosenhouse's book, page 54, and is attributed to Steven Krantz (1997), Techniques of Problem Solving, Publisher: American Mathematical Society, Providence. It's also in Krauss and Wang, 2003, and they have yet another reference to this solution. Richard Gill (talk) 13:27, 19 November 2012 (UTC)

The first sentence of the "conditional probability" section had a frequentist slant: "assuming the host chooses evenly" means, is think, that we imagine many repetitions and an "unbiased" (non-signalling) host. I added the Bayesian version: because we know nothing of how the host chooses, either choice is equally likely for us.

it's my opinion that a lot of the confusion around MHP is due to the fact that different authors have a different preferred interpretation, though often only implicit, not explicit. EG Kraus and Wang, and Morgan et al., always use a frequentist story in explaining probabilities. For subjectivists a different story has to be told. And for a subjectivist, the irrelevance of door numbers can be argued in advance, before looking at probabilities. Hence so many people's mystification about the need for a conditional solution or difference with a simple solution.

Well: this is my personal opinion; I think there is a gap in the literature on this point. Nobody (except yours truly, who therefore has a conflict of interest) has written explicitly about the matter. Anyway, I hope the subjectivist line, which I added, makes the conditional solution more easy to understand for everyone. Richard Gill (talk) 08:56, 19 November 2012 (UTC)

I agree with adding the Bayesian perspective but I think the 'conditional' solution would best be explained from a frequentist perspective showing the average result of 6 plays of the game in which the player has chosen door 1.

"The average result of 6 plays of the game" makes no sense to me. 6 plays can have all kinds of different outcomes including 6 times the same outcome and the player losing by switching. Or are you supporting the "six equally likely cases" solution? Or do you mean the approximately expected result of 600 plays?

I mean the average of many sets of six plays, but you know the words better than I do. My point is simple, that whole numbers are easier to understand than fractions. The reader has to get a 2 out of 3 chance from the diagram somehow. You know how to do that and so do I but it may not be so obvious to everyone. Martin Hogbin (talk) 17:57, 19 November 2012 (UTC)

The diagram shows that there are two ways to arrive at the situation with door 1 chosen and door 3 opened: one, having probability one sixth, with the car behind door 1, the other, having probability one third, with the car behind door 2. One third is twice one sixth: given door 1 chosen and door 3 opened it is twice as likely that the car is behind door 2 than door 1.

If therefore you think of many, many repetitions of the game, then on average out of every six repeats there will be three with door 2 opened and three with door 3 opened. Of the latter three, the car will be behind door 1 once and behind door 2 twice.

This is exactly what the "six equally likely cases" table introduced by TotalClearance is saying. And giving the argument with simple whole numbers (indeed, just counting) instead of arithmetic with fractions. Which according to Kraus and Wang is the best way to build the right visual picture of the problem (the right one, is the one which makes the right answer obvious).

Hopefully anyone who is interested in the conditional solution will find an explanation which they can easily grasp. Richard Gill (talk) 17:43, 20 November 2012 (UTC)

I think the article has been improved significantly. Will it be possible to find somewhere Marilyn's first "answer" after 9 Sept. 1990, and before 17 Feb. 1991? Gerhardvalentin (talk) 17:56, 20 November 2012 (UTC)

I agree, huge improvement! I'm interested in how accessible you find the conditional part! Does it help people to understand the issues? And to understand what the solution *means*?

Jason Rosenhouse says in his book that MHP was discussed by MvS on September 9, 1990; December 2, 1990; February 17, 1991; and on a fourth date (unspecified). He got much of his information from her book "The Power of Logical Thinking". Maybe you can find what you are looking for in that book?

The article itself cites: Parade magazine p. 16 of issue of 9 September 1990, p. 25 of issue of 2 December 1990, p. 12 of issue of 17 February 1991, and p. 6 of issue of 26 November 2006.

In any case, if the conditionalists want to convince the simplists of the value of their solution, they had better bear in mind that most simplists are probably using probability in a very intuitive way, and therefore most likely Bayesian way. Richard Gill (talk) 14:29, 19 November 2012 (UTC)

As you know, I do not like these solutions anyway and wonder who the audience is for these diagrams. I have made some constructive suggestions but I am happy to leave the decisions on the best way to explain the 'conditional' solutions to those who consider them important. Martin Hogbin (talk) 13:32, 19 November 2012 (UTC)

In my opinion, the audience for these diagrams are the people who were perfectly happy with simple solutions but who want to find out what the fuss is all about, about conditional solutions. The aim should be that such intelligent and inquisitive readers can find out what the conditional solutions mean, so that finally they can draw their own conclusions as to which solutions they actually prefer. Richard Gill (talk) 15:04, 19 November 2012 (UTC)

My first question is, 'What fuss?', most of the fuss has been in these pages; there is much less of a dispute in the literature. If you want to explain to interested amateurs why the conditional solutions are considered necessary or desirable by some people then I would do what most sources do and consider the (variant) possibility in which the host does not choose evenly. There are very few sources that talk about the need for a conditional solution if the host is taken to choose evenly. If we start from the Bayesian (or host chooses evenly) perspective the requirement for a conditional solution is somewhat pedantic, and not even that rigorous (as discussed). Having considered the variant case you can then go in to say that mathematicians?/some people? consider the conditional solution is required even in the standard case, although there are not many references for that. Martin Hogbin (talk) 17:57, 19 November 2012 (UTC)

Rosenhouse's book seems to me to be the definitive resource on MHP at the moment and he pays a lot of attention to the matter.

Fuss or no fuss, most people trained in probability theory will "automatically" produce the conditional solution. Because it is follows automatically from following basic principles. On the other hand, most popular sources automatically produce simple solutions. Wikipedia pages on MHP will be visited just as much by students learning probability (where they will automatically meet MHP as a popular pedagogical example) as by ordinary people who heard about the puzzle from an aquaintance. The *fuss* is not important, but it is important that there are two common and differing approaches to the problem and the article has to cover both. If the article is good, it will also show that both have their merits, that the conditional solution is also easy to understand, and hopefully even that the two approaches are linked in numerous ways. Richard Gill (talk) 09:43, 21 November 2012 (UTC)

Intro to Conditional Solution: biased host

I added to the introductory text to the conditional solutions the derivation of the conditional probabilities of 1/2 and 1 corresponding to known, maximal (in either direction), host bias. So the section now begins with an elementary derivation of conditional 2/3 by consideration of the expected outcomes of 600 repetitions in the standard situation, as well as illustration that if we have certain knowledge of host bias, we get different answers. The same motivation as many writers, e.g. Rosenthal, give for the conditional solution. Richard Gill (talk) 19:49, 26 November 2012 (UTC)

The section goes on to give three more derivations of conditional 2/3: by a conditional probability diagram, by a decision tree, and by extension of Vos Savant's table to 6 equally likely cases. I propose that the conditional probability diagram is deleted - it seems superfluous among all the other graphical and tabular methods, and it takes a huge amount of space. Any opinions on this? Richard Gill (talk) 19:54, 26 November 2012 (UTC)

By all means! Delete all traces of any accessible conditional solution - it is of course merely an academic contrivance! We find your recent work here to be most excellent, although it is easily seen that the authors should perhaps read Wikipedia:Manual of Style/Mathematics.

In case there's anyone clueless enough to miss the sarcasm, I'm saying no, the diagram should not be deleted. -- Rick Block (talk) 05:08, 27 November 2012 (UTC)

Rick, you think the conditional probability diagram is more accessible than anything else? I really do wonder what "first time readers" find the most enlightening. Under the motto "less is more" the conditional probability part of the article would be more accessible (I think) if it was trimmed of superfluous fat, started with the a really good motivation of *why* a conditional solution, and clearly showcased the most accessible/appealing solution. Alternative derivations can then be relegated to a collection of "alternative approaches". (PS thanks for the tip regarding manual of style and thanks for the compliment, which I take as being seriously meant!)

The article was reorganized by Martin a few weeks ago and this left the conditional probability parts rather disjointed. There was the big diagram alongside some other solutions but no explanation of where you "see" that the answer is 2/3 in that diagram. I have tried hard to remedy this, but I don't know if I succeeded (I added a row in the table, splitting 1/3 into two times 1/6).

At the moment my feeling is that the discussion of the expected numbers of different outcomes in 600 repetitions, and the table of six equally likely cases extending vos Savant's table of three, are the most accessible derivations for people who never saw conditional probability before, can't read maths formulas, and are scared of arithmetic with fractions. If you have any familiarity with probability and conditional probability at all, then the decision tree is useful visual aid. Maybe I'm wrong. Maybe the big conditional probability diagram is a wonderful pedagogical tool for most readers. Richard Gill (talk) 15:55, 27 November 2012 (UTC)

Richard, I did not (intentionally) change the organisation of the 'conditional' section, I moved it as it was.

Rick, as you know, I do believe that the 'conditional' solutions are 'merely an academic contrivance', but the proposal was to give them equal prominence so, as far as I am concerned you can edit them in whatever way you think showcases them well and makes them accessible to your intended audience. This is what Richard was trying to do. I asked Richard to think about the intended audience for this section and would make the same suggestion to you. Martin Hogbin (talk) 23:45, 28 November 2012 (UTC)

The example of 600 repetitions explains the "conditional approach" that is viable with a biased host, but never does address the direct approach. Rick's mapping is pointless and incorrect. The article must stop to be kept blurry and unclear. Gerhardvalentin (talk) 21:50, 29 November 2012 (UTC)

What do you mean: the example with 600 repetitions never does address the direct approach? It tries to explain what conditional 2/3 means. The 600 repetitions are imaginary, not real. Later there is an equivalent description using 6 equally likely possibilities. Do you understand it? Richard Gill (talk) 06:42, 30 November 2012 (UTC)

What do you mean by "the direct approach"? The simple solution? 600 repetitions ilustrates the simple solution too. Richard Gill (talk) 06:44, 30 November 2012 (UTC)

The "600 repetitions", explaining all possible outcome, never belong to the section "Criticism of the simple solutions", where Rick did rearrange it. It belongs where it was, before Rick disarranged it. Gerhardvalentin (talk) 07:41, 30 November 2012 (UTC)

Rick made some big changes. The "criticism" section is now a mess. He had better revise it, too. Richard Gill (talk)

Is now cleaned up again. Richard Gill (talk) 08:11, 1 December 2012 (UTC)

The problematic choice of goats

I believe much of the confusion this problem has generated has been caused by an incorrect choice of animal. If the goats are replaced by Schrödinger's cats, everything becomes much clearer. Rumiton (talk) 07:14, 28 November 2012 (UTC)

Your suggestion is too late, a quantum version of the puzzle has already been proposed. Martin Hogbin (talk) 09:36, 28 November 2012 (UTC)

Gosh, what are the chances of that? Rumiton (talk) 10:32, 28 November 2012 (UTC)

Several different quantum versions. Ours (me and my friends') is the best (d'Ariano et al). Richard Gill (talk) 08:16, 1 December 2012 (UTC)

Source of confusion

The Sources of confusion section has this confusing sentence: "However, if a player believes that sticking and switching are equally successful and therefore randomizes their strategy, they should, in fact, win 50% of the time, reinforcing their original belief." It could be just due to English not being my native language, but I read that as going against the entire point of the paradox: that the player somehow suddenly has the intuitive 50/50 odds of winning at their last choice, should they simply choose to believe it so. I suggest that this be mended or expanded to better express whatever it is that it's supposed to mean. -Uusijani (talk) 10:56, 28 November 2012 (UTC)

You are right, Uusijani. The article is a mess, never clearly saying what it purports to say. It explicitly should say that (50% x 1/3) + (50% x 2/3) gives "1/2". And in presenting the paradox it does not clearly say what Marilyn vos Savant herself did say concerning the correct "scenario" where the paradox arises. Gerhardvalentin (talk) 12:16, 28 November 2012 (UTC)

It seems clear to me. If a player believes that there is no advantage in switching, and therefore randomises their strategy (in other words sticks half the time and switches half the time) they will win the car half the time. This will (illogically) reinforce their belief that it makes no difference whether you stick or switch. In this respect pigeons do better than humans.Martin Hogbin (talk) 23:31, 28 November 2012 (UTC)

The last sentence in the "Sources of Confusion" appears to be original research (it doesn't seem to have a reference), and it makes no sense. For example, if we had a game where switching would ALWAYS lead to the prize and not-switching would NEVER lead to a prize, and if someone believed it made no difference, so they randomly switch or not switch on a series of trials, they would win 50% of the time, but why would they not notice that they win 100% of the time when they switch and 0% when they don't? That simply makes no sense. That experience in no way confirms their prior belief that switching makes no difference. Quite the opposite, that sequence of trials shows that switching makes ALL the difference. So I don't believe this silly idea is a source of confusion. If someone can find a reference for that daft idea, then I guess it could be left in the article (with the reference added please), but if no one can find a reference, it should be deleted as original (not to mention dumb) research.My3scents (talk) 16:38, 3 December 2012 (UTC)

Intro to conditional solution

Some days ago, I wrote an intro to the conditional solution, motivating it (as many authorities do) by showing that if the host has a preference, then the conditional probability answer is different from 2/3. This intro has now got spit over two distant sections, I think by Rick. Please Rick do things properly or not at all. The "criticism" section is now a mess. Richard Gill (talk) 19:15, 30 November 2012 (UTC)

The previous intro brought up the issue of biased host behavior - which is distinctly unnecessary to motivate the conditional solution. Most sources presenting conditional solutions do not (directly) bring this up. IMO, the biased host scenario is directly related to the criticism of the simple solutions - since these solutions produce the same "2/3 chance of winning by switching" result regardless of any host bias. The sources that are critical use the biased host variant to justify their criticism. -- Rick Block (talk) 01:10, 1 December 2012 (UTC)

Many sources would disagree with you, Rick. Rosenhouse and Rosenthal use the host bias variant to explain and motivate the conditional solution. This forces people to realise that by ignoring door numbers they are implicitly assuming that the host's choice gives no information to the player.

Traditional probability textbooks don't bother to motivate their approach, since in that context, one automatically figures out the conditional solution, because one works from first principles. First step: write down a probability model for the history of events leading up to the moment the player is asked whether he wants to switch. Second step: write down the probabilities of possible locations of the car given what the player has seen so far, using the definition of conditional probability. No need to be clever or creative. Just follow the rules and blindly (but carefully) calculate.

Why start half way through the problem? Martin Hogbin (talk) 23:32, 1 December 2012 (UTC)

Writers of probability texts start where they find it convenient and natural and sensible to start. It seems to me a very natural choice to start with (1) the car is hidden, (2) the player picks a door, (3) the host opens a door, (4) the player picks. Anyway, that's what everyone does, and it works fine. Next, since we will assume the player's choice is independent of the location of the car and later we want to condition on it anyway, we might as well consider the player's choice fixed. So we now have: (1) the car is hidden, (2) player picks door 1, (3) host opens a door, (4) player decides what to do. There is no point in making the story more complicated. We can't make it less complicated.

Having decided the sequence of steps leading up to the player's choice and specifying the probability of each outcome of each step given the preceding history, we are in business to routinely calculate what we want to know: the probability the car is behind any particular door given everything that has happened up to the moment the player must decide. Richard Gill (talk) 08:06, 2 December 2012 (UTC)

This may be what some sources do but it is incomplete. We are given a game in which the player can pick any of three doors. As we are given no information on how this is done we should take is as uniform at random. Does it matter how this initial choice is made? Of course it does. As you know, if the choice is made uniformly then the player's door choice and the location of the car are irrelevant. If the choice were not uniform, we would have to take account of these other two distributions (which, as it happens, we also take as uniform). So for the game, in general, the unconditional game, we must consider how the player initially chooses.

This is what many sources do, it is completely standard and seems to me completely reasonable too. If you are using probability in the Bayesian sense to measure the beliefs of the player at the moment when he must make his decision, then you are interested in the probability that the car is behind door 2 given the player chose door 1 and the host opened door 3. How you, the player, chose your initial door is irrelevant. Richard Gill (talk) 14:26, 2 December 2012 (UTC)

Some people (somewhat perversely) consider there to be two conditions implied by the question. These two conditions are: that the player initially chooses door 1; and that the host then opens door 3. The routine way to tackle this problem is to start with the sample space of the unconditional game and then to condition it according to the conditions that we wish to apply. To start half way through the problem with a hand-waving argument about starting where convenient is not acceptable if we want to be thorough. Martin Hogbin (talk) 10:14, 2 December 2012 (UTC)

This is not perverse, it is eminently sensible. If you want to decide rationally on a course of action you need to know the probabilities of the different possible states of the world, given your information at the moment the decision is needed.

The perverseness is in assuming that the problem intends to specify door numbers just because it mentions them as examples. We know it was not intended that way by either Selvin or vS. Martin Hogbin (talk) 19:12, 2 December 2012 (UTC)

What Selvin or vos Savant intended is irrelevant - what's relevant is the literature which grew from their problems. It's also not perverse to introduce door numbers. Introducing door numbers is an eminently sensible way to solve the problem in a principled way. That's why all the probability texts do it that way. That's why Selvin wrote his second paper (because his first was sloppy, incomplete, and many people wrote to him and told him so). In his second paper he writes that his solution is based on the assumption that the host's choice is random. If you rely on this assumption, you need to use it, or your logic is wrong. He implicitly admitted that his first solution was wrong, by his own criteria. Richard Gill (talk) 07:26, 3 December 2012 (UTC)

There was no handwaving in my description of the solution. I told you explicitly that we assume that the choice of the player is independent of the location of the car and therefore that we can, and shall, condition on the choice of the player, since anyway we are only interested in the probability of possible locations of the car at the moment the player has to make his decision and given the information which he then possesses. Absolutely sensible, absolutely normal. Richard Gill (talk) 14:26, 2 December 2012 (UTC)

Not exactly clear or simple. Why not start with all the possibilities and the exclude those that do not meet the specified conditions, just as you do with the host's door choice. Martin Hogbin (talk) 19:12, 2 December 2012 (UTC)

Clear and simple if you've learnt some formal probability theory. Why not? Because it would be a complete waste of time. Conditioning on the choice of the host is not completely trivial. Conditioning on the choice of the player is completely trivial. Why? Because the choice of the host is dependent on the location of the car and the initial choice of the player. Some careful thinking and/or calculating is required. Richard Gill (talk) 10:31, 3 December 2012 (UTC)

PS This is the principled way to solve the problem from the perspective of probability theory. From that perspective, giving a simple solution is simply plain wrong. It only coincidentally happens to give the right numerical answer, and anyway, it does not target what the probabilist wants to know, according to his own "rule book". Note: this is nothing to do with Morgan and nothing to do with biased hosts. Selvin 1975b gave the conditional solution in response to complaints from correspondents that his initial 1975a solution was wrong (that is: the derivation was wrong). Amusingly though, he does not admit it was wrong, and in the same article reproduces the "combined doors" simple solution provided by Monty Hall himself. So he just includes the conditional solution to satisfy the nit-pickers. On the other hand he emphasizes that all his solutions use the unbiased host assumption. Richard Gill (talk) 08:14, 2 December 2012 (UTC)

Satisfying nit pickers is fine but we must make sure that we pick all the nits. Martin Hogbin (talk) 10:14, 2 December 2012 (UTC)

No. No own research. We have to survey the literature. It Martin Hogbin has problems with how modern scholars teach probability theory to their students, he had better get his objections published. If anyone takes any notice of them, then in about 10 years Wikipedia can start to take account of them too. Richard Gill (talk) 14:32, 2 December 2012 (UTC)

There is no OR her it is standard practice to start with and unconditional case and the add in the conditions but I am happy to leave these solutions to those who feel that they are necessary.

Popular sources for conditional solutions do seem to feel a strong need to motivate the approach. Especially since they have already, typically, explained some simple solutions.

Oh well. Wikipedia is not a text book. We just survey what's out there: motivation is not relevant. Richard Gill (talk) 08:08, 1 December 2012 (UTC)

Rick, I have challenged you on several occasions to give sources that say the conditional solutions are necessary in the case where the host is stated to choose evenly between legal doors. I think you may have found one, maybe two. Most other sources which give conditional solutions explain and justify them by considering the case where the host is biased or they do not mentioning the host's strategy. Martin Hogbin (talk) 23:30, 1 December 2012 (UTC)

Just about every beginner's text book in probability theory gives the conditional solution without motivation and only for the symmetric case because it is the natural principled way to solve the problem from the point of view of probability theory. See Selvin (1975b). Richard Gill (talk) 08:18, 2 December 2012 (UTC)

All taken from the same original source. Martin Hogbin (talk)

That's what you say. Everyone who has studied probability theory can solve MHP quite on their own without any need for any source, and they'll almost all solve it the same way that Selvin did in his second, 1975b, note. It's the result of routinely writing down the natural assumptions and then routinely calculating the thing you have to calculate, from first principles. Richard Gill (talk) 14:32, 2 December 2012 (UTC)

But it was a good idea to put the "300 repetitions" into the conditional probability diagram, and take out my splitting into 6 equally likely cases! Richard Gill (talk) 08:13, 1 December 2012 (UTC)

I do not think so, Why 300 and not 6 as I suggested? Martin Hogbin (talk) 23:31, 1 December 2012 (UTC)

There are two common ways to make probability calculations intuitive, both of them are based on converting fractions to whole numbers, one goes for large numbers, the other goes for small.

Large numbers: people think of probabilities in terms of what would happen in many (imaginary) repetitions. Emphasis on many. For this example, thinking of 300 or 600 repetitions is convenient. How many times would each outcome, roughly, occur? About 100 times this, about 100 times that. The story has to have large numbers to become realistic. Talking about what we would expect in 6 repetitions doesn't make sense. Anything could happen. We certainly wouldn't expect the four distinct probability 1/6 outcomes each to occur exactly once and the single probability 1/3 outcome to occur exactly twice.

Yes I know that, but 600 is not a large number it is a quite small and arbitrary number. We give the impression to our readers that if we repeated the game 600 times we would get the exact results shown, which is , of course, wrong. After a large number of plays the results would tend towards the fractions shown but there is nothing special about 600. To show a total of 600 plays with exact figures is misleading. Martin Hogbin (talk) 10:41, 2 December 2012 (UTC)

The text talks about 300 now, and says "about 100 ..." and so on. Not misleading at all, and easy to undertand. Richard Gill (talk) 14:16, 2 December 2012 (UTC)

Small numbers: when all distinct outcomes (cases) have the same probability, then probabilities are found by counting all cases favourable to the event in question. This is the "table of six" solution which TotalClearance discovered some weeks ago, and which I subsequently found in a number of notable literature references. I recall that Martin and Nijdam both thought it was completely crazy. But anyway, it is now in the article. Richard Gill (talk) 07:55, 2 December 2012 (UTC)

The table of six would be a good idea if it were indeed a table of six equally likely possibilities, but it is not. It shows exactly the same event twice in order to make the table work.

It shows one event of probability 1/3 split in two equal parts, e.g. by imagining a fair coin toss being added to the story. This makes the table work and it is both intuitively and mathematically correct. It helps to have a bit of imagination. Richard Gill (talk) 14:16, 2 December 2012 (UTC)

If you think that adding a coin toss, the result of which is completely ignored, to the problem makes it easier to understand go ahead and add it. Martin Hogbin (talk) 19:40, 2 December 2012 (UTC)

I did already add that, seems you didn't notice. Moreover I gave a number of authoritative sources. Seems the solution is useful to some people. It does require a bit of imagination. Richard Gill (talk) 07:20, 3 December 2012 (UTC)

So both approaches are now included. Richard Gill (talk) 07:55, 2 December 2012 (UTC)

What we have is neither fish nor fowl. It does not show the results of a large number of games or an number of equally likely events. Martin Hogbin (talk) 10:41, 2 December 2012 (UTC)

Martin, you clearly haven't read the relevant section recently. It's all there, it's all carefully explained. Hopefully other people than you will find it illuminating, even if you don't. Richard Gill (talk) 14:16, 2 December 2012 (UTC)

Vos Savant, Morgan, and the media furore

I did not get much response last time I mentioned this subject but it is quite a large part of the subject and should have a section in the article. I will start one. I am not fussed about the section title. Martin Hogbin (talk) 12:38, 2 December 2012 (UTC)

I have added a section. What do you think? Do we have a useable picture of vos Savant anywhere?Martin Hogbin (talk) 14:21, 2 December 2012 (UTC)

There is Vos Savant and the media furore. A completely different affair was Morgan's fuss about vos Savant, which is already covered extensively in the article, and which (IMHO) needs to be downplayed. Richard Gill (talk) 14:35, 2 December 2012 (UTC)

But anyway - I like the new section. Richard Gill (talk) 15:26, 2 December 2012 (UTC)

No picture of Marilyn vos Savant on the Wikipedia page devoted to her. Richard Gill (talk) 15:31, 2 December 2012 (UTC)

No, I looked on the vS page too.

I agree that the Morgan argument is not really media but vS did have a fairly long and public argument with Morgan which followed on from her 'Parade' arguments. This may be of interest to our readers. I did not think the vS/Morgan argument was worth a section on its own so I put it in the vS/media section. There is quite a lot about vS/Morgan in the review literature.

The vS/Morgan section also helps explain what all the Simple/Conditional solutions are all about to the general reader who is not interested in the maths at all. I end the section by saying that Morgan consider conditional solutions to be necessary even when the host is defined to choose evenly, which is the position of the conditionalists, and you I believe. I do not think I say anything contentious or biased. Martin Hogbin (talk) 18:01, 2 December 2012 (UTC)

You misunderstand my position. The fact is that different solutions get different conclusions under different assumptions. The numbers 2/3 may be the same but what it means is different. 2/3 of what? My opinion is that it is up to the consumer to choose. The simple solutions and the conditional solutions talk about different things. The decision theoretic solutions yet again. If you want to deduce the probability of winning by switching given you chose door 1 and the host opened door 3 you have more work to do and need to make more assumptions than if you want to deduce the overall probability of winning by switching. I do not say you have to go for one or for the other. I say you legitimately can go for one or the other. The consumer does need to be correctly informed. And false logic has to be exposed.

If your solution is implicitly relying on a fair host assumption but you do not use it explicitly, then your logic is sloppy, your argument, at best, in incomplete. Thesimple solutions don't use this assumption and don't need it either. They draw a weaker conclusion than the conditional solutions. These are mathematical facts of life, like it or not. Richard Gill (talk) 07:16, 3 December 2012 (UTC)

Do you agree that the Morgan argument forms some kind of introduction to the conditional solutions? It was how things happened historically with the vS question.

I agree that the Morgan argument can be used as an intro to the conditional solutions. It's a good way to motivate the general reader, and is used by various authorities precisely for this purpose (Rosenhouse, Rosenthal, write popular accounts of the MHP story, and are anxious to motivate and to educate their readers). Historically: look at Delvin 1975a, 1975b. Devlin insisted in 1975b that his solution is based on a "fair host" assumption (or from the Bayesian perspective, from symmetry in our prior knowledge about the hosts 's behaviour). And he presented the conditionsl solution, showing where he uses this assumption. His earlier 1975a unconditional solution did not use that assumption. Clearly, he was criticized for that.

Wikipedia is not a textbook. Our task is to inform, not to educate. I agree that the conditionalists do a poor job in explaining the rationale behind their approach (both in the literature, and here on wikipedia). The preferred motivation seems to be "you have to do it that way, because that's how we mathematicians/probabilists/... do it". A very sad state of affairs. That's why I wrote my own paper on the subject. Richard Gill (talk) 10:14, 3 December 2012 (UTC)

That is the most important thing, to get the article right. Let me know if I am going too far or you think I am pushing my POV there. I want to add a bit more about what vS did get right, along the lines suggested by Gerhard. Martin Hogbin (talk) 09:40, 4 December 2012 (UTC)

Looks OK to me. But I wouldn't say "false proof", I would say "incorrect proof". And I wouldn't say "false simulation" but "inappropriate simulation". But going back to her proofs, which are not formal mathematical proofs, but informal arguments: I would not say that her arguments are incorrect, either: they are correct arguments for showing that the unconditional probability of winning by switching is 2/3. She simply shows no interest in the conditional probability of winning by switching. Richard Gill (talk) 12:19, 4 December 2012 (UTC)

Those were not my words and I agree that they a bit strong, I really do not want to start up the whole 'conditional' argument again. I would suggest the wording, '...given a certain assumption, her answer is correct; her methods of proof however, are not'. This is again pretty much a direct quote from Morgan but a little less inflammatory than Nijdam's wording. Martin Hogbin (talk) 14:07, 4 December 2012 (UTC)

Inflammatory or not, it was bad use of English. Richard Gill (talk) 06:33, 5 December 2012 (UTC)

Then let's quote Morgan verbatim. Nijdam (talk) 10:33, 5 December 2012 (UTC)

Regarding solutions which rely on the fair host assumption, I do understand your position but you do not seem to understand mine. I will try to explain on my talk page. Martin Hogbin (talk) 09:54, 3 December 2012 (UTC)

You already tried, but you did not convince me. I explained why, but you didn't accept my explanations. How about we leave it at that (agree to disagree)? Richard Gill (talk) 17:11, 3 December 2012 (UTC)

If you like. Martin Hogbin (talk) 09:40, 4 December 2012 (UTC)

Dispute resolution: The next step

(Please try to keep the conversation in this section on-topic instead of duplicating discussions already underway in other sections.)

OK, it has been a little over a month since my RfC was closed (See Talk:Monty Hall problem/Archive 33#Conditional or Simple solutions for the Monty Hall problem? for details). I would like to ask all involved to give me an unbiased assessment of where we are now. I am going to individually sign each question so you can easily place your answers below the question. --Guy Macon (talk) 17:23, 12 December 2012 (UTC)

Firstly, thanks for your help, it was appreciated. Martin Hogbin (talk) 18:53, 12 December 2012 (UTC)

Did the RfC resolve the specific issue it was written to address (the content dispute between Martin and Rick)? Has that issue been put to bed for good, or do I need to take it to the next step in content dispute resolution? --Guy Macon (talk) 17:23, 12 December 2012 (UTC)

The dispute between two ways of resolving a longstanding dispute between two groups of editors has, in my opinion, been resolved. We now start with the unencumbered simple solutions and then have the 'conditional' solutions. There is no assertion in the article that either type is 'correct' or better so both sides of the original dispute can edit cooperatively and the article can cater for all readers from those with no interest in mathematics to academic experts. Martin Hogbin (talk) 18:53, 12 December 2012 (UTC)

Are there other content disputes involving other editors that need to be addressed? Please note that if the discussion on the talk page is productive, then more talk on the talk page is the best answer. Dispute resolution is only for when the discussion has stalled. --Guy Macon (talk) 17:23, 12 December 2012 (UTC)

No significant disputes that I am aware of. Martin Hogbin (talk) 18:53, 12 December 2012 (UTC)

So far we have been focusing on article content. Are there any user behavior issues that need to be addressed? If you are not comfortable making a public comment, feel free to email me, and please include diffs. If I agree that there is a problem I will take it from there. --Guy Macon (talk) 17:23, 12 December 2012 (UTC)

I have no complaints about anyone. Martin Hogbin (talk) 18:53, 12 December 2012 (UTC)

Is there anything else that I have missed? Anything that I can do to help the editors of this article to reach consensus? Is there anything I could have done better that I might be able to correct? --Guy Macon (talk) 17:23, 12 December 2012 (UTC)

The only problem that I can see is that the supporters of the 'conditional' solutions have not shown much interest in improving that section. It seems odd to me that those who think the conditional solutions are the best ones are not interested in presenting them well. Perhaps you could try to encourage the 'conditionalists' back to the article. Martin Hogbin (talk) 18:53, 12 December 2012 (UTC)

I have very similar answers to Martin on all of these questions. I think the article is now "liberated" from the straight-jacket of the old conflict. The form which was determined by the RfC gives freedom to the content. That's a great step forward.

By the way, I think the problems we had on this article reflect a big gap in the literature on MHP. There is no "reliable source" making a decent synthesis of popular and academic approaches to MHP. The closest is Rosenhouse's book: he presents both approaches side by side but can't go beyond that. There is no reliable source giving careful, principled accounts of the various approaches which people have used to solve the problem, corresponding to different understandings of the meaning of probability. Richard Gill (talk) 22:07, 12 December 2012 (UTC)

The controversy continues

I'm not particularly happy with this title. It suggests a continuation of the controversy between vos Savant and the readers who refuse to belief her. This section, however, introduces the mathematical and logical criticism on vos Savant. Nijdam (talk) 10:57, 5 December 2012 (UTC)

I reverted uncoordinated edit. Please respect the corporate endeavor. Gerhardvalentin (talk) 13:21, 5 December 2012 (UTC)

How about the title "Controversy continues"? After all it is a different controversy, now. And how about a link to the section where this criticism is discussed at length, further down the page? Richard Gill (talk) 08:13, 6 December 2012 (UTC)

Or even, 'A new controversy'. It was not my intention to try to mislead readers into thinking that the Morgan issue was a continuation of the original disbelief in vS. Martin Hogbin (talk) 09:13, 6 December 2012 (UTC)

Call it: "A more serious controversy". Nijdam (talk) 11:06, 6 December 2012 (UTC)

Nijdam, I think your latest addition is a bit strong and would better in the 'Conditional' section of the article where it could be explained more fully. I am also not sure 'more serious' is justified, 'more academic' might be better, but I am happy with Richard's "Controversy continues". This does not imply that it is the same issue continuing. Martin Hogbin (talk) 16:51, 6 December 2012 (UTC)

The text had "and suggested a false simulation as a way of experimental verification". I can't find any statement like that in the paper of Morgan et al. So I deleted this remark.

Anyway, there is nothing wrong with the simulation. It certainly isn't "false". And it illustrates exactly the point which Marilyn's arguments prove and that she wanted to make, namely that always switching gives the car 2/3 of the time. Morgan et al. proved that there is no way to do better. In that sense they added something interesting and important to Marilyn vos Savant's analyses.

A different point is that one can use the same simulation to study the conditional probabilities. It simply suffices to study the success rates (when always switching) separated out over the possible doors opened by the host. Morgan et al. already mentioned this in their paper. Richard Gill (talk) 19:09, 6 December 2012 (UTC)

Yes, the section is meant to be about the controversies which surround the problem. I have written it is a balanced way which does not one side or the other but gives the general reader a feeling for the strong feelings that this problem seems to generate. Some people may find this interesting even if they are not interested in the maths. There is a place to address the actual point raised by Morgan and this is in the conditional section. Martin Hogbin (talk) 19:51, 6 December 2012 (UTC)

Question: what is the stupidest way to "sell" conditional solutions? Answer: how it is done in the article. I think that if you want to "sell" the conditional approach you need to present it in a positive way as adding something new and interesting to the simple solutions, complementing them and strengthening them. If you present it by saying that some professors of mathematics found some problems with vos Savant's solutions, that is the surest way to make sure that no-one will take any notice of it at all. Richard Gill (talk) 19:18, 6 December 2012 (UTC)

I agree. Some kind of introduction to conditional solutions that indicates what conditional probability is, why it is necessary in some cases, and how it could make an unexpected difference to the answer if the host is biased for example seems like a good idea to me. Rick, at least, objects to this introduction occurring before the solutions themselves because, I presume, he sees this as reducing the impact of the solutions. I think the opposite way; a good introduction makes them more appealing and accessible. Martin Hogbin (talk) 19:51, 6 December 2012 (UTC)

I have two observations on the addition of this "Controversy" section to the present article (which is in addition to the preexisting "Criticism of the simple solutions" section).

1. I find it remarkable that a section about "furore" does not clearly cite it's sources, and the subsection on "Controversy" appears to contain errors of attribution.
   • Item: "Acting purely as the agent of chance", attributed to vos Savant, looks like what Richard G. Seymann wrote, "viewed as nothing more than an agent of chance", in a comment published with the Morgan, et al. paper. Is there a different, un-cited source in which vos Savant says this?

M. vos Savant (1991), "Marilyn vos Savant's reply", Letters to the Editor, The American Statistician, vol 45, p. 347. Well worth reading! "Pure probability is the paradigm, and we published no significant reason to view the host as anything more than an agent of chance who always opens a losing door and offers the contestant the opportunity to switch, as Seymann states". Marilyn is saying that her MHP is intended to be a pure probability puzzle which is to be solved in the classical ("objective Bayesian") fashion by allocating equal probabilities to events which can be judged "equally likely" under obvious symmetries. It is not to be solved by speculations on the psychological mechanisms of door choice by a real host in a real show. Richard Gill (talk) 09:26, 7 December 2012 (UTC)

1. • Item: "Morgan et al responded by saying that ... a conditional solution was still strictly necessary", does not accord with what they wrote in their rejoinder published with Seymann's comment: "Certainly the condition p = q = 1/2 should have been put on via a randomization device at this point. It could also have been mentioned that this means that which of the unchosen doors is shown is irrelevant, which is the basis for solving the unconditional problem as a response to the conditional one." Is there a different, un-cited source in which Morgan, et al. contradict themselves saying, in a response to vos Savant, that a random agent of chance still requires a strictly "conditional" solution?

Morgan et al. answered vos Savant on the same page as her letter to the editor. In particular they say "One of the ideas put forth in our article, and one of the few that directly concerns her responses, is that even if one accepts the restrictions that she places on the reader's problem [this means: the usual disambiguation and the usual symmetric probability assumptions on location of car and host's choice of door to open] it is still a conditional probability problem". So: they insist that Whitaker's question requires a conditional probability solution, though of course they admit that it can be easily obtained from the unconditional solution plus symmetry. And we have that solution later in the article. Richard Gill (talk) 09:26, 7 December 2012 (UTC)

1. More broadly, the article appears to have taken a turn in the opposite direction from that indicated by two of the three closing statements in the recent RfC.
   • Closing statement by I Jethrobot: "Long discussions of how a limited number of sources have criticized vos Savant's approach seem unwarranted."
   • Closing statement by Churn and change: "Criticism of Savant's solution does not meet our neutral-point-of-view policy."
   • Closing statement by Eraserhead1 does not specifically address the issue of criticizing solutions.

I fear my prediction, that framing the RfC in terms of the structural ordering of solutions would not settle the problems of non-neutral point of view and undue weight, has been realized. I encourage contributors to reconsider the wisdom of playing fast and loose with the sources, and of devoting major portions of the article to discussion about what amount to quibbles over ambiguities and formalisms or, mostly, poorly contextualized observations about solving generalizations of host behaviors. ~ Ningauble (talk) 20:14, 6 December 2012 (UTC)

The RfC led to a conclusion about how the article should be structured, and that is what we are doing now. What is the status of the closing statements of a couple of "outside" editors? I have the impression that some of those editors are ill-informed on the extent of the literature, and I think that some of their arguments false. For instance: I Jethrobot talks about a limited number of sources with criticism, but the fact remains that a huge number of sources present different solutions from vos Savant and many explicitly criticise her solutions. I simply don't understand Churn and change's remark. Not reporting the criticism of vos Savant's solution would also not meet NPOV. If mathematics educationalists have produced a huge literature on MHP, is the material from that field to be excluded from a wikipedia article on MHP because "wikipedia is not a text book"?

The fact is, that there is a large literature devoted to quibbles over ambiguities and formalisms. These quibbles are discussed at length, critically, in what seem to me to be the authoritative reliable sources on MHP: Jason Rosenhouse's book, Jef Rosenthal's book chapter. What Wikipedia calls tertiary sources. (Primary equals original research articles, secondary equals research surveys concerning the latest developments written by researchers in the field, tertiary equals standard textbooks, written by reputable academic authors and published by reputable mainstream publishers). The wikipedia page surely has to report the existence of these quibbles, in a neutral way. Wikipedia can not take sides. It should surveys the disputes neutrally so that the reader can judge for themselves.

Secondly, the fact remains that almost every student who meets MHP in their introductory statistics class is going to meet the conditional solution there, presented in a matter of fact and direct (constructive) way, "look, this is how we can now solve this famous problem", usually taking it absolutely for granted that the question of whether or not the player should switch should be answered by calculating the probability of the different possible locations of the car given the information which the player has at the moment they are asked whether or not they want to switch.

So like it or not, the article has to present both popular solutions suitable for ordinary folk to discuss in casual conversation at a pub or a party, and academic solutions suitable for students of probability theory, where naturally both a higher level of precision in use of concepts, and rigour in argumentation with them, is required.

My guess is that readers of wikipedia and editors of wikipedia come roughly equally divided from these two communities. Notice that since the RfC the content of the article is little changed, what has been going on is reorganisation of the material that was already there. The editors you quote apparently don't like to read criticism of simple solutions. Well, they don't have to. It is now well separated from the solutions themselves; the two kinds of solutions are well separated as well, so people who are happy with simple solutions don't have to struggle with unfamiliar concepts or with arithmetic with fractions. Richard Gill (talk) 09:48, 7 December 2012 (UTC)

It does seem that I have accidentally sparked off a re-run of the original debate. This was most certainly not my intention, which was to show some of the issues which made the problem so infamous. I would be quite happy to delete or move some of the 'Furore' section if you think it will prevent the original argument from re-starting. The Vos Savant sections are,I think, not so contentious. The fact that thousands of people wrote in telling vS that she was wrong is a major part of the background to the problem.

To repeat, this section was intended to be an introduction to the history of the problem, which will be of more interest to some people than the mathematics, rather than a detailed discussion of the issues involved.

Some suggestions I would make are: remove the Furore section entirely, or move it to later in the article. Remove or move just the 'Controversy continues' section. Rewrite the section to have very little mathematical content and to concentrate on the media, social, and psychological issues. I would go with any of these. what do you both suggest? Martin Hogbin (talk) 09:50, 7 December 2012 (UTC)

As my new section appears to have stirred up such a hornet's nest I have deleted it pending discussion. Martin Hogbin (talk) 09:58, 7 December 2012 (UTC)

I think the first part of the section on the media furore was excellent; but the "controversy continues" subsection completely out of place. Richard Gill (talk) 10:04, 7 December 2012 (UTC)

I agree that the initial parts of the section, if more fully sourced, would make a good addition to the article. ~ Ningauble (talk) 16:04, 7 December 2012 (UTC)

Regarding several points in Richard Gill's lengthy post of 09:48, 7 December:

• The status of the entire RfC is "archived, gone, forgotten", but what relevance may be found in the closing statements may be gleaned from the larger context of the quoted statements. I, Jethrobot goes on to say "This is especially true as one editor [apparently a reference to myself] noted that much criticism of that approach has not been properly contextualized." To wit: sources that actually say vos Savant is wrong generally do so with respect to generalizations of the problem that arise from ambiguities in its informal statement. Churn and change remarks at some length that "the criticism is mostly in journals catering to undergraduates or math teachers, and are written with them in mind. These are not refutations of correctness, they are more objections from a math pedagogical point of view. Lack of math precision and lack of generalizability seem the main objections, and neither matter for the average Wikipedia reader. ... The context of WP is different from that of a journal intended for math pedagogy." I appreciate that you disagree with this point about the audience Wikipedia addresses, but it reflects the comments of many participants in the RfC.

Sorry, this is Churn and change's opinion, and moreover it is simply untrue. The fact is that the critics argue (1) vos Savant solves the wrong problem; (2) the coincidence that her answer includes the same number 2/3 does not retrospectively make her solution correct.

Important point to notice: it is very difficult to get these points across to the general reader. I think it is a major weakness of the critics that they generally give few arguments for (1), or no arguments at all. This is the challenge to editors who want to present the conditional solution, a triple challenge: motivation, explanation, deduction. Why approach the problem differently? Explanation of new concepts. Derivation of answer.

It is nothing to do with generalizability! Considering different host behaviours is merely a powerful and commonly used device to expose errors in an argument. It seems the general wikipedia reader is not familiar with common methods of analysis of alleged logical deductions. Richard Gill (talk) 08:24, 8 December 2012 (UTC)

• Yes, there are many quibbles in the literature but, to the extent they are of interest to some portion of our audience, I think it is important to contextualize them as treating variant questions arising from ambiguities, questions that do not take it as given that the host must open a door revealing a goat, and does so as an impartial agent of chance. Contextualizing them as controversy about the answer when they are generally addressing different questions is a disservice to our readers, placing undue emphasis on the contentious tone of some sources and obscuring the substance of the points they make.
• I do like it that the article has to present academic solutions suitable for students of probability theory. I believe it is entirely appropriate and beneficial for the article to explain the application of conditional probability to MHP. I did not say, nor intend to imply, anything to the contrary.

Finally, I find your closing remark about people who struggle with arithmetic, whether directed at me or at Wikipedia readers in general, singularly undignified for collegial discourse, on a par with the catty ad hominem remarks in the rejoinder of Morgan, et al. to Seymann. You have made many valuable contributions to the discussions of this article and to the academic literature, but this is the sort of thing that poisons the well. ~ Ningauble (talk) 16:09, 7 December 2012 (UTC)

My remark about arithmetic is not about anyone personally. I believe that the general reader will "switch off" as soon as calculations start being done with fractions, or when a formula is mentioned. It's a fact that if one wants to communicate probability data to ordinary people it is wise to "translate" them into statements about whole numbers (one of the main messages of Kraus and Wang). Those editors who think MHP *must* be solved by using conditional probability, presumably want this message also to reach the general reader. It follows that they should pay a lot of attention to motivation and explanation, and that they should derive the conclusion they want to get without numerical calculations but through intuitively appealing ideas. Through logic, not through computations. It's not about calculating the right answer, it's about *seeing* the right answer.

I'm not saying whether or not the general wikipedia reader can do arithmetic with fractions. I'm saying that 99% of readers will skip the calculation. It follows that these readers gain no intuition, no understanding of the solution, from the calculation. It's wasted. Richard Gill (talk) 07:31, 8 December 2012 (UTC)

The MHP seems to generate some very interesting points about the meaning of probability itself and how natural language problems should be interpreted and expressed in mathematical terms. I think it would be beneficial to add something along these lines (supported by sources of course) to the latter part of the article. Martin Hogbin (talk) 10:33, 9 December 2012 (UTC)

Agree! Richard Gill (talk) 18:36, 10 December 2012 (UTC)

What I was trying to do

Just to explain, in this section I was trying to cover the social and media aspects of the problem rather than the mathematical aspects. I agree that there was far less (if any) media attention to the argument between vS and Morgan but there certainly was a strongly worded argument between them which was well covered in the books on the subject.

I was trying to frame the argument in simple and neutral terms which did not make any assertions about who was right or even exactly what the argument was about. Just like in these pages, I suspect that vS and Morgan might have had different approaches to probability which remained unresolved.

So, my question now is, 'Should we have a section on the Morgan/vS argument before the solutions section and, if so, how should it be worded?'. Regardless of the answer to that question, I think there is a place for a more mathematical treatment of the subject later on in the article.

I would ask everyone to try to forget the old battle. My addition was not a tactical manoeuvre to get my opinion on the simple/conditional argument into the article but an attempt to move on to the more social aspects of the problem which have helped to make it so infamous. Let us work together on this. Martin Hogbin (talk) 10:14, 8 December 2012 (UTC)

My answer is a clear yes. It helps the reader to "see" the basics. It helps him to understand what the article tries to be aimed at. Gerhardvalentin (talk) 00:06, 9 December 2012 (UTC)

Gerhard, although I agree that readers should be helped to understand the basics, this still was not the intended purpose of the section. Some people have little or no interest in probability or mathematics but still may be interested in the social and psychological aspects of the problem. The most interesting thing about this problem is that, although it is in principle, very simple, the majority of people from all walks of life not only get the answer wrong but will not believe the right answer when it is explained to them. Many people find this fact interesting in itself. Martin Hogbin (talk) 10:29, 9 December 2012 (UTC)

My answer is a clear no. The media controversy is about the issue which makes MHP interesting: everyone thinks the chance of winning by switching is 1/2 (so don't) but it's 2/3 (do!). The vS/Morgan "controversy" was from the start a non-controversy, a storm in a tea-cup. Muddled ideas, mixed up contexts. Under the natural objective Bayes assumptions the difference between unconditional and conditional solutions is so tiny as to be of little interest to any non-specialist. Richard Gill (talk) 18:54, 10 December 2012 (UTC)

I could not agree more with your assessment of the mathematical importance of Morgan's paper but there was a significant social/personal aspect. The tone of Morgan's paper was highly unusual for an academic paper, I called it 'arrogant and patronizing' and Rosenhouse refers to it as 'their bellicose and condescending essay'. Vos Savant's reply was sharp and Morgan's response to that was puzzled and defensive. Surely there is a story to be told, not of the maths but of a rather strange and public argument. Martin Hogbin (talk) 00:02, 12 December 2012 (UTC)

To continue, from vos Savant's perspective, she gave the right answer, thousands of people told her she was wrong but she eventually managed to convince most of them that she was right. Then another group of people told her that she was wrong, or at least that her method of solution was wrong. Vos Savant again claimed that she was correct and and another fierce argument ensued which reached no universally accepted conclusion. Martin Hogbin (talk) 14:15, 12 December 2012 (UTC)

Yes, I agree that there is an interesting social/personal aspect which might indeed interest the general reader. Rosenhouse tells this part of the story right in the beginning of his book and implies that the mathematical community had been humiliated by the Parade affair, and that the Morgan article showed the retaliation of the mathematical community: at least they could attack her for wrong arguments, even if her answer could be admitted to be more or less correct. By the way Morgan et al's point of view in their 1991 article is that the answer "2/3" is wrong! The answer is, according to them, 1/(1+p) where p is the unknown chance that the host will open door 3 when the car is behind the door chosen by the player, door 1. Thus according to them, the chance of winning by switching is unknown! Fortunately, we do know it is at least 1/2, so switching is always right. And they have hereby shown that the overall chance of winning by switching, 2/3, can't be improved on. Richard Gill (talk) 10:03, 13 December 2012 (UTC)

Just checking

Can I just be sure that nobody objects to the first two subsections of the 'Furore' section that I added. If there is no objection from anyone I will restore them. Martin Hogbin (talk) 01:32, 8 December 2012 (UTC)

No-one has objected! Please put the furore back in. Richard Gill (talk) 18:55, 10 December 2012 (UTC)

Done. Martin Hogbin (talk) 19:07, 10 December 2012 (UTC)

How to introduce the conditional solutions

As Ningauble pointed out, Morgan et al. admitted (just as earlier discussants of their paper had already pointed out), under the standard or default assumption that either choice of the host is equally likely when he has a choice, the answer to the conditional probability question is obviously the same as the answer to the unconditional probability question. "Certainly the condition p = q = 1/2 should have been put on via a randomization device at this point. It could also have been mentioned that this means that which of the unchosen doors is shown is irrelevant, which is the basis for solving the unconditional problem as a response to the conditional one."

In other words, no difficult calculations or definitions are needed to figure out the success rate of switching on those occasions when the host opened door 3. It has to be equal to the rate when door 2 is opened, hence equal to the overall successrate, 2/3.

Why does it all have to be made so difficult, so confrontational? Richard Gill (talk) 15:32, 8 December 2012 (UTC)

I have just noticed that what I, and others, have been saying for years is said clearly in a good reliable source (M. Bhaskara Rao, my bold)

The point to make is that AGG and D3 turn out to be independent events, thus P(AGGID3) = P(AGG), and the answer to the conditional and unconditional problems is the same. This is formally demonstrated by observing that P(D3|AGG) = 1/2 by assumption and that P(D3) = P(D3|AGG)P(AGG) + P(D3|GAG)P(GAG) + P(D3|GGA)P(GGA) = (1/2)(1/3) + 1(1/3) + 0(1/3) = 1/2. If the independence of AGG and D3 is judged intuitively obvious, then the use of a solution to the unconditional problem (e.g. the solution offered by vos Savant that the authors label F2) is valid .

I have made this exact point to Richard and Nijdam several times.Martin Hogbin (talk) 22:33, 9 December 2012 (UTC)

@Martin: So? My point is, the vos Savant solution still is not valid - in spite of what's written in this article - , as a solution, because it lacks the above argumentation. I've made this point several times, and also in particular to you. (NB I've replaced sveral 1's by "!", for correctness) Nijdam (talk) 16:26, 10 December 2012 (UTC)

The statement I have said above starts, 'If the independence of AGG and D3 is judged intuitively obvious'. It is by me and, I suspect, by many other people. No comment about this fact is therefore required. It then continues, '...then the use of a solution to the unconditional problem (e.g. the solution offered by vos Savant that the authors label F2) is valid.' Note that it just refers to the unconditional solution, it does not say the unconditional solution plus a a further explanation or justification of any kind is required.

To put this more plainly, it says that the the simple solutions are correct as they are, provided that you accept that AGG and D3 are obviously independent events. We all agree that, given the usual assumptions, AGG and D3 are independent events, the only point on which we might disagree is that this fact is obvious and therefore, in common with all other obvious facts, does not need to be mentioned.

If you want to argue against this obviousness the tell me, would you expect it to be more likely that the car was behind door 1 or less likely that the car was behind door 1 had the host opened door 2 rather than door 3? Martin Hogbin (talk) 19:22, 10 December 2012 (UTC)

I have also been saying for years that the conditional solution follows from the simple solution plus symmetry. And long ago pointed out that another discussant of Morgan et al., William Bell, made the same point.

Richard, the reference does not refer to a simple solution plus symmetry it refers the the simple solutions alone, which are considered correct, provided only that the independence of AGG and D3 is judged intuitively obvious. That would be my judgement. Martin Hogbin (talk) 20:05, 10 December 2012 (UTC)

I agree: if independence is obvious then it's obvious too that the simple solution can be considered correct.

We agree on that point then. Martin Hogbin (talk) 17:26, 11 December 2012 (UTC)

First point: (1) if challenged, one has to be able to give an argument for independence. That is where symmetry comes in, it seems to me. A priori, D2 and D3 are equally likely by symmetry. Given AGG, D2 and D3 are equally likely, by symmetry. Hence P(D3)= 1/2 = P(D3|AGG). Therefore D3 and AGG are statistically independent, because the equality of P(D3) and of P(D3|AGG) is a necessary and sufficient condition for statistical independence of D3 and AGG.

Yes, we should be able to give an argument for independence if challenged. Martin Hogbin (talk) 17:26, 11 December 2012 (UTC)

Second point: (2) suppose someone just writes down the simple solution with no mention of independence at all. How can we tell whether (A) they are very smart, and realize that the specific door-numbers are irrelevant, and understand why that is the case, or (B) they are not smart at all and they haven't realized that there is a possible complication here?

I agree again and ask if there even is a clear distinction between A and B. Martin Hogbin (talk) 17:26, 11 December 2012 (UTC)

In reality there is a continuum of degrees of understanding between A and B. Our task is merely to help readers understand the issue; they have to decide for themselves whether or not it is interesting.

My guess is that most people understand this intuitively and hence simply show no interest in the probability of winning by switching given the door opened by the host. In this case, common sense agrees with the results of careful analysis. Hence the careful analysis itself becomes rather dull. Much more interesting to look at situations where common sense and careful judgement give different answers.

Absolutely. It is more interesting and instructive to consider cases where here common sense and careful judgement give different answers. Three such cases spring to mind: the car is not placed uniformly, the player's initial choice is not uniform, and the host's legal door choice is not uniform. The first two are not so trivial and have different effects. Martin Hogbin (talk) 17:26, 11 December 2012 (UTC)

The decision theoretic approach considers all three variations at the same time. If the player wants to maximize his chance of getting the car and if the host wants to minimize it, then von Neumann's minimax theorem applies. The player's optimal play is initial choice completely random and then always switch. He is guaranteed of a 2/3 chance of getting the car. The host's optimal play is hide car completely at random and open goat door completely at random, when there's a choice. He is guaranteed that the player does not have a better than 2/3 chance of getting the car. Richard Gill (talk) 19:15, 11 December 2012 (UTC)

And do we find all this particularly important? I think that depends strongly on the context. The criteria of logical completeness and logical correctness in a mathematics class are different from those in a popular magazine column. Should Wikipedia consider MHP as being only a popular brain teaser with no academic interest at all, or should Wikipedia also cover MHP in academia? I think the story of MHP in academia is also a notable story. Richard Gill (talk) 13:59, 11 December 2012 (UTC)

Marilyn was apparently not interested in a conditional probability. She was interested in giving a good reason why switching was better than staying which everyone would understand. After all, she was writing in a column of a popular weekly family magazine, not in a mathematics research journal. The simple solution gives a good reason for switching. Nijdam thinks (like Morgan and like some others, but not like everyone) that her correspondent Whitaker's question needs to be answered by giving a conditional probability. If you agree with Nijdam then vos Savant's solution is wrong: she does not give a correct derivation of the relevant probability. But very many readers of Wikipedia won't agree. Saying vos Savant was wrong will make readers angry and they'll ignore what else you write, from that point onwards.

Yes. I think the structure we have now is conducive to making the article useful for academics and the general reader. Martin Hogbin (talk) 17:26, 11 December 2012 (UTC)

Anyway, great that there is a simple derivation of the conditional probability which everyone can understand and which depends on simple and intuitive ideas, not on calculations or formulas. Now we just need a simple and convincing argument why Whitaker should be interested in the chance of winning the car by switching given the door opened by the host. I think it's obvious: by postponing his decision as late as possible, and taking account of everything he knows at that moment, he has the best chance of getting the car. There is a mathematical theorem which says exactly that.

So: if the player wants to be sure of having the very best chance of getting the car, he needs to choose the door which has the highest probability of hiding the car given his initial choice and given the door opened by the host.

If you want to sell the conditional solution you have to make it simple and make it natural and compelling. Sell it in a positive, constructive way. Not as criticism of vos Savant. But as refinement, strengthening, of her analysis.

If on the other hand you want everyone to ignore the conditional solution, then sell it by saying that some maths professors wrote a paper saying that vos Savant got it wrong, and then go on to solve the problem using formulas or arithmetic with fractions. Richard Gill (talk) 16:56, 10 December 2012 (UTC)

I have boldly added an introductory section to the conditional stuff (a) motivating the conditional approach (b) getting the answer 2/3 by using symmetry as a bridge from the simple solutions. ie building on what the reader already knows. Richard Gill (talk) 17:55, 10 December 2012 (UTC)

I think that this is an excellent ideas but others might argue that it reduces the impact of the 'conditional' solutions. I think it enhances their impact and accessibility. Martin Hogbin (talk) 19:23, 10 December 2012 (UTC)

One problem though, you correctly say, 'That information consists, in the example given by Marilyn vos Savant, of the fact that it was door 1 which was initially chosen by himself, and that it was door 3 which was opened by the host to reveal a goat' but you the proceed to consider only half of this 'information'. Martin Hogbin (talk) 19:27, 10 December 2012 (UTC)

I think I use both parts of the information! I fixed the choice of door 1 as initial door chosen by the player throughout, since I *can* do so. It's independent of the location of the car. So given the player chooses door 1, the car is still equally likely behind any of the three doors. The host's choice of door to open is made after the hiding of the car and after the player's initial choice. I take it as equally likely that the host opens door 2 or door 3 given that the player chose door 1 *and* the car is behind door 1. Richard Gill (talk) 13:32, 11 December 2012 (UTC)

I've made the text more explicit on this point: the whole discussion is already "given the initial choice of the player". Richard Gill (talk) 14:22, 11 December 2012 (UTC)

It does not seem very explicit to me. You need to show that the 'information' that the player has originally chosen door 1 does not affect the probability of winning by switching. Of course this is obvious but we are intentionally being pedantic. Martin Hogbin (talk) 16:37, 11 December 2012 (UTC)

I am not being pedantic, I am being careful and methodical. I don't have to show that the chance the car is behind door 1 given the player chose door 1 is 1/3. I assumed this. And I did that, because I'm solving the problem from the point of view of the player, using objective Bayesian principles to determine prior probabilities. The player knows nothing about where the car is located. He picks door 1. His knowledge about the location of the car is unchanged. The chance is 1/3 that it's behind door 1.

The next thing that happens is that the host opens a different door to reveal a goat. Since the player knows nothing about how the host makes a choice of door to open when he has a choice, either choice by the host (when he has one) is equally likely. Any other probabilities in this whole story have to be deduced from the ingredients which I have mentioned so far. We can now *deduce* from the prior probability distributions which have been fixed so far that AGG is independent of D3. I write out the argument using the invariance of these distributions under exchange of door numbers 2 and 3, in the text. Richard Gill (talk) 19:03, 11 December 2012 (UTC)

You say here, 'He picks door 1. His knowledge about the location of the car is unchanged. The chance is 1/3 that it's behind door 1' but you do not say that in the article. It may be obvious but it still needs to be said.

There is another point though. If the player picks any door, which we must take to be uniformly at random according to our Bayesian perspective, then we need make no assumptions about the host's door policy or the independence of AGG and D3; the probability of winning by switching is 2/3 regardless. Once we are told that the player has actually picked door 1 we need to show that AGG and D3 are independent. You cannot ignore these facts and claim to be careful and methodical. Martin Hogbin (talk) 23:48, 11 December 2012 (UTC)

Disagree. I'm taking the Bayesian perspective and I'm looking at everything from the point of view of the player. He starts off knowing nothing about the location of the car. So the car is equally likely behind any of the three doors. Now he picks a door. Say he picks door number 1 because he happens to be standing close to it at the moment. He still doesn't know anything about the location of the car so for him it is still equally likely behind any of the three doors.

You cannot just start of by saying, 'Say he picks door number 1' the player, under the rules of the game can pick any door. If all we are told is that the player has picked a door then we need not concern ourselves with the car placement or the host's door opening policy. Once we are given the additional information that the player has, in fact, chosen door 1 we must then make use of our Bayesian perspective of the producer's choice of car location and the host's choice of door.

There are two different questions: the player chooses a door and the player chooses door 1, and they have different methods of solution. To treat them as equivalent is sloppy. Martin Hogbin (talk) 13:53, 12 December 2012 (UTC)

Now I work forwards from this time. I did take care of your main concerns, in particular, I did very carefully show that the events AGG and D3 are statistically independent conditional on the player having picked door 1. (They are not independent given that he picked a different door!).

So I am being careful and methodical but I am not explaining everything at maximal length since I don't want to be pedantic and I want to concentrate on the tricky bits. I am not forced to assume that the player's choice is uniform, because I am working from the player's point of view. I want to make probability statements from his perspective and posterior to both his choice and the host's choice. In particular, posterior to the player's initial choice. I am only interested in the player's state of knowledge about the location of the car and about the opening of a goat door posterior to his initial choice. Richard Gill (talk) 08:58, 12 December 2012 (UTC)

But I did go back and edit the text some more to make it even more clear that I am working throughout given the initial choice of the player.

That is just a restatement of the problem. You might as well say 'Given the initial choice of the player and the choice of the host the answer is 2/3'. That is a correct statement but does not show how the two stated conditions affect (or in fact do not affect) the answer. You laboriously show the (non) effect of the second condition but leave the reader to work out the (non) effect of the first. As this seems to be how much of the literature treats the problem we should leave your section as it is. Martin Hogbin (talk) 09:27, 13 December 2012 (UTC)

Why does the literature do it this way? The literature is written by people who care about probability theory and who know about probability theory. For these writers, the non-effect of the first condition (how the player chooses his door) is immediate, while the non-effect of the second condition is not completely obvious. It's intuitively to be expected, sure, but it is not entirely trivial to give a rigorous argument. You call my derivation "laborious". Maybe you have a better derivation? If I would just write "obviously, by symmetry", then you would be happy but other editors will complain that "symmetry" is a much too expensive word and that they are need to see a complete formal derivation. I can do that, but sometimes the translation of a verbal argument into mathematical formulas can make the argument harder to follow, not easier. I tried to find a middle way: keep the argument verbal, but make it complete. That's what you call "laborious". Which steps can I skip? Maybe you should spend some time learning elementary probability theory, so you can begin to understand the point of view of the people whose writings you are criticising. Richard Gill (talk) 13:10, 13 December 2012 (UTC)

I am not suggesting that you skip any steps, I am suggesting that you do not skip any steps, obvious or not.

You say that the effect of the first condition is obvious but the effect of the second is not second is not. Who is to say exactly what is obvious and what is not? Are we writing only for those people who find the effect of the first condition obvious but the effect of the second not?

In any case, the effect of the first condition is not so trivial. Given just that 'the player has picked a door', it turns out that we can ignore the second condition in calculating the probability of winning by switching. Once we are told that the player has picked door 1, we must then consider the second condition. The existence of the first condition makes the second condition significant. That at fact is not obvious and should be explained here. That would be most easily shown by treating both conditions equally by starting with a sample space based on the basic game rules and then conditioning it according to the two conditions. Martin Hogbin (talk) 18:41, 13 December 2012 (UTC)

I disagree. There is no "must". There are simply a lot of choices. You keep writing " the probability of winning by switching" but all probabilities are relative. Relative to an imagined collection of repetitions. Outside of probability theory we are free to choose a "reference class". Outside of probability theory we must argue why a certain choice is most relevant. Having chosen one, we see if we can do the calculations, we see what is needed to do them. I argued for a particular reference class and then I did the calculations as carefully as I could. Having decided what I wanted to calculate, it was unnecessary to say anything about the player's initial choice, except that *after* his choice is made, the car is equally likely... , the host is equally likely.... Richard Gill (talk) 09:25, 16 December 2012 (UTC)

Richard, I have already agreed that we are not going to put what I would consider a 'complete' solution to the problem in the article. Very few sources give such a solution and no other editor seems to want to improve the article in that way. I also accept that it is possible to contrive scenarios or perspectives where the 'conditional' solution given in this article is justified.

All I am asking you to do is to understand my point, which is that, under most natural approaches to and interpretations of the problem, there are missing steps from both the simple solutions and the 'conditional' solutions given here. We may disagree about the relative obviousness of these missing steps but it seems from what you have written above that you do, at last, recognise that, in principle, there are missing steps from both solutions. If you want to discuss this further I suggest that we do so on one of our talk pages. Martin Hogbin (talk) 12:42, 16 December 2012 (UTC)

I can agree that there are missing steps in all the solutions we presently have in the article, but there is a good reason for this: no-one ever publishes anything like a "full" solution.

Just about every published "solution" is working within some traditional or conventional context of background assumptions. The author silently assumes the reader shares these background assumptions. This allows the author to jump over a lot of preliminaries and "cut to the chase". Nijdam for instance, knows for sure exactly what conditional probability Whitaker is asking for, but he can never explain to us *why*. It's become a dogma. Presumably he was taught by his teacher how you *have* to solve problems like this, and now he teaches his students to do the same.

We've already agreed that a careful solution of MHP has to include making clear what we mean by "probability" and it has to include motivation of any probability assumptions which are used in the subsequent analysis. It should also include a careful motivation of which probability we are going to calculate. You keep saying the answer is 2/3, but two thirds of what? Different solutions tell us it is 2/3 of different things. Which choice is most relevant? Why?

The mathematicians who write for students of traditional probability theory don't bother to explain things which for them are almost axiomatic. I think they have become so accustomed to the traditional approach that they don't even realise that it can be questioned. They just adopt a dogmatic stance and say "this is how you have to do it". No wonder they don't succeed in getting their point across to stubborn outsiders. Richard Gill (talk) 13:24, 16 December 2012 (UTC)

What or whose probability?

This discussion again emphasizes that if you solve MHP by probability reasoning there are a lot of choices which have to be made and not every writer makes the same choices. Many writers do not motivate their choices, but from the way they explain things one can often deduce, I think, that they have a particular perspective on probability. Jason Rosenhouse, conclusion of chapter 3 in which he introduces conditional probability and Bayes' rule, writes: "The most natural approach to the basic Monty Hall problem involves the classical view of probability. In seeking empirical verification for our solution, it was best to think in frequentist terms. And in analyzing more complex behaviours, a Bayesian view was indispensable. Three different interpretations useful in three different contexts. Exactly as it should be."

I think the article ought to have some material on this topic but unfortunately the literature is almost silent on it. Each writer has silently adopted a particular point of view and expects his readers to have the same point of view. Richard Gill (talk) 09:09, 12 December 2012 (UTC)

I agree that we should have something, but we need sources. Martin Hogbin (talk) 13:53, 12 December 2012 (UTC)

Morgan

@Richard: Reread Morgan, or buy glasses! Nijdam (talk) 21:56, 9 December 2012 (UTC)

Sorry, you were right! Morgan and friends do say that vos Savant give a false proof and illustrated it with a false simulation. However they were widely criticized for being so harsh. Seymann says "That she didn't offer a rigorous, mathematical proof in a popular Sunday supplement does her no discredit". Rosenhouse criticises Morgan et al. for altering vos Savant's wording to their advantage. Morgan et al. point out how vos Savant's simulation can be used to get information on the conditional probabilities which they find so important. Richard Gill (talk) 16:38, 10 December 2012 (UTC)

Incidently, Morgan et al. also say that Selvin's 1975b conditional solution 2/3 is false too, because it is not given that the host chooses randomly! Perhaps it's better not to quote them at all. Richard Gill (talk) 18:31, 10 December 2012 (UTC)

They don't say so, at least not in these words, in their 1991 paper. Nijdam (talk) 00:29, 12 December 2012 (UTC)

They give a list of "solutions" which they heard from students in their class or from their colleagues and they say that *all* of them are false. One of them is Selvin's (1975b) solution. Their point of view is that the right answer is not 2/3 but 1/(1+p) where p is the chance the host will open door 3 when the car is behind the door chosen by the player, door 1; it is unknown. Since 1/(1+p) is at least 1/2, whatever the value of p, the answer is switch but the probability is not 2/3. The probability is unknown, just like p, but at least we do know it supports always switching. Hence "always switching" is the optimal strategy - one cannot improve on its overall success-rate of 2/3. Richard Gill (talk) 09:48, 13 December 2012 (UTC)

Richard, where does the quotation, "Certainly the condition p = q = 1/2 should have been put on via a randomization device at this point. It could also have been mentioned that this means that which of the unchosen doors is shown is irrelevant, which is the basis for solving the unconditional problem as a response to the conditional one" come from? Martin Hogbin (talk) 10:29, 10 December 2012 (UTC)

In Morgan et al's (19991) "rejoinder" to the discussion contribution by Seymann to their original paper. The American Statistician vol. 45, p.289 Richard Gill (talk) 16:38, 10 December 2012 (UTC)

Happy New Year

Best wishes for all the discussants. As it's said: Wisdom comes with age. Nijdam (talk) 12:09, 2 January 2013 (UTC)

Correctness of vos Savant's numerical answer.

The article says, 'Vos Savant correctly responded in her column (vos Savant 1991a) that the player should switch and that the first door has a 1/3 chance of winning, but the second door offered to switch on has a 2/3 chance as the host always opens a losing door on purpose'

There are practically no sources which challenge the correctness of the answer that you have a 2/3 chance of winning by swapping. Martin Hogbin (talk) 09:14, 13 December 2012 (UTC)

The problem as 'presented' on the show is not 'complex'. Ignoring all the theoretical bull-shit about randomness, pre-existing conditions, etc., and presuming the show is not 'rigged' ( there are several who will now argue that issue ), the contestant is offered 3 doors, and picks one. I don't care if she does this based on the day of the week, her shoe size, or her perception of a sign from God. The host opens one of the other doors ( it matters not if he is forced or chooses to, as that is not within THIS problem's issue; that is a different scenario than the one presented on the show ). Before the host's reveal, the odds of winning the car are 1/3rd, and losing 2/3rds; restated, the chance the car is behind one of the 'other' two doors is 2/3rds. After the reveal, THOSE ODDS DO NOT CHANGE, and the contestant, if thinking clearly, will switch. The biggest problem is having the other contestants screaming at you as Monty's eyes twinkle at you. -- anon — Preceding unsigned comment added by 66.81.255.177 (talk) 09:13, 15 December 2012

"These odds do not change". True, assuming the host is equally likely to open either door if he has the choice. The issue, as always, is not the number, but (a) to what it refers (two thirds of what?) and (b) the correctness and completeness of the argument used to deduce it.

The required standards of correctness and completeness depend on the context. Are we in the pages of a popular family magazine or are we in a university mathematics class? Readers of Wikipedia need to be able to find the encyclopedic facts on MHP appropriate to each context, well organised so as to make the context clear, for the simple reason that the readers themself have different needs, different backgrounds. Richard Gill (talk) 07:46, 22 December 2012 (UTC)

it matters not if he is forced or chooses to: This is the widespread crucial error about MHP; accompanied by the error THOSE ODDS DO NOT CHANGE. --Albtal (talk) 18:06, 12 January 2013 (UTC)

Yes, exactly. It is true that "these odds did not change" in the clear context of the scenario intended and repeatedly described in detail by Marilyn vos Savant. These odds remain 1:2. Marilyn vos Savant made clear that the host just from the outset had intended to open one of his two unselected doors to show a goat, and to offer a switch to his second still closed door, regardless of whether the contestant coincidentally did select a goat or did select the car. Yes, the host did show a goat with intent.

And as it is just only "one" game show Marilyn vos Savant reported on, we have absolutely no knowledge about whether the host – in case he exceptionally should have disposed of two goats – would have preferred one "special" door to open and thus would have avoided to a certain extent to open his other door. As such possible bias is totally unknown to us, it is a wrong illusion to think that we could profit of such "unknown bias" in order to determine any "actually closer rate of odds" than 1:2. Never in the context Marilyn vos Savant did report on. So any of such endeavor is unnecessary from the outset, because it clearly would lead to wrong values. Any such endeavor a priori is superfluous and never constructive. Nobody is interested in inaccurate data.

For us, in the said scenario, the odds impossibly can have changed after the host has shown a goat, whichever door he might have opened. Any illusion of "a closer rate" of odds, in the given context that MvS did present, forever just must be plain wrong. For us, the odds are still 1:2, and probability to win by switching still is 2/3. Any other, any "closer" values are an illusion and inevitably are plain wrong. Never helpful for the contestant in making the required decision that he was asked for. Once more: Nobody is interested in inaccurate data. Gerhardvalentin (talk) 19:52, 22 December 2012 (UTC)

Under any reasonable interpretation of the problem, regardless of whether you are a popular magazine reader or a university professor, the odds that the car is behind the originally chosen door do not change after the reveal. Anyone who disagrees is challenged to state a reasonable interpretation of the problem that is both soluble and in which the odds of the originally chosen car change after the reveal. It is, of course, possible to contrive circumstances in which this is the case but only by proposing an inconsistent approach to the probabilities involved.

Yes. MvS did base her "simple" solution on her repeatedly specified scenario. She clearly laid out what we effectively do know and what we never will know. MvS repeatedly presented her clearly intended scenario.
Some sources however were addressing quite different scenarios, and the article actually still doesn't make it clear enough that all of such criticism ("shaky, misleading, false, ... answering a different question, ... disregarding which one of the two host's doors actually had been opened" etc.) still is addressing quite different scenarios. Scenarios of never ever given but eventually possible additional knowledge, that neither the contestant nor the audience of the one-time show presented by MvS ever will have (like Morgan et al.)
Please help the article, in reporting on "such" criticism, to clearly show that such criticism does address quite differing, quite other scenarios, but not the scenario intended and repeatedly given by MvS. Thank you. Gerhardvalentin (talk) 11:58, 31 December 2012 (UTC)

Regarding the 'correctness and completeness of the argument', please somebody show me the 'correct and complete' way to solve the problem. Martin Hogbin (talk) 23:37, 22 December 2012 (UTC)

The article has a number of correct and complete proofs (complete relative to commonly accepted standards in applied mathematics). Richard Gill (talk) 08:28, 23 December 2012 (UTC)

I'm glad you bring this up. I think the word "correctly" should be removed. It's a value judgement, an opinion, of the writer editor (Martin in this case). Moreover it apparently refers to vos Savant's complete response which gives more than just a number but also some kind of reasoning as well. The reader can read the rest of the article and judge for themselves whether or not they agree. Most people will agree that the number 2/3 occurs in a correct answer, but some will disagree with Vos Savant's reasoning. The phrase "chance of winning" is moreover ambiguous (chance of winning, given what? Whose chance, when? What is the imaginary population of repetitions?). The article goes on to discuss these subtlties, so using the word "correctly" here is inconsistent with the rest of the article. Richard Gill (talk) 09:40, 13 December 2012 (UTC)

The answer is not wrong, but it relies on some assumptions which were not clearly stated in the original letter and which do not apply to the actual Monty Hall show (it is assumed that the moderator has to open a door and that he chooses a door with equal probability). Thus it is not the correct answer. It is the correct answer for the most common formalisation of the problem, but every elaborate treatise of the problem takes other possibilities into account. See for example [2]. --Chricho ∀ (talk) 10:37, 13 December 2012 (UTC)

I am not the one calling vos Savant's answer of 2/3 'correct' all the sources say this. Even Morgan originally said, 'given certain assumptions, her answer is correct', and later said '...had we adopted conditions implicit in the problem, the answer is 2/3, period'. That is fairly definite. K&W clearly assume a 'correct' answer of 2/3 throughout their paper as does nearly every other source. Can you give any reliable source which says that the answer (probability of winning the car by switching) is not 2/3?

The article does not refer to vos Savants reasoning, only her numerical answer which is undoubtedly correct for any reasonable interpretation of the problem.

Sorry. Read the text which you composed yourself. The article does refer to reasoning, not just to a number. Vos Savant correctly responded a whole heap of things. She did not repond "2/3". Richard Gill (talk) 12:59, 13 December 2012 (UTC)

Looks like it was not clearly worded

We know what vos Savant intended the exact problem to be, we know how most people interpret it (from vos Savant and K&W) and we know how it is interpreted in all the literature. To say that the answer of 2/3 is anything other than correct is just perverse. As writers of an encyclopedia, it is not our job to let readers make up there own minds on a subjects when all the sources say one thing. Our job is to give the answer given by the sources and this is clearly 2/3. Any details can be discussed later in the article. Martin Hogbin (talk)

We know that vos Savant had no clue as to the distinction between conditional and unconditional probability. We also know that in the exact circumstances by which she meant everyone to understand the problem, the distinction is unimportant, more or less irrelevant. Richard Gill (talk) 12:59, 13 December 2012 (UTC)

Yes, we know her conception of the problem, but this is not necessarily exactly reflecting the given problem text or the show. Is the current formulation ok? (it is a little bit inprecise, since we either have to consider the over-all-probability not taking into account which door has been opened by the moderator, or assume that the moderator is unbiased when choosing a door) --Chricho ∀ (talk) 11:53, 13 December 2012 (UTC)

(1) Yes I can name a source which says that 2/3 is wrong: Morgan et al. (1991) state that the answer 2/3 is *wrong*. They state that the answer is 1/(1+p) where p is the unknown chance that the host would open door 3 if the car is behind door 1, the door opened by the host. Later they retract that somewhat, admitting that it is quite natural to assume p=1/2. They seem to have a frequentist point of view of probability, somehow *know* that the car is hidden completely at random, somehow *don't know* how the host chooses a door to open.

The Morgan scenario is indeed bizarre and cannot be called a reasonable interpretation of the problem in my opinion. They now seem to agree with me as my quote from them above shows, here it is again, '...had we adopted conditions implicit in the problem, the answer is 2/3, period' . You cannot get any clearer than that. Martin Hogbin (talk) 17:01, 13 December 2012 (UTC)

(2) I agree that almost everyone agrees that 2/3 is part of the answer, but in the text you wrote, Martin, the word "correctly" appears to refer not only to the number but also to the argumentation. The article has "Vos Savant correctly responded" and what she responded is several sentences, not just a number. Anyway, it seems to me that the word "correctly" can safely be deleted. As Cricho says, what vos Savant says is imprecise. Depending on how you interpret her meaning, it can be correct or it can be incorrect. It's not important.

What everyone does agree on is that the answer is "switch". The question asked for a decision, not for a number. Richard Gill (talk) 12:52, 13 December 2012 (UTC)

It was not my intention to suggest anything other than the that the numerical answer given by vS was correct. I have changed the text to make this clearer.

“Switch” is only the answer assuming the moderator has to open a door. And then when repeating the game you will definitely get a 2/3 performance by switching, no further assumptions needed—but in a single round when considering a bias of the moderator you have the $1/(1+p_{i})$ chance, where $i$ is the door opened by the moderator, and $p_{i}$ defined as you said above ($p_{2}+p_{3}=1$). --Chricho ∀ (talk) 14:41, 13 December 2012 (UTC)

It seems generally agreed on this talk page and also in the literature on the Monty Hall problem, that the problem is about the situation where the host knows where the car is, and so can open a different door and reveal a goat, and moreover that this is all certain to happen. The literature is not agreed on specifications of probabilities. It's typically agreed that the car is equally likely behind any of the three doors and that the choice of the player is independent of the location of the car (but some writers take the location of the car fixed, and the door chosen by the player as random). Either no explicit assumption is made about how the host opens a door, but if anything is said at all, it is usually the assumption that his choice is completely random too.

Most authors just write down some assumptions (or make some assumptions implicitly) but almost no-one tries to reason what the probabilities should be, from the problem statement. If we are an objective Bayesian then probability is about our lack of knowledge. We know nothing about the location of the car so we take its three possible locations as equally likely. We know nothing about how the host chooses a door to open when he has a choice so we take its two possible locations as equally likely. I don't know any author who writes this explicitly except for yours truly in a recent paper which I wrote, inspired by the discussions held here over many years, actually several papers, see [3].

If we are not Bayesians, then it seems to me we are stuck. We are not given any information on the basis of which we can determine the (frequentist) probabilities of the different possible initial locations of the car, nor any information on the basis of which we can determine the (frequentist) probabilities of the doors opened by the host when he has a choice. But there is a way out. We can decide ourselves to choose our own door initially completely at random. This is the basis of the decision theoretic approach. Choose completely at random and then switch is the minimax solution. Guarantees us 2/3 chance of getting the car, whatever the probability distributions used to hide the car and to open doors.

I have the impression that the infamous Morgan et al. are frequentists, but somehow seem to know in advance that the car is hidden completely at random. Their solution is mathematically interesting but from the point of view of "real life modelling" rather peculiar.Richard Gill (talk) 16:54, 13 December 2012 (UTC)

Short summary: Choose the door randomly, then switch. Assuming that the moderator has to open a door and ignoring all other information, which might be given, you get a 2/3 overall performance. This seems to be the interpretation with the simplest assumptions leading to the 2/3 value. --Chricho ∀ (talk) 01:47, 14 December 2012 (UTC)

Summary of what?? Nijdam (talk) 11:42, 14 December 2012 (UTC)

Short summary of a simple and well-founded frequentist solution. Common in the popular literature and common in the decision theoretic literature. Often spontaneously proposed by new editors.

Choose at random and switch and you'll get the car with probability 2/3. Don't even ask the question what is the chance the car is behind door 2 given you chose door 1 and the host opened door 3. Ask a stupid question, get a stupid answer: the chance is 1 if the car is behind door 2 and zero if it is behind door 1. Richard Gill (talk) 12:56, 14 December 2012 (UTC)

Well, the problem is not about strategies. The player has chosen door 1, and nothing is known about the distribution of their choice. This knowledge is anyway unimportant, as the door has already been chosen. Nijdam (talk) 08:57, 16 December 2012 (UTC)

That's editor Nijdam's personal opinion. Not all sources agree. Many think the problem is all about strategies. Wikipedia reports what sources say, whether true or not. If you want to promote a particular point of view you should publish in the literature and then wait 10 years till you are an often cited authority. Richard Gill (talk) 09:29, 16 December 2012 (UTC)

@Martin: For the so many-th time: the number 2/3 is not the solution to the problem. If you still do not understand the difference between what has to be calculated and the numerical outcome, years of discussion has passed in vain. Nijdam (talk) 13:24, 13 December 2012 (UTC)

The VALUE of overall probability to win by switching  is  2/3,  full stop. And for every distinct constellation that complies with the intended specifications of MvS, all what anyone can say is that probability to win by switching must be within the range of at least  "1/2"  (if you assume the "unknown q" to be 1) to "1" at max., if you assume the "unknown q" to be 0.

As "q" is completely unknown (0? to 1/2? to 1?) any VALUE of conditional probability to win by switching, differing from "2/3",  inevitably is plain wrong,  just from the outset.

Conditional probability is out of position to give any other actual VALUE than 2/3. For any distinct constellation: silence or 2/3. Gerhardvalentin (talk) 14:37, 13 December 2012 (UTC)

I think some people read my words in the spirit if the old conflict. I meant only to say that the numerical value of winning by switching is 2/3. I appreciate that trying to suggest any more than this at an early stage of the article is unnecessarily provocative. I hope what I meant is clear now. Martin Hogbin (talk) 16:51, 13 December 2012 (UTC) If the problem is that my 'correctly' appears to refer to vos Savant's reasoning or method of solution then I will change it the wording. The word 'correctly' was intended to refer only to the figure of 2/3, nothing else was to be implied. Can anyone suggest a way to make this clear. Martin Hogbin (talk) 16:43, 13 December 2012 (UTC)

Solution: delete the word "correctly"! This has the added advantage of maintaining some mystery for a little longer as to whether or not vos Savant was right. The reader is enticed to study the solutions. Then the reader can decide for themselves. And this is how we should do it according to official Wikipedia policy, too. Richard Gill (talk) 16:57, 13 December 2012 (UTC)

I hope what I meant is clear now. Martin Hogbin (talk) 16:51, 13 December 2012 (UTC)

As far as I know WP policy prefers giving straight answers according to reliable sources rather then trying to build up an air of mystery. Martin Hogbin (talk) 17:07, 13 December 2012 (UTC)

Indeed, WP articles are not for building up an air of mystery. However, WP articles should not give answers oversimplifying the situation. The problem is not well-posed and all elaborate sources take this into account when talking about “correctness”. @Nijdam Regarding your revert: What was your reason? “Own research?”—no, definitely not, the “standard understanding” is defined in the previous section. And assuming this “standard understanding” and assuming that we are talking about the overall performance or assuming that the host chooses the door at random, it is definitely correct that switching gives a winning probability of 2/3. --Chricho ∀ (talk) 01:47, 14 December 2012 (UTC)

Chricho, I hope you are happy with my new wording, which makes it clearer what I mean. I apologise to all those who thought that I was trying to stir up the old argument; that was not my intention. Martin Hogbin (talk) 15:09, 14 December 2012 (UTC)

The second controversy

I made a new attempt at composing a subsection on the human/sociological side of the Morgan et al controversy (or if you prefer, storm in a tea cup) immediately after the material on the media storm around the vos Savant column. Richard Gill (talk) 13:25, 14 December 2012 (UTC)

It looks good to me. Hopefully it will not start any unintended arguments. Martin Hogbin (talk) 15:40, 14 December 2012 (UTC)

It is of avail for the article and it's beneficial for the reader to show what the sources say. +1.   Gerhardvalentin (talk) 19:50, 14 December 2012 (UTC)

Removed text

Nijdam, why have you removed, 'Perhaps it was because the mathematical community had been so shamed by so many of its members getting the answer wrong (Rosenhouse, 2009)'? As you can see, a reference to a well known book on the MHP is given. Martin Hogbin (talk) 20:24, 15 December 2012 (UTC)

Exactly. Here we are reporting on the human drama, the emotions. And citing Rosenhouse's authoratitive work. ie we report Rosenhouse's opinion. Not our own. Nijdam's argument was "own opinion? + speculation". But the article reports Rosenhouse's opinion/speculation. Rosenhouse thoroughly documents Morgan et al's tactics to discredit vos Savant which he finds quite unfair, over the top. His speculation about what lies behind it is his speculation. It's part of the sociological/human story. Richard Gill (talk) 09:11, 16 December 2012 (UTC)

I put the text back, modifying it to make clear that we are reporting Rosenhouse's speculation as to the reason behind Morgan et al.'s aggressive stance and unfair tactics. Richard Gill (talk) 12:20, 16 December 2012 (UTC)

In its present formulation it is okay. Nijdam (talk) 11:24, 19 December 2012 (UTC)