Talk:Monty Hall problem/Archive 15

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FYI

Hate to spoil the conversation but thought y'all might be interested in this. hydnjo (talk) 04:16, 30 January 2010 (UTC)

Yet another source supporting Morgan's POV

See Lucas, Rosenhouse, and Schepler Lucas, Rosenhouse, and Schepler (it's the same Rosenhouse as the recent book). Published in Math. Magazine, 82(5) 332-342 (Dec 2009) - another refereed journal (this is targeted at the undergraduate level, see http://www.maa.org/pubs/mm-guide.html). Some selected quotes:

  • "The general principle here is that anything affecting Monty’s decision-making process is relevant to updating our probabilities after Monty opens his door."
  • "This [the "high numbered Monty" variant - Monty picks opens the highest numbered door without revealing the car] shows that any proposed solution to the MHP failing to pay close attention to Monty’s selection procedure is incomplete."

They reference vos Savant and Selvin, but not Morgan et al. -- Rick Block (talk) 06:04, 30 January 2010 (UTC)

I saw nothing in the Lucas and Rosenhouse paper that remotely supports Morgan. In fact, their problem description uses no door #s. I couldn't find those quotes, or references to Selvin or vos Savant. Is that the right link? Glkanter (talk) 10:44, 30 January 2010 (UTC)
The paper mainly addresses a more general form of the problem including variants where the host is known to pick non-randomly. Obviously in that case 'any proposed solution to the MHP failing to pay close attention to Monty’s selection procedure is incomplete'. The paper has little to do with this article. Martin Hogbin (talk) 12:20, 30 January 2010 (UTC)
Sorry, it was the wrong link (I've corrected it above). Like Morgan et al., this paper addresses 'Classic Monty" conditionally, even though their problem statement does not use door numbers (!), and addresses variants (including both "High-Numbered Monty" and "Random Monty") to draw attention to the fact that "any proposed solution to the MHP failing to pay close attention to Monty’s selection procedure is incomplete". And, to be clear, the way I read it this quote pertains to ANY version of the MHP - not just variants where the host is known to pick non-randomly. Again, like Morgan et al., they use these variants to show the truth of this statement. -- Rick Block (talk) 14:48, 30 January 2010 (UTC)
Fine, but you cannot use a variant to prove that a solution relating to the standard problem only is wrong. That would be exactly like using a non-right-angled triangle to prove Pythagoras wrong. Martin Hogbin (talk) 16:52, 30 January 2010 (UTC)
Yes, but if Pythagoras's logic never used the fact that the triangle is a right triangle, his proof would presumably apply to all triangles not just right triangles. We've been here before and it's really not a very productive argument. Rather than argue about this, I'll pose a question to you (and Glkanter). What is your suggestion for how, in an NPOV manner, to include the POV expressed by multiple reliable sources that unconditional solutions are missing something? In particular, I'm talking about the following (bold added):
  • Morgan et al.: Ms. vos Savant went on to defend her original claim with a false proof and also suggested a false simulation ...
  • Morgan et al.: Solution F1: If, regardless of the host's action, the player's strategy is to never switch, she will obviously will the car 1/3 of the time. Hence, the probability that she wins if she does switch is 2/3. ... F1's beauty as a false solution is that it is a true statement! It just does not solve the problem at hand.
  • Morgan et al.: Solution F2: The sample space is {AGG, GAG, GGA}, each point having probability 1/3, where the triple AGG, for instance, means the auto behind door 1, goat behind door 2, and goat behind door 3. The player choosing door 1 will win in two of these cases if she switches, hence the probability that she wins by switching is 2/3. ... That it [F2] is not a solution to the stated conditional problem is apparent in that the outcome GGA is not in the conditional sample space, since door 3 has been revealed as hiding a goat.
  • Gillman: Marilyn's solution goes like this. The chance is 1/3 that the car is actually at #1, and in that case you lose when you switch. The chance is 2/3 that the car is either at #2 (in which case the host perforce opens #3) or at #3 (in which case he perforce opens #2)-and in these cases, the host's revelation of a goat shows you how to switch and win. This is an elegant proof, but it does not address the problem posed, in which the host has shown you a goat at #3.
  • Grinstead and Snell: This very simple analysis [as a preselected strategy, staying wins with probability 1/3 while switching wins with probability 2/3], though correct, does not quite solve the problem that Craig posed. Craig asked for the conditional probability that you win if you switch, given that you have chosen door 1 and that Monty has chosen door 3. To solve this problem, we set up the problem before getting this information and then compute the conditional probability given this information.
  • Lucas, Rosenhouse, and Schepler: This [the "high numbered Monty" variant - Monty opens the highest numbered door without revealing the car] shows that any proposed solution to the MHP failing to pay close attention to Monty’s selection procedure is incomplete.
  • Rosenthal Monty Hall, Monty Fall, Monty Crawl: This solution [what he calls "Shaky Solution" which basically says your original chance of selecting the car is 1/3 and this doesn't change since you knew the host would open a door revealing a goat] is actually correct, but I consider it "shaky" because it fails for slight variants of the problem.
Let's see. So far we have 2 peer reviewed academic papers (Morgan et al., and Lucas, Rosenhouse and Schepler), a textbook (Grinstead and Snell), an article appearing in American Mathematical Monthly by a past president of the Mathematical Association of America (Gillman), and an article in Math Horizons (another publication of the Mathematical Association of America). All these sources are basically saying the same thing, which is that solutions that don't address the conditional probability of winning by switching are shaky, or incomplete, or don't quite address the problem, or are (most bluntly) false solutions. If these sources aren't sufficient for you, I could find more (but, really, 5 impeccably reliable sources ought to be enough for anyone).
To repeat, my question is how would you like, in an NPOV manner, to include the POV expressed by these sources that unconditional solutions are missing something? -- Rick Block (talk) 02:31, 31 January 2010 (UTC)

Rick, in the 15 months we have been discussing this, I find that my conclusions upon reading the identical material, Wikipedia policies or MHP sources, are almost never the same as the conclusions you reach. Nothing you cited above causes me to think those sources are capable of telling either Selvin or vos Savant (as proxy for Whitaker) what they 'really' meant, when they both made it so clear in so many ways. In Selvin's MHP, the contestant's original 1/3 can never change, and no outcomes get removed. Only with a different problem, with different premises, can Monty reveal the car, causing the contestant's 1/3 to go to 0, which he cannot do in Selvin's or vos Savant's problem. And the combining doors solution does show an open door #3 with a goat. I could go on extensively, but for what purpose? Glkanter (talk) 05:36, 31 January 2010 (UTC)

Let's take this one step at a time.
Do you agree the 5 sources I've cited above are reliable sources by Wikipedia's standards? If not, why not? -- Rick Block (talk) 05:56, 31 January 2010 (UTC)
Rather than repeat myself here my comprehensive criticism of the Morgan paper can be found here. The paper also has its own criticism published in the same journal in the form of a commentary by Seymann, and many editors here think that it is unreliable. It was, however, published in a peer-reviewed journal so we are obliged to have some reference to it in the article. It is essentially an academic diversion.
Gillman and G&S just seem to be repeating Morgan with adding anything. They should be included as references to the academic diversion that is described by Morgan.
Lucas, Rosenhouse, and Schepler show that a different problem requires a different solution. This should be in the variant section.
Rosenhouse says the simple solution is actually correct, but he considers it shaky because it fails for slight variants of the problem. This is a useful reference as it goes some way to settling some of the debate here. For the symmetrical problem an unconditional solution is correct. Once we add possible variations, such as the host is known to choose a legal door non-randomly, the unconditional solution fails. This might form the basis of a link section between two sections of the article. Martin Hogbin (talk) 12:48, 31 January 2010 (UTC)
Lucas, Rosenhouse, and Schepler start off by formulating what they consider to be the Monty Hall problem. They are teachers of elmentary university mathematics, and they are anxious to show the power of probability theory. The question they pose is not "would you switch?" but "what would you do to maximize your chances of winning the car?". They make the explicit assumption that when Monty has a choice "he chooses his door randomly". Later it becomes clear that by "randomly" they mean "completely randomly" or "at random, with equal probabilities". They later give the doors equal probabilities 1/3 to be hiding the car, with no reasoning why, except at some point for a side remark "since the doors are identical..." In my opinion these are defects to the paper. They have a tool, probability theory, and they (re)formulate the MHP in order to show off their tool. True, the paper contains a lot of useful references, history, and variants. It takes a long time to discover that they are writing about the conditional version of Monty Hall, even though initially they don't mention any door numbers. It's clear to me why they choose the conditional version: otherwise there wouldn't be much to write! They don't discuss why they have formulated the problem this way; they just make lazy and conventional choices, many of them only implicitly; so they add a whole lot more restrictions to the formulation of the problem as it became famous (in Parade magazine). I don't think the existence of their paper changes the fact that there are several points of view as to "what" the MHP is. Gill110951 (talk) 16:13, 31 January 2010 (UTC)
Agreed. Many here think the Monty Hall problem is a mathematical puzzle and as such should be formulated in a way that keeps it simple, just like the Three Prisoners Problem is. Martin Hogbin (talk) 16:37, 31 January 2010 (UTC)
I'm now reading Rosenhouse's book. It's nice, in fact much deeper than the Lucas, Rosenhouse, and Schepler (LRS) paper, but still has some strange features. On page 35, at the beginning of Chapter 2: "Classical Monty" he writes down what he calls the canonical version of the problem. As in LRS he doesn't name any doors. His first sentence is "you are shown three identical doors". A lot of pages later, after having discussed many informal solutions and talking a bit about elementary probability, he turns to presenting "the solution". At this point he numbers the doors. We are now 12 pages further on and he wants us to write down probabilities for all the possible initial configurations of (door chosen by, you, door opened by Monty, door hiding the car). For simplicity he focusses on those situations in which you have chosen door 1, so there are only four possibilities (1,2,1),(1,2,3),(1,3,2),(1,3,1). In two of those four, the car is behind door 1. At about this point point he says "it is built into the statement of the problem that the car is equally likely to be behind any of the three doors". Together with the assumption that when the quizmaster has a choice, he chooses with equal probabilities, this forces the four probabilities to be 1/6,1/3,1/3,1/6. Rosenhouse finally computes the probability that switching will give the car, given that you initially chose door 1. Thus he neither solves the unconditional problem, nor the conditional problem (he has not conditioned on the door opened by the quizmaster). Only in the penultimate chapter does he consider explicitly the unconditional problem. BTW he also acknowledges wikipedia for some of his alternative solutions. It seems to me that the authoritative literature shows that there are a lot of different opinions out there, and that it is the charm of the MHP that it allows so many formulations. BTW he has some rude things to say about the Morgan et al paper, in particular, their dogmatic style. Gill110951 (talk) 17:04, 31 January 2010 (UTC)
@Martin, where does Rosenhouse say the unconditional solution is correct, but shaky? In the situation of total symmetry (all permissable choices by all parties uniform random) it is clear that the conditional and unconditional probabilities must be the same. Prob(switching will give you the car)=Prob(switching will give you the car|Your initial choice)=Prob(switching will give you the car|Your initial choice and door opened by quizmaster; moreover (in the case of total symmetry) Prob(switching gives you car)=1-Prob(not switching gives you car)=1-1/3=2/3. (This symmetry argument was Boris Tsirelson's genial observation). There is absolutely nothing shaky about all this. Without symmetry, conditional probabilities can vary, and we can't know much about them; it is only in pedagogical articles (for students and teachers of elementary probability) by pedantic mathematicians who choose dogmatically to fill in the missing details in such a way that the pretty answer remains true, that we are told dogmatically that everything is "random" or "identical". My POV is that a sensible player chooses her initial door uniformly at random; on being asked whether or not to switch, she'll certainly switch. It doesn't matter which door she happened to choose first, and which door Monty happened to open. Her behaviour is the same. In her situation, it helps not one d*** s*** to answer the question "what is the probability the car is behind the other door, given my initial choice and the door opened by Monty?". The value of this probability depends on a load of things which she doesn't know. And her (or our) "ignorance" does not mean that suddenly various probabilities get fixed at 1/3, 1/2 etc etc. No: her ignorance means that she cannot use her conditional probability (since she has no way of knowing it) as a guide to her choice. Gill110951 (talk) 18:43, 31 January 2010 (UTC)
Martin meant Rosenthal, not Rosenhouse. A reference above he presumably accidentally deleted that I just restored. -- Rick Block (talk) 19:00, 31 January 2010 (UTC)
You are quite right Rick. Sorry for the accidental deletion. Martin Hogbin (talk) 20:02, 31 January 2010 (UTC)

Rick, I see none of those sources state that an 'unconditional solution' cannot solve the problem. That is your interpretation. They state that they don't solve the exact problem, e.g. because they don't use the right condition, or don't make the right assumptions. That's what they all say. Heptalogos (talk) 21:36, 31 January 2010 (UTC)

Right, they state unconditional solutions don't solve the exact problem. Glkanter has brought up the version as of the last FARC (claiming it is full of POV and has been improved since then). Re-reading this I'm actually not seeing the problem. In this version there are two paragraphs between an initial unconditional solution and a conditional solution that say:
The reasoning above applies to all players at the start of the game without regard to which door the host opens, specifically before the host opens a particular door and gives the player the option to switch doors (Morgan et al. 1991). This means if a large number of players randomly choose whether to stay or switch, then approximately 1/3 of those choosing to stay with the initial selection and 2/3 of those choosing to switch would win the car. This result has been verified experimentally using computer and other simulation techniques (see Simulation below).
A subtly different question is which strategy is best for an individual player after being shown a particular open door. Answering this question requires determining the conditional probability of winning by switching, given which door the host opens. This probability may differ from the overall probability of winning depending on the exact formulation of the problem (see Sources of confusion, below).
This structure is very close to how Grinstead and Snell handle this issue (they present an unconditional solution first and then say how this addresses a slightly different question). I assume these are the paragraphs that Glkanter most objects to. Is there any way we can work on these two paragraphs to satisfy everyone's goal of making the article NPOV? -- Rick Block (talk) 22:09, 31 January 2010 (UTC)
Well, I'm not sure what you mean by 'unconditional solution', but it's good to see your pragmatism by bringing in some article text. I think the first paragraph is POV. The Morgan link probably refers to solution F3, which they call wrong because it does not use the number of the door. However, if the equal goat assumption would have been made, I think Morgan would not have called this wrong. And in the article, this assumption is actually made. That's why it is IMO not correct, but a specific interpretation of Morgan. Heptalogos (talk) 22:34, 31 January 2010 (UTC)

Proposal to complete the popular solution

I only just realized that the "popular solution" just needs the one word "symmetry" added in order to turn it into a mathematically complete and rigorous solution of what we call here the conditional problem, under the further assumptions that many people (but not me) consider the canonical extra conditions. I wrote:

In order to convert this popular story into a mathematically rigorous solution, one has to argue why the probability that the car is behind door 1 does not change on opening door 2 or 3. This can be answered by an appeal to symmetry: under the complete assumptions made above, nothing is changed in the problem if we renumber the doors arbitrarily, and in particular, if we switch numbers 2 and 3. Therefore, the conditional probability that the car is behind door 1, given the player chose 1 and Monty opened 2, is the same as the conditional probability that the car is behind door 1, given the player chose 1 and Monty opened 3. The average of these two (equal) probabilities is 1/3, hence each of them separately is 1/3, too.

Is the missing word "symmetry" the reason that Rosenthal found the unconditional argument "shaky"? If you just say "opening door 3 doesn't change the chance the car is behind door 1" you are certainly being shaky. *Why* doesn't it change the probability? Intuition can so easily be wrong in probability puzzles! So let's make this step in the argument rigorous, rock-solid. Symmetry does that for you.

Mathematicians love using symmetry, since they love to make results obvious, they hate calculations; they are looking for beauty (two important sources of beauty are chance, and symmetry). I notice that non-mathematicians are often not entirely convinced by the symmetry argument. They feel tricked, suspicious. Of course, they are not used to it. Using symmetry is using a meta-mathematical argument, ie, a mathematical argument about mathematical arguments. And such arguing can be tricky, think of Gödel! Gill110951 (talk) 10:31, 1 February 2010 (UTC)

Looks good to me. I have been arguing here for years that symmetry is a valid reason to accept the unconditional solution, using much the same argument as you. It is also worth noting why Rosenthal found the unconditional solution shaky, this was only because it did not apply to slight variants (such as host chooses non-randomly). Nobody finds Pythatgoras' theorem shaky because it applies only to right angled triangles.
There is also the argument that random information is no information. That is a standard point of information theory. Again, this argument needs to be carefully made. Martin Hogbin (talk) 13:17, 1 February 2010 (UTC)
The issue is what source would you use for this? -- Rick Block (talk) 13:34, 1 February 2010 (UTC)
I won't object the symmetry argument on the ground of lacking source. And to Martin: Gill solves the CONDITIONAL problem! Nijdam (talk) 13:58, 1 February 2010 (UTC)
I am not the one who is hung up on the conditional/unconditional issue. If you just want to use the word 'conditional' that is fine with me, so long as you do not try to complicate the problem and solution by insisting that it matters which door the host opens, or say that the solution does not address the exact problem 'as asked'. Martin Hogbin (talk) 23:17, 1 February 2010 (UTC)
I am not the one who is hung up on the use of the word conditional. Never said so. But I do insist that the simple solution, and equivalently the combined doors solution, is not addressing the exact problem 'as asked'. And I want the article to make this clear to the reader. Nijdam (talk) 23:43, 1 February 2010 (UTC)
Is that not what this thread is all about? If the host chooses randomly then the simple solution does exactly answer the question as asked (after the host has opened a door). The reason - symmetry. Martin Hogbin (talk) 23:49, 1 February 2010 (UTC)
No, it doesn't.!Nijdam (talk) 17:33, 2 February 2010 (UTC)
If i understand that correctly defining symmetry in such a way basically forces the host to pick between 2 goat doors at random with p=1/2 (it seems to be an equivalent assumption). I don't mind to modify the article in such way, but i agree with Rick to the regard that we would need a source for that.--Kmhkmh (talk) 18:00, 1 February 2010 (UTC)
@Kmhkmh. It's the other way round. Suppose we want to solve the conditional problem. IF the host chooses *uniformly* at random when he has a choice, and IF the car is initially equally likely behind every door, then all the conditional probabilities Prob(car is behind door x| You chose x, Monty Hall opened y), where x,y are any two different door numbers, are the same. In particular,
Prob(car is behind 1| you chose 1, MH opened 3)
=Prob(car is behind 1| you chose 1, MH opened 2).
We have been told already that
Prob(car is behind 1|you chose 1)=1/3.
But by the law of total probability,
1/3 = Prob(car is behind 1|you chose 1) =
Prob(MH opened 2|you chose 1) x Prob(car is behind 1| you chose 1, MH opened 2)
+ Prob(MH opened 3|you chose 1) x Prob(car is behind 1| you chose 1, MH opened 3)
Denoting the unknown probability by p, we have
1/3 = Prob(MH opened 2|you chose 1) x p + Prob(MH opened 3|you chose 1) x p = p
Gill110951 (talk) 09:35, 2 February 2010 (UTC)
Would Selvin's letters satisfy the requirement? Where he says this:
"The basis to my solution is that Monty Hall knows which box contains the keys and when he can open either of two boxes without exposing the keys, he chooses between them at random." Glkanter (talk) 19:08, 1 February 2010 (UTC)
Strictly speaking "choosing at random" is not enough, since that does not define the distribution (i.e. you need p=1/2 for each goat door). However "choosing at random" is often meant to implicitly assume a uniform distribution, which is how i would read Selvin's description. Conclusion : If you are picky, there some ambiguity in Selvin's description as well.--Kmhkmh (talk) 20:39, 1 February 2010 (UTC)
That is getting overly picky. I am sure Selvin meant that the box choice was 'uniform at random'. Martin Hogbin (talk) 23:09, 1 February 2010 (UTC)

How about the following. I'm not exactly happy with the wording and somebody will have to find appropriate references, but I think this more or less captures what folks are saying.

<unconditional solution here>
The reasoning above directly addresses the average probability across all possible combinations of initial player choice and door the host opens (some reference). This means if a large number of players randomly choose whether to stay or switch, then approximately 1/3 of those choosing to stay with the initial selection and 2/3 of those choosing to switch would win the car. This result has been verified experimentally using computer and other simulation techniques (see Simulation below).
A subtly different question is which strategy is best in a specific case such as that of a player who has picked Door 1 and has then seen the host open Door 3. This difference can also be expressed as whether the player must decide to switch before the host opens a door or is allowed to decide after seeing which door the host opens (Gillman 1992). Because of the symmetry of the problem, the average probability as determined above applies to any specific case as well (this must have a reference). The probability of winning by switching in a specific case can also be determined as a conditional probability, given which doors the player picks and the host opens. Although this is the same as the average probability of winning by switching for the unambiguous problem statement as presented above, in some variations of the problem the conditional probability and average probability may be different, see Variants below.
<symmetric conditional solution here>

The idea would be to have ONE solution section, sort of like the version as of the last FARC [1] but with these two paragraphs between the unconditional and conditional solutions. -- Rick Block (talk) 04:32, 2 February 2010 (UTC)

The simple symmetrical solution does address the question 'which strategy is best in a specific case such as that of a player who has picked Door 1 and has then seen the host open Door 3?'. If the host action is random, it does not matter that the host has opened a specific door, because we know that the door opened makes no difference, by reason of symmetry. Can any one else explain this better? Martin Hogbin (talk) 09:15, 2 February 2010 (UTC)
I have tried Martin, above Gill110951 (talk) 09:35, 2 February 2010 (UTC)
To be specific: "The (simple) symmetrical solution" is just a way of calculating the CONDITIONAL probability, without using Bayes' formula. I put simple between brackets, because it is not simpler than the solution using Bayes'.Nijdam (talk) 17:33, 2 February 2010 (UTC)
On the other hand if, between the two solutions we had something like, If it is considered that the host might not choose between the two available doors randomly, the door actually opened by the host may give information which changes the probability that the player has originally chosen the car and thus it becomes important whether the player must decide to switch before the host opens a door or is allowed to decide after seeing which door the host opens (Gillman 1992) I would be happy. Martin Hogbin (talk) 09:37, 2 February 2010 (UTC)

Sources present three different kinds of solutions

Does anyone disagree that most solutions presented in sources are one of the following three types:

1) Completely unconditional, i.e chance of initially picking the car is 1/3 and a goat 2/3, and if you switch these flip.

2) Assuming the player has picked (for example) door 1, i.e. vos Savant's table:

Door 1 Door 2 Door 3 result if switching
Car Goat Goat Goat
Goat Car Goat Car
Goat Goat Car Car

3) Assuming the player has picked (for example) door 1, conditional given the host has opened (for example) door 3, e.g. any of the "conditionalists". These end up as (1/3) / (1/3 + 1/6) = 2/3.

There is clearly conflict among sources about these solutions and clearly conflict among editors about these solutions, so how about a single solution section somewhat like this:


Solution
Different sources present solutions to the problem that directly address slightly different mathematical questions using a variety of approaches.
The average probability of winning by switchingSimplest approach
This is the simplest kind of solution. The player initially has a 1/3 chance of picking the car. The host always opens a door revealing a goat, so if the player ignores what the host does and doesn't switch the player has a 1/3 chance of winning the car. Similarly, the player has a 2/3 chance of initially picking a goat and if the player switches after the host has revealed the other goat the player has a 2/3 chance of winning the car. (some appropriate reference, perhaps Grinstead and Snell)
What this solution is saying is that if 900 contestants all switch, regardless of which door they initially pick and which door the host opens about 600 would win the car.
The probability of winning by switching givenEnumeration of all cases where the player picks Door 1
If the player has picked, say, Door 1, there are three equally likely cases.
Door 1 Door 2 Door 3 result if switching
Car Goat Goat Goat
Goat Car Goat Car
Goat Goat Car Car
A player who switches ends up with a goat in only one of these cases but ends up with the car in two, so the probability of winning the car by switching is 2/3. (some appropriate reference, perhaps vos Savant)
What this solution is saying is that if 900 contestants are on the show and roughly 1/3 pick Door 1 and they all switch, of these 300 players about 200 would win the car.
The probability of winning by switching given the player picks Door 1 and the host opens Door 3
Tree showing the probability of every possible outcome if the player initially picks Door 1
This is a more complicated type of solution involving conditional probability. The difference between this approach and the previous one can be expressed as whether the player must decide to switch before the host opens a door or is allowed to decide after seeing which door the host opens (Gillman 1992).
The probabilities in all cases where the player has initially picked Door 1 can be determined by referring to the figure below or to an equivalent decision tree as shown to the right (Chun 1991; Grinstead and Snell 2006:137-138 presents an expanded tree showing all initial player picks). Given the player has picked Door 1, the player has a 1/3 chance of having selected the car. Referring to either the figure or the tree, if the host then opens Door 3, switching wins with probability 1/3 if the car is behind Door 2 but loses only with probability 1/6 if the car is behind Door 1. The sum of these probabilities is 1/2, meaning the host opens Door 3 only 1/2 of the time. The conditional probability of winning by switching for players who pick Door 1 and see the host open Door 3 is computed by dividing the total probability (1/3) by the probability of the case of interest (host opens Door 3), therefore this probability is (1/3)/(1/2)=2/3.
Although this is the same answer as the simpler solutions for the unambiguous problem statement as presented above, in some variations of the problem the conditional probability may differ from the average probability and the probability given only that the player initially picks Door 1, see Variants below. Some proponents of solutions using conditional probability consider the simpler solutions to be incomplete, since the simpler solutions do not explicitly use the constraint in the problem statement that the host must choose which door to open randomly if both hide goats (multiple references, e.g. Morgan et al., Gillman, ...).
What this type of solution is saying is that if 900 contestants are on the show and roughly 1/3 pick Door 1, of these 300 players about 150 will see the host open Door 3. If they all switch, about 100 would win the car.
A formal proof that the conditional probability of winning by switching is 2/3 is presented below, see Bayesian analysis.


Car hidden behind Door 3 Car hidden behind Door 1 Car hidden behind Door 2
Player initially picks Door 1
Player has picked Door 1 and the car is behind Door 3 Player has picked Door 1 and the car is behind it Player has picked Door 1 and the car is behind Door 2
Host must open Door 2 Host randomly opens either goat door Host must open Door 3
Host must open Door 2 if the player picks Door 1 and the car is behind Door 3 Host opens Door 2 half the time if the player picks Door 1 and the car is behind it Host opens Door 3 half the time if the player picks Door 1 and the car is behind it Host must open Door 3 if the player picks Door 1 and the car is behind Door 2
Probability 1/3 Probability 1/6 Probability 1/6 Probability 1/3
Switching wins Switching loses Switching loses Switching wins
If the host has opened Door 3, these cases have not happened If the host has opened Door 3, switching wins twice as often as staying

I'm not overly attached to any of the specific wording used, but I think presenting these as three different types of solutions and including with the last one the essence of the controversy is an NPOV approach. -- Rick Block (talk) 15:41, 2 February 2010 (UTC)

Responses

It is just the same old thing again. Nobody is interested in the average probability of winning by switching. The only question to be answered is the probability after the player has chosen a door (say door 1) and the host has opened another door to reveal a goat (say door 3). If the host chooses randomly which legal door to open then this probability is always exactly 2/3 and, by reason of symmetry, the simple table at the top of this section is a mathematically valid solution. The, so called, condition, that the host has opened a specific door is irrelevant because it can be shown that the probability of interest is independent of the door opened in the symmetrical case.

That's not quite true for 2 reasons in particular:
  • 1.) the average probability provides a reasonable heuristic (in particular if no exact knowledge regarding the influence of the conditions is given). Generally due to the lack of better information people assume the total probability (average) to hold for subsets as well (=being approximately identically with the true conditional probability) unless you have specific data for a subset telling you so or you have some very good reason to believe that this subset differs from the total set regarding a property influencing the probability.
  • 2.) Many (probably even most or all) unconditional approaches in reputable literature do make an argument based on the average (total) probability (this includes in particular vos Savant, Devlin, Henze, Behrens). and we need to summarize what the sources do and not how we might improve or extend on them as the latter would be WP:OR.
--Kmhkmh (talk) 14:09, 4 February 2010 (UTC)
If he host chooses a legal door randomly, the probability of winning by switching given the host has opened door 3 is 2/3, given the host has opened door 2 it is 2/3, and the average is also 2/3. There is no condition. Martin Hogbin (talk) 22:41, 5 February 2010 (UTC)
I don't quite see what this has to do with the 2 points above.--Kmhkmh (talk) 01:18, 6 February 2010 (UTC)
I assumed that the subsets that you wre referring to above are the sets where the host opens door 3 and the set where the host opens door 2. If this is not what you mean, per hap you could explain further. Martin Hogbin (talk) 23:21, 6 February 2010 (UTC)
Somehow it looks as if you, like others, desperately are seeking a way of reasoning, to avoid conditioning. But there is none! Try to formulate your above reasoning in formulas, and you will see. BTW. what do you mean with: "the probability of interest is independent of the door opened in the symmetrical case". Nijdam (talk) 23:25, 2 February 2010 (UTC)
Is this not the same discussion that we are having elsewhere? If the host chooses a legal door uniformly at random then you agree that the probability of interest (probability of winning by switching) is exactly 2/3 whichever door the host opens. In other words the probability of interest is independent of the door number opened by the host. I can show this more formally if you like.
Since the probability of interest is independent of the door number opened by the host, the door number opened by the host need not be a condition of the problem, by your own argument. Martin Hogbin (talk) 23:35, 2 February 2010 (UTC)

Regarding what sources say, the sources that give the simple solutions do not in any way say that their solutions apply to a different problem or that they are only average solutions. The sources that present simple solutions present them as complete solutions to the question as asked. Martin Hogbin (talk) 16:09, 2 February 2010 (UTC)

Are you denying that the simplest (fully unconditional) solutions literally address the average probability of winning by switching, or that (say) vos Savant's solution literally addresses the probability of winning by switching given the player picks Door 1? The solution vos Savant presents (for example) includes cases where the host opens Door 2, so it is clearly not directly addressing the conditional probability where the player has chosen a door (say door 1) and the host has opened another door to reveal a goat (say door 3). As far as I know, that the sources offering these solutions present them as "complete" solutions is only implied by these sources. I'm attempting to do the same thing here - they're in the "Solution" section, hence these are by implication complete solutions.
I think it's fairly clear these are the three typical sorts of solutions that are presented. If you find the words I've used to describe them POV, can you suggest a way to describe them that you don't find POV? -- Rick Block (talk) 21:24, 2 February 2010 (UTC)
I do agree that the simplest solutions address the average probability of winning by switching, it is just that nobody cares about this. Vos Savant, in common with many editors here, obviously believes that, in the symmetrical case, the door numbers are of no relevance to the problem. Thus a solution presented in which the host opens door 2 is equally valid if the host opens door 3, which of course it is, by symmetry. This especially true as it is not clear that Whitaker even intends to specify door numbers.
I am not sure what you mean by your second point. Several reliable sources present simple solutions without any form of reservation or restriction. Thus they clearly are asserting that they are answering the question as asked. Some other sources say, to varying degrees, that they are wrong. Martin Hogbin (talk) 23:26, 2 February 2010 (UTC)
The second point? About vos Savant's solution including cases where the host opens Door 2? The question asks about the case where the player has picked, say, Door 1 and the host has opened, say, Door 3. The table shows the (unconditional) probabilities where the player picks Door 1 and the host opens Door 2 and (not or) the host opens Door 3. What this table literally shows are the unconditional probabilities where the player has initially selected Door 1 - i.e. the player is not looking at either door the host has opened but is thinking about all cases. To analyze the situation after the host has opened a door without caring about the door numbers, call them A, B, and C. The player has selected Door A and the host has opened Door C while Door B remains closed. With this set up, the probability of winning where the host has opened Door B is "this case didn't happen", not 1/3. There's only one case, the player picked Door A and the host opened Door C. This is a conditional case. This is the point Nijdam keeps making that you keep ignoring. The probability of the car being behind A,B, or C is initially 1/3 for each. After the host opens Door C its probability is now 0 (no longer 1/3). What are the other door's probabilities? The question is asking what are their conditional probabilities. -- Rick Block (talk) 01:13, 3 February 2010 (UTC)
This is about where we started some years ago. In the symmetrical case, the fact that a specific door has been opened reduces the options available to the player but does not affect the probability of interest. It is therefore not a condition. I suggest that we continue this discussion in the section that I have set up on the arguments page. Martin Hogbin (talk) 09:46, 3 February 2010 (UTC)
Indeed, and you still don't get it. You're still talking about the value 2/3, where we have explained over and over it is not the value that counts, but the nature of the probability. We demonstrated that opening of a door is a condition, and still you persist in saying it is not. What is your intention? Nijdam (talk) 12:53, 3 February 2010 (UTC)
'demonstrated that opening of a door is a condition' is exactly what you have not done. You have asserted this several times and also suggested reasons why the event should be considered a condition. I have shown all these reasons to be invalid. I suggest we continue on the appropriate section on the arguments page. Martin Hogbin (talk) 13:58, 3 February 2010 (UTC)
This perhaps should go elsewhere, but Martin if I've picked Door 1 and the host has opened Door 3 what lines of the table do I look at to figure out my probability of winning by switching? The table looks wrong to me since it says the probability of the car being behind Door 3 is 1/3. It clearly isn't anymore. The probabilities inferred from this table are prior probabilities - those that are in effect before the host opens a door. To convey what you're saying it conveys, this table should really only have two lines, a Door 1 line and a "Door 2 or Door 3" line. This makes the probabilities inarguably unconditional, which I think is what you're saying - but then it's clearly showing the probability of winning given the player has picked Door 1 (which you seem to object characterizing it as). -- Rick Block (talk) 18:23, 3 February 2010 (UTC)
Thus we should, in the article, present the simple solutions in the same way that they are presented by the sources that give them, as complete solutions without reservation or restriction to the exact question that was asked. Martin Hogbin (talk) 23:37, 2 February 2010 (UTC)
How about the changes as above? Better? -- Rick Block (talk) 01:13, 3 February 2010 (UTC)
Better, but you still have, 'so if the player ignores what the host does', and 'What this solution is saying is that if 900 contestants all switch, regardless of which door they initially pick and which door the host opens about 600 would win the car'. Later on, in the conditional solution section, you say, 'This is a more complicated type of solution involving conditional probability. The difference between this approach and the previous one can be expressed as whether the player must decide to switch before the host opens a door or is allowed to decide after seeing which door the host opens (Gillman 1992)'.
My first two quotes should go completely, they are your POV concerning sources that give simple solutions, there are several sources giving a simple solution that do not say what you say.
Regarding, 'This is a more complicated type of solution involving conditional probability....'. We are now talking about the link between the sections and sources. This should therefore start with something like, 'Some sources treat the problem as requiring the use of conditional probability...' preferably including 'this is important if the host may not choose randomly which door to open, when he has a choice...' Martin Hogbin (talk) 09:46, 3 February 2010 (UTC)
I've struck "player ignores what the host does". I'm reluctant to strike the 600 out of 900 win bit. You agreed above this solution is literally addressing the average probability. Is it the numbers you object to or the "regardless of which door ..." wording?
I am not sure what you mean by average probability here. What you say is quite correct, but why mention it?. In the symmetrical case the probability of winning by switching is 2/3 if the host opens door 3, 2/3 if the host opens door 2, and 2/3 on average. Martin Hogbin (talk) 22:44, 3 February 2010 (UTC)
And the chance is also 2/3 that a random picked child of one of my neighbours is a boy. Nijdam (talk) 23:53, 5 February 2010 (UTC)
The "difference between this approach" needs to go somewhere. The point is not to say conditional probability is required, but to explain what it means. The tone I'm trying to hit is "an alternate approach", not "the correct approach". Your suggested addition about the importance is your POV, which as far as I recollect no one's ever offered up a source for (just WP:OR) - and addressing the problem using conditional probability is certainly a valid approach whether the host chooses randomly or not. -- Rick Block (talk) 18:23, 3 February 2010 (UTC)
Your last point is the one to focus on. The simple solution only applies if the host is taken to choose a legal door randomly, it fails if the host chooses non-randomly. Martin Hogbin (talk) 22:44, 3 February 2010 (UTC)
To make that argument you actually need a conditional approach.--Kmhkmh (talk) 14:13, 4 February 2010 (UTC)
Yes I agree, but you do not need the conditional approach just to show that the simple argument is valid in the symmetrical case. See my discussion with Nijdam on the arguments page. Martin Hogbin (talk) 15:06, 4 February 2010 (UTC)
No, no, no. The simple argument is never valid, i.e. for the MHP where the offer to switch is given AFTER the goat door is opened. The probability of interest is the conditional probability. It may be calculated in a lot of ways, with Bayes', with symmetry, etc., but it IS a conditional probability! Nijdam (talk) 23:49, 5 February 2010 (UTC)
This remains just an assertion on your part. We are currently discussing this subject on the arguments page. Martin Hogbin (talk) 00:44, 6 February 2010 (UTC)
I entirely agree with Martin here against Nijdam: there is nothing in the Marilyn vos Savant question saying "you must use a conditional probabilty". It is also not the case that because you are to decide *after* this and that event, you *must* decide by computing a conditional probability. I realise that if the question is posed as an examination question in Probability 101, this might be "understood" by the student eager to get full marks. But I am not interested in Probability 101 questions. I am interested in a famous question posed in a popular weekly magazine in which any technical code words like probability, equally likely, identical... did not occur. Gill110951 (talk) 05:39, 7 February 2010 (UTC)
I will get along with Gill in so far the academic discussion goes about what is meant by the MHP. But I don't follow his reasoning about the decision to be made. If an event has happened, it is almost compulsery to take this into account. In most cases it is even of vital importance, as I showed in examples. And anyone I gave the choice to decide on basis of the unconditional (average) probability, or on the specific conditional, did not hesitate to choose the conditional probability. Only because the answers i the symmetric MHP will be the same, one might have the idea it is not needed to consider the occurence. But ... to know the answers do not differ, one has to consider the conditioning on the ocurence of the event. Nijdam (talk) 21:48, 8 February 2010 (UTC)

Nice discussion. My present point-of-view is written up (for mathematicians) in http://arxiv.org/abs/1002.0651 (eprint/prepublication archive). I have also submitted it to a peer-reviewed journal. But I already see that I want to improve the symmetry solution, and to add equivalent verbal proofs next to each mathematical proof, so that ordinary people can read it too. Suggestions for improvement are welcome. Gill110951 (talk) 08:09, 4 February 2010 (UTC)

Rick wrote: "this table should really only have two lines, a Door 1 line and a "Door 2 or Door 3" line. This makes the probabilities inarguably unconditional". Rick, given that "door2 or door3" is opened, it's a conditional probability just the same? Heptalogos (talk) 21:54, 5 February 2010 (UTC)

Well, sort of, but with only two lines (the 2nd line is "car behind Door 2 or Door 3") the condition "door 2 or door 3 opened" (more) obviously does not reduce the sample space. A table with number of samples might help clarify what I'm saying:
Door 1 Door 2 Door 3 result if switching total cases cases if host opens Door 2 or Door 3
Car Goat Goat Goat 100 ?
Goat Car Goat Car 100 ?
Goat Goat Car Car 100 ?
What "in two out of three equally likely cases switching wins" means is in 200 out of 300 cases switching wins - it's not saying anything at all about the column with the question marks. The cases are only obviously equally likely before the host opens a door. Looking at this table can you fill in the ? entries based on knowing the host opens "Door 2 or Door 3"?
I'll give you a hint: focus on the row where the car is behind Door 1. In how many of these 100 cases does the host open Door 2 or Door 3?
The point is if all you know is the host opens "Door 2 or Door 3", the numbers are ALL still 100 (!). The host opens Door 2 or Door 3 in ALL 300 cases. Asking what happens after the host opens "Door 2 or Door 3" is the same thing as asking what the situation is before the host opens a door. It's the same set of cases, i.e. the same sample set. All the before and after probabilities are the same.
If we do the same thing with the 2 line table, here's what happens
Where's the car result if switching total cases cases if host opens Door 2 or Door 3
Behind Door 1 Goat 100 ?
Behind Door 2 or Door 3 Car 200 ?
The ? entries in this table are (more) clearly 100/200 (right?). Wherever the car is, the host always opens Door 2 or Door 3. The confusion about whether the ? in the row where the car is behind door 1 should be 100 or 50 pretty much goes away. This confusion is exactly what Morgan et al. are referring to when they say "The distinction between the conditional and unconditional situations here seems to confound many".
In my opinion (some WP:OR), a better table would be like this:
Door 1 Door 2 Door 3 total cases host opens Door 2 host opens Door 3
cases result if switching cases result if switching
Car Goat Goat 100 50 Goat 50 Goat
Goat Car Goat 100 0 N/A 100 Car
Goat Goat Car 100 100 Car 0 N/A
And the explanation becomes, "in two equally likely cases (host opens Door 2 or host opens Door 3), switching wins twice as often as staying". The key to the solution is not that the car is equally likely to be behind each door before the host opens a door (this is obvious to everyone), but that it's equally likely that the host open either door and in each case you're twice as likely to win by switching. -- Rick Block (talk) 04:26, 6 February 2010 (UTC)
The sample space is anyway reduced by opening a door:
Door picked Door unopened Door opened result if switching total cases cases if door is opened
Car Goat Goat Goat 100 50
Goat Car Goat Car 100 100
Goat Goat Car Car 100 0
Heptalogos (talk) 12:36, 6 February 2010 (UTC)
You're so close! However there's a problem in your table. In the "Door opened" (3rd) column, the last row says it has a car. This cannot happen. So, in this row, the number in the "total cases" column must be 0. The basic problem is if you want to show a 100/100/100 column (before the host opens a door) and a 50/100/0 column (after the host opens a door) you have to individually identify the doors. You could call them doors A,B,C if you'd like, but you might as well just stick with 1,2,3. The 50/100/0 column can only exist if you know which door the host has opened. It is a conditional case. I suspect what you're really going for is something sort of like this
Door picked Unpicked door A Unpicked door B result if switching total cases cases if host opens Door B
Car Goat Goat Goat 100 50
Goat Car Goat Car 100 100
Goat Goat Car Car 100 0
The point is the so-called condition "host opens door 2 or door 3" does not have an "exclusive or" sort of sense (meaning a specific one of these doors) but only a "union" sort of sense (meaning all cases where the host opens either door 2 or door 3, i.e. all cases). This is why the ? entries in the first table above are all 100. If you want a conditional case (a 50/100/0 column), the condition has to be "host opens a specific door". Given the problem statement, you might as well stick with Door 3 - it's representative of any other specific case.
If you're really aiming for the Door A/Door B table, what is your description for why the probability of winning by switching is 2/3 and why the top number in the "cases if host opens Door B" (last) column is 50 and not 100? I assume your explanation doesn't start with "in 2 out of 3 equally likely cases ..." And, BTW, if this had been vos Savant's table I'd imagine the Morgan et al. paper would have never been written. -- Rick Block (talk) 17:46, 6 February 2010 (UTC)
It's hopeful to see that you don't like 'impossibilities' in the sample space. For now, let's keep focused on the given method. In your example door3 has a car, which is of course just as impossible. You are confusing the situation before with the situation after. Heptalogos (talk) 18:20, 6 February 2010 (UTC)
In which example? My WP:OR table above? I thought it was obvious the first several columns are the before state, and the last were the after state. My point all along has been the 100/100/100 (or 1/3:1/3:1/3) distribution is obviously the case only in the before state. We could explicitly label the columns if that helps. This makes the table equivalent to the large figure at the end of the Probabilistic solution section of the article.
Situation BEFORE the host opens a door Situation AFTER the host opens a door
Door 1 Door 2 Door 3 total cases host opens Door 2 host opens Door 3
cases result if switching cases result if switching
Car Goat Goat 100 50 Goat 50 Goat
Goat Car Goat 100 0 N/A 100 Car
Goat Goat Car 100 100 Car 0 N/A
Anything talking about "2 out of 3 equally likely cases" must be talking about the before state. Are you disagreeing with this? -- Rick Block (talk) 18:40, 6 February 2010 (UTC)
Yes. But what I mean is that my "door opened" which has a car is the same as your door3 with a car. This is only impossible after we know it has been opened. Before that, I don't identify doors 2 and 3 individually. One can look at the number 3 and always open it, or one can open a door and always number it as 3; there's no relevance in the difference. Unless you assume that a "number bias" may exist. Heptalogos (talk) 19:07, 6 February 2010 (UTC)
But then, to be consistent, the before condition must be 100/200, not 100/100/100. The door you're opening is one of the 200 that is not door 1, not one of either of the 100 that is door 2 or door 3. You're talking about the 2-line table. -- Rick Block (talk) 19:52, 6 February 2010 (UTC)
It is known that the 200 cases are equally divided over the two doors without the need to identify them individually. It's conditional because opening one of the doors reduces the sample space. Heptalogos (talk) 20:17, 6 February 2010 (UTC)
Are you calling the door that is opened (whichever one it is) door 3 or not? If yes, you're not reducing the sample space. We start with 300 cases with an obvious 1/3:1/3:1/3 distribution, one of the doors is opened, whichever door is opened is now called door 3 - leaving us still with 300 cases (like the table below).
Situation BEFORE the host opens a door Situation AFTER the host opens a door
Door picked Unpicked door A Unpicked door B total cases result if switching cases if host opens a door
Car Goat Goat 100 Goat 100
Goat Car Goat 100 Car 200
Goat Goat Car 100
The opened door3 is either A or B, not both. The doors are unique, although we don't identify them statically BEFORE. So AFTER we do have 150 cases instead of 300. If you state door3 may be both A or B, it makes no sense to create 2 columns for them with certain, different distributions. Heptalogos (talk) 22:26, 6 February 2010 (UTC)
BEFORE they are identified vertically; AFTER they are also identified horizontally (statically). Vertically means that each door has a certain distribution (GCG, GGC). Horizontally means that any opened door (last column) is one of those vertical columns. Heptalogos (talk) 23:02, 6 February 2010 (UTC)
If you are reducing the sample space, you're talking about the Door A/Door B table above (where the first 5 columns are the before state and only the last column is the after state).
If we cut through all the pointless arguing here, what the simple solutions mean to be saying is that the average probability of winning by switching is 2/3, and assuming any particular case (for example the case where the player has picked Door 1 and the host has opened Door 3) has the same probability as any other particular case (which is true only if the host picks randomly between two goats) then in all cases the probability must be the same as the average (this is the "symmetry" argument). The issue is most sources presenting simple solutions don't include the "and assuming" part of this - and certainly never mention that how the host picks between two goats matters. They say what the average probability is (which is NOT affected by how the host chooses between two goats), and then stop. Morgan et al. (and others) are saying because they stop after saying what the average probability is they haven't quite answered the question.
You (and Martin and Jeff) are saying any idiot should understand what they mean, but since they're not explicit about it many idiots walk away with the misunderstanding that "what the host does can never affect the player's initial probability" - vos Savant said it ("The odds that your choice contains a pea are 1/3, agreed? Then I simply lift up an empty shell from the remaining other two. As I can (and will) do this regardless of what you've chosen, we've learned nothing to allow us to revise the odds on the shell under your finger.") it must be true! It is true, but only if we're talking about unconditional probabilities. Conditional probabilities (like what are asked about in this problem) are always affected by what the host does. The conditional probability of the player's door MAY end up numerically the same as its initial probability, but it's always a different probability (in the tables above it's out of a sample set of 150 cases, not 300 cases). The only conditional probability that's obvious from the problem statement is the one for the opened door. Both of the others are always a direct function of how the host chooses between two goats. -- Rick Block (talk) 21:24, 6 February 2010 (UTC)
Why is the average probability not affected by the host's goat-door prevalence? It may be. Heptalogos (talk) 22:35, 6 February 2010 (UTC)
No, it may not be. If we're talking about 300 cases where the player has picked Door 1, and no player switches, about 1/3 of them win regardless of what the host does (assuming the host opens a door). Similarly, about 2/3 of them win if they all switch. This is the average chance of winning by staying and by switching and it is independent of any host goat-door preference. If the host picks randomly the 300 cases split into two equivalent samples of 150 cases. The stickers and switchers win 1/3:2/3 in each sample, i.e. 100 out of 150 win by switching in each case. With an extreme host preference, say the host always opens Door 3 if possible, the 300 cases split unevenly - 100 players will see the host open Door 2 and 100% (all 100) of these players win if they switch, while 200 players will see the host open Door 3 and only 50% (also 100) of these players will win if they switch. The average chance of winning by switching is (100+100)/300 = 2/3. Using formulas, if the host preference is p, (1/3)+(1/3)p players see the host open his preferred door and (1/3)+(1/3)(1-p) see the host open his non-preferred door. The first group has a 1/(1+p) probability of winning by switching, the second group has a 1/(1+1-p) probability. The composite probability (which is the average) is
( (1/3)+(1/3)p )( 1/(1+p) ) + ( (1/3)+(1/3)(1-p) )(1/(2-p) )
which simplifies down to
1/3 + 1/3 = 2/3
-- Rick Block (talk) 00:56, 7 February 2010 (UTC)
You assume several players. In the problem statement you are the only player, and you may have a bias with a significant relation to the host's bias. Heptalogos (talk) 12:57, 7 February 2010 (UTC)
Conditional probabilities are always affected by the conditions, yes. The condition is the opened door, not the number. Heptalogos (talk) 22:35, 6 February 2010 (UTC)
I have no idea what distinction you're trying to make here. -- Rick Block (talk) 00:56, 7 February 2010 (UTC)
@ Rick, please try to understand the distinction. It is vitally important. Gill110951 (talk) 05:42, 7 February 2010 (UTC)
I think I understand the "indistinguishable doors" POV, and that in this POV the host opening a door does not change the sample space (so the prior probability of the car being behind the player's door is unchanged by the host opening a door since the posterior sample space is identical to the prior sample space). I believe this is the POV Jeff, Martin, and Glkanter (and possibly Heptalogos, but I'm not sure) think is "correct".
I think I understand the "distinguishable doors" POV, and that in this POV the host opening a door does change the sample set (i.e. there are different prior and posterior probabilities, where the posterior probability depends on how the host selects between two goats). I believe this is the POV Nijdam thinks is "correct". I think I also understand that in the "distinguishable doors" POV, the host's preference may or may not be accessible to the player, and if it's not accessible the most sensible assumption is to take it as random.
If somebody could please clarify the distinction Heptalogos is making between the condition being the "opened door" as opposed to "the number" I'd appreciate it. -- Rick Block (talk) 19:51, 7 February 2010 (UTC)
Of course what a wikipedia editor *thinks* is strictly speaking irrelevant, but just for the record, I do not think that it is a sensible assumption at all, to take something as uniformly at random when you don't know it. That's the whole point of the game theoretic approach: what is sensible to do, when you don't know the other's strategy!!! Regarding the distinction: Prob(car is not behind your door|you chose 1,MH opened 3) might well be different from Prob(car is not behind your door|you chose 1, MH opened a door). Isn't that obvious???? Gill110951 (talk) 22:48, 7 February 2010 (UTC)
Yes, of course P(not your door|you chose 1, host opens 3) is not the same as P(not your door|you chose 1, host opens a door) - that's the exact point I've been trying to explain this entire thread. Have you read this whole thread? In addition P(not your door|you chose 1, host opens a door) is exactly equivalent to P(not your door|you chose 1) - the conditions "you chose 1, host opens a door" and "you chose 1" define the same sample space, so all probabilities are the same with regard to these two conditions. My point is that vos Savant's solution addresses probabilities in this sample space, i.e. she's addressing P(not your door|you chose 1) (whether you add "host opens a door" or not changes nothing), not P(not your door|you chose 1, host opens 3). Martin and Heptalogos seem to be objecting to characterizing vos Savant's solution as addressing P(not your door|you chose 1). This is what it literally does, doesn't it? -- Rick Block (talk) 23:44, 7 February 2010 (UTC)

Table showing the host opening an unnamed door

Thread moved to /Arguments#Table showing the host opening an unnamed door. -- Rick Block (talk) 17:54, 7 February 2010 (UTC)

I love it

Can I just say Monte Halperin turns 89 this summer and I think he would never have imagined the role he has assumed in this fascinating mathematical back-and-forth. I've watched this page for some time, and very much enjoy the discussions. My humble suggestion to involved editors is to remember what brought you here in the first place, which I'm guessing are the deliciously captivating counter-intuitive qualities of the Monty Hall problem. I hope you can edit collaboratively to communicate and elucidate this to readers! Respectfully, RomaC (talk) 17:54, 6 February 2010 (UTC)

Was Monty the first in the world to reveal small prices and offer to switch? Or did he copy the act? I heard the puzzle on television and indeed had my 50/50 answer ready. But that was because of the problem statement which didn't make clear that the host would always do that. So I checked it out on Wikipedia and understood the paradox, but there I met some people who were convinced that the number of the door had any significant information or relevance. I am still trying to show them that it hasn't, and for sure wasn't meant to have any relevance. Heptalogos (talk) 18:31, 6 February 2010 (UTC)
That's what I think too, @ Heptalogos! But lots of authoritative sources have said that the numbers do have significance and that the problem *has* to be solved by conditional probabilty and that you *have* to make a load of extra assumptions too (their arguments are often pretty shaky or absent)... One is not obliged to agree with these sources. One can find equally authoritative sources who think otherwise. Both points of view have to be covered by wikipedia since they are all out there, part of the living Monty_Hall_problem problem. We the editors have to look down on all these little ants running around in circles dispassionately, and give a wise and clear overview so that future new ants who fall into this sand-hill don't have to run in all the same circles all over again. Gill110951 (talk) 06:10, 7 February 2010 (UTC)

I love it too

The discussions going on here are nothing else than the eternal rerun of the Monty_Hall_Problem problem. The problem to decide what is the problem. That is what the wikipedia page should be about. There are authoritative sources for all kinds of particular versions. There are authoritative sources that people see the problem in many different ways. A small problem here is that many authoritative sources, especially those by mathematicians, are authoritarian, dogmatic. And often the more ignorant the mathematician, the more dogmatic.

Can't wikipedia present a number of points of view, can't it present a history?

I too have a point of view and as a mathematician I often appear to write in dogmatic authoritarian style. I strongly want us to structure the article by starting with "Craig's problem" as quoted by Marilyn. If that can be agreed, then we can move on to a next step, structuring the material that we agree needs to be put on the page. Gill110951 (talk) 05:58, 7 February 2010 (UTC)

I agree with you except for one thing, The order of presentation. I, at one time, suggested historical ordering of the article, but the problem is that the history of the subject is mainly one of argument and dispute. This is not conducive to writing a good article.
A more logical order, in my opinion, is one of complexity, start with the simplest understanding and solution of the problem first and then proceed to add complications. This is how most good text books work, they start with the basics, possibly glossing over some of the subtleties for the sake of clarity, then, when the reader is assumed to understand the basics, they go on to say that the initial exposition was, maybe, a little over simplified and deal with the complicating factors.
The logical order for me would be:
  • Simple (non-conditional) problem and solution - fully explained with aids to understanding and no confusing mention of conditions.
  • Conditional problem
  • Game theory
  • Variants
Agreed, to order by increasing complexity is very wise. Seems we two agree about almost everything. Those who disagree, seem to want to erase some parts of the Monty Hall story because of their personal taste or because of the dogmatic/authoritarian outloook of their preferred source.Gill110951 (talk) 22:39, 7 February 2010 (UTC)

The justification given does not justify the original problem but only justifies the revised problem

When i saw this i originally thought the math was ridiculous. If a newcomer came onto the game show the chance of the car being behind the doors would be 50:50 so why not with the original contestant! To explain the problem wiki extrapolated the problem and likened it to 1 in a million chance. This altered my thinking for a moment. But only for a moment. I quickly realised that this was a different argument. True what are the chances of your initial prediction being the 1 in a million. In this scenario there would be definately an advantage to switching. But if this argument is extrapolated again and we say what if monty did this simuteaneously with 1 million other contestants clearly it would not always be advantageous to switch even in these evidentally advantageous conditions. One may argue that this is not fair and not the same problem - in this case its the literal truth - but then that is my point. Exaggerating the problem changes the problem and it is no longer the same problem.

Matt Cauthon —Preceding unsigned comment added by 194.106.220.83 (talk) 12:58, 11 February 2010 (UTC)

You're talking about two different issues.
1) In the standard problem a newcomer comes onto the game show after the door has been opened and does not know the situation, this newcomer has a 50:50 chance of correctly picking the door hiding the car but the original contestant has a 2/3 chance of winning by switching.
Read this statement very carefully. It says the newcomer has a 50:50 chance of correctly picking the door. This does not mean the "fully aware" probability of the car being behind the two doors is 50:50, but if you make a random choice between them you end up with the car 50% of the time. What's 50:50 here is not the distribution of the car behind the doors, but the result of a random choice. Think about 150 shows where the original contestant picked door 1 and the host opened door 3. We'd expect the car to be behind door 1 about 50 times and behind door 2 about 100 times reflecting the contestant's 1/3:2/3 chances. If the newcomer randomly picks door 1 or door 2, she ends up with the car how many times? The answer is if she picks between the doors randomly she'll pick each door about 75 times and win the car about 1/3 of the times she picks door 1 and 2/3 of the times she picks door 2. So, the number of times she'll win the car is (about) 75*1/3 + 75*2/3 = 75(1/3+2/3) = 75, which is 1/2 of 150.
2) What if different contestants pick different doors - how can it be advantageous for them all to switch?
I think this is the other issue you're talking about. For example in the 3 door case what if there were two contestants, one picked door 1 and the other picked door 2 and the host opens door 3. This is a completely different problem and requires some extra rules. To make it possible for the host to always open a door that doesn't reveal the car and isn't one of the doors one of the players picked, there must be at least two more doors than the number of players - so this 3-door example doesn't really work (what happens if the car is behind door 3?). You could say if there are m contestants there must be n >= m+2 doors and, after each contestant picks a unique door, the host (randomly) opens all but 1 of the unpicked losing doors - leaving the m doors the contestants picked and 1 more. In this case the contestants should all switch to the unopened unpicked door, but switching between themselves doesn't make any difference. The probability for each contestant's door is 1/n and the probability for the other door is (n-m)/n.
Your point might be that the million door example requires extra rules also. This is true, but I think the general assumption for the 1-player, n-door case is that the host (randomly) opens n-2 losing doors none of which is the door the player initially picked, leaving the player's door and one more. Under these rules, the player's door has a 1/n chance and the other door has a (n-1)/n chance. -- Rick Block (talk) 17:05, 11 February 2010 (UTC)
Yes, anything correct and clear, just want to once more underline this one detail:
"The newcomer chooses randomly between the two still closed doors" says: The newcomer can not distinguish these two still closed doors. He does NOT know which one had been chosen by the guest (out of three doors with a chance of 1/3), and so he does NOT know which one has been offered as an alternative for switching with a chance of 2/3. He has no information about that, and so has to choose "randomly". But if he could distinguish, e.g. if the doors were marked, he could decide for the originally chosen door with a chance of 1/3, or for the alternatively offered second closed door with a chance of 2/3. Just wanted to underline that detail once more. Regards, -- Gerhardvalentin (talk) 18:17, 11 February 2010 (UTC)

2b) What if different contestants pick different doors - 3 contestant, 3 door variation

As in the original style MH game, there are 3 doors, only one of which has a new car.

But here, Monty has a large crowd of potential contestants; each one has picked one of the 3 doors.

For each of the 3 doors, Monty randomly selects one contestant who has picked that door.

Monty separately isolates the 3 contestants so that none of them know the others are playing.

Monty randomly opens one of the doors with a goat, and the contestant for that door looses.

Monty separately allows each of the 2 remaining contestants to switch to their previously unchosen door. (Davrids (talk) 00:55, 16 February 2010 (UTC))

(If both contestants pick the winning door, they both get a new car.)

Should each remaining contestant switch to their previously unchosen door?

Each contestant's situation seems to be the same as if he were the only contestant in the original style (single contestant) MH game.

If so, the arguments that prove that the player in the original style MH game should switch, would also prove that both remaining contestants in this game should switch to get their 2/3 chance of winning.

But they can't both have a 2/3 winning chance if they switch, can they? Wouldn't the combined chance, for the two remaining doors together to have a car, be 4/3? What is wrong?

Davrids (talk) 00:34, 16 February 2010 (UTC)

Monty randomly opens one of the doors with a goat, and the contestant for that door looses.
Let's say you pick Door 1 and the player who picks Door 3 is eliminated (the same analysis applies for any particular door you pick and door the host opens). The prior probability of each door (before the host opens a door) is 1/3. Like in the normal MHP the posterior probability that Door 1 hides the car (after the host opens Door 3) is
P(car is behind Door 1 and host opens Door 3) / ( P(car is behind Door 1 and host opens Door 3) + P(car is behind Door 2 and host opens Door 3) )
Because of the rule I've put in bold above, P(car is behind Door 1 and host opens Door 3) is (1/3)(1/2) = 1/6 and this is the same as P(car is behind Door 2 and host opens Door 3). So the posterior probability the car is behind Door 1 is (1/6) / (1/6 + 1/6) = 1/2. There are only two unopened doors, so the probability the other one hides the car must also be 1/2.
In the normal MHP the host MUST open Door 3 if the car is behind Door 2, so P(car is behind Door 2 and host opens Door 3) is 1/3, not 1/6. -- Rick Block (talk) 03:11, 16 February 2010 (UTC)

Proposal

Mediation seem to be proceeding painfully slowly and many of us here want to get on with improving the article. I have suggested this before but I am going to propose it again here as a compromise between differing viewpoints. I therefore ask all those on both sides of the fence to give their support for this basic proposal to try to reach an end to the argument on this page.

Andrevan, if you would like to move this discussion to your mediation page and act as a mediator for it please feel free to do so.

I propose changing the order of sections to be as follows:

  • Basic (non-conditional) problem and solution
Much as it does now this section should deal with the MHP (mathematical puzzle) as it is understood by the vast majority of the general public (who are aware of it) and the majority of sources (of all types) on the subject. It is the notable MHP and almost certainly what the reader has come here to read about. No mention of conditional/unconditional probability should be made here.
  • Aids to understanding

This section should, as it is currently, be aimed a helping the reader to see why the answer is 2/3 and not 1/2. No mention of conditional/unconditional probability should be made here.

  • Conditional solution

Here the view of those who think the problem is essentially one of conditional probability (such as Morgan) should be put forward, including the reasons that they believe simple solutions do not address the true problem.

This is essentially an academic section (although in the interests of neutrality I do not call it that) is which what some sources say what about other sources, and the value of studying the problem in this way for students of probability, can be discussed in a proper manner. By fully discussing these issue here we ensure that the academic credentials of WP are maintained.

  • Game theory

This a new section which I think should be added. Once you get past the simple (puzzle) problem and consider the characters to be acting as more than agents of chance it is reasonable to ask what happens if they try to maximise their gain or minimise their loss.

Sources of confusion

Here we discuss what sources say about why people cannot answer the simple problem correctly. The tests on pigeons might fit in here

  • Variants

Intentional complications, generalisations etc as currently.

  • History

Much as now.

I would request all editors, on both sides of the fence, to subscribe to this proposal. Either way it would help if people try to restrict their replies to 'support' or 'oppose' at this stage. There will always be plenty to discuss later. I strongly believe that these changes will help us all to work together. Martin Hogbin (talk) 12:15, 7 March 2010 (UTC)

Support

Oppose

Comments

Please try to keep these at a minimum.

  • My counter proposal, which is a compromise, is to include a single solution section presenting both unconditional and conditional solutions in an NPOV manner, more or less like the draft still currently above (permanent link). IMO, Martin's proposal creates a structural NPOV violation (see WP:NPOV#Article structure). -- Rick Block (talk) 16:16, 7 March 2010 (UTC)
While your proposal might be more npov or better than Martin's, I really fail to see how it is a compromise in anyway (the same somewhat holds for Martin's suggestion). You both simply reiterate your point of view, you've been arguing the whole time. Imho there's nothing compromising here at all.--Kmhkmh (talk) 16:24, 7 March 2010 (UTC)
I completely agree Martin's suggestion is no compromise at all (in any respect). The proposal I'm making is a compromise between presenting on the one hand simple solutions as correct and sufficient vs. on the other hand saying the problem is inherently conditional and that these solutions fail to address it. The tone I'm aiming for is "here's a solution ..., here's another kind of solution ..." - not "here's a solution ..., that solution is wrong here's the real solution ...". In a personal sense, the compromise is keeping my individual POV out of the article - which I think is the compromise we must all make. -- Rick Block (talk) 17:15, 7 March 2010 (UTC)

Firstly, I suggest that you both sign in the oppose section above.

Several editors here, including myself, believe that the Morgan paper makes a very limited contribution to the subject in that it discusses a very specific, indeed perverse, formulation of the problem that is of little interest to anyone except students of elementary statistics. It is just one source and, as such, deserves limited space within the article. Indeed the whole subject of conditional probability is not what the MHP is all about from the POV of the majority of editors here, the majority of sources on the subject, and the vast majority of readers. It therefore also deserves limited mention in an article about an infamous but simple mathematical problem. As a compromise, I am proposing that we keep the section on conditional probability including the Morgan paper, complete with its arrogant criticism of other sources but the we just move it to a place where it will not confuse the general reader. If you do not like this, please just sign in the 'oppose' section above. Martin Hogbin (talk) 17:33, 7 March 2010 (UTC)

You are continuing to mischaracterize the issue and I would very much appreciate it if you'd stop. The notion that conditional probability is central to the MHP is NOT based on a specific, perverse formulation published by a single paper (Morgan et al.), but a common, mainstream (mostly academic) POV. Yes, Morgan et al. is "just one source" - but it is one of many similar sources. I've provided at least 5 specific high quality references. You continue to focus on this one paper which you don't like and ignore the far bigger issue. What is your claim about the POV of the majority of sources based on, exactly? Have you surveyed sources and counted? What you seem to be saying is you really want no mention of conditional probability anywhere in the article but you're willing to "compromise" by including a small section discussing this issue (as close to an appendix as you can manage), so it doesn't "confuse" the general reader. If that's not virtually the definition of POV pushing, I don't know what is.
I haven't signed in the 'oppose' section above because polling is not a substitute for discussion. We're in formal mediation already. "Voting" about this is pointless. -- Rick Block (talk) 18:20, 7 March 2010 (UTC)

Bayes section is REALLY bad: propose to replace by application of Bayes' law in odds form

I think the article is not bad at all at present. However the section on the Bayesian approach to it is totally out of proportion to the rest. It seems to imply that a philosophical Bayesian viewpoint, ie the viewpoint that all probabilities are subjective and represent degrees of belief (as reflected by your willingness to take bets at certain odds) helps understand the problem. Whereas what is actually done here is simply to apply the standard definitions of conditional probability, which hold for anybody's probability, whether subjective or "objective" ("frequentist"). Having defined the basic events and written out the probabilties which we are told in the full problem statement, we simply calculate Prob(car is behind door 2 | player chose 1, host opened 3).

In other words, there is a confusion between use of Bayes theorem aka Bayes formula (which is proved in one line from the definition of conditional probability), and a Bayesian philosophy. The philosophy is irrelevant here, since the calculation is the same whatever philosophy of probability you adhere to. You don't even have to have a unique philosophy, you can just apply probability calculus to whatever kind of probability seems most appropriate in a particular context.

Secondly, *if* we are going to do this by a calculation by Bayes theorem/formula (it is not the only way to get to the answer), *then* it is more attractive to use Bayes' theorem in its form: posterior odds equal prior odds times likelihoods.

In case you didn't know this before, I'll briefly run through a numerical example, then apply the technology to MHP.

We want to decide about some proposition A and we are given some data D. A priori, we think of A as having a certain probability p, the probability that it was not true was therefore 1-p, the odds on its being true were p:1-p. For instance, if A had a priori probability 1/10, then initially the odds were 9 to 1 against it. (Which is the same as odds of 90 to 10 against, or 180 to 20 against..). If the odds are 90 to 10 against, then the probabilities for and against are 10/(10+90) and 90/(10+90), ie (of course) 1/10 and 9/10.

OK, now we get some data D. The data D can be more or less likely to be produced if A is true, or if A is not true. Suppose there are probabilities that D is observed in both situations: Prob(D|A) and Prob(D| not A). These two numbers are called "the likelihood for/against A, given the data D". Again, only their ratio is important. For instance, it could be that Prob(D|A)=0.05 while Prob(D|not A)=0.0025. In both cases (A true, A not true), D is pretty unlikely, but D is 20 times more likely to occur if A is true than if A is not true.

So we have so far: prior odds of A versus not A are 1 : 10

Likelihood ratio for A versus not A, given data D are 20 : 1

Bayes formula states that the posterior odds for A versus not A are now 1 x 20 : 10 x 1 , or 20 to 10, or 2:1. In other words, having seen D, A is now twice as likely as not A, while before it was 10 times less likely. Simply because the data D is 20 times more likely to happen if A is true than when it is not true.

Three door problem. Suppose we choose door 1. The data is going to be "host opens 3"

Prior odds on car was actually behind door 1 : door 2 : door 3 = 1:1:1 (initially all doors equally likely)

Likelihood ratio for doors 1, 2, 3 given door 3 opened = 1/2 : 1 : 0

The 1/2 for door 1, because if the car is behind door 1, the probability is half that door 3 (and not door 2) is opened

The 1 for door 2, because if the car is behind door 2, the probability is one that door 3 will be opened (you chose 1)

The 0 for door 3, because if the car is behind door 3, the host is not going to open door 3 and show you a goat!

The posterior odds are therefore 1/2 : 1 : 0 which is the same as 1 : 2 : 0 which means probabilities 1/3, 2/3 and 0 on doors 1, 2 and 3. Gill110951 (talk) 18:43, 6 March 2010 (UTC)

I disagree with the proposed change. The Bayesian analysis section in its current form is (a) referenced, (b) self contained (c) accessible to a reader with a basic understanding of probability and/or logic (d) typed in good style. The proposal makes reference to a non-primitive concepts (odds and likelihood ratios, rather than straight probabilities). Further, the section's content makes no implication that the Bayesian interpretation of probability theory is in general preferable. However, the use of Bayesian notation to solve the MHP is referenced and appropriate.--glopk (talk) 01:11, 9 March 2010 (UTC)
It is kinda interesting that towards Nijdam you argued the other way around, claiming "low accessibility" or if you will avoiding non-promotive concepts doesn't matter and would be only "conservpedia style bigotry".
You missed my point entirely. "Simple" is not the same of "short, intuitive and wrong (or incomplete)"--glopk (talk) 20:10, 9 March 2010 (UTC)

I agree with however the ratios suggestion is not making it easier. However the usual encylopedic approach should be to give the most simple formally correct & complete approach first.

A Linus Torvalds said: "Show me the code". Seriously, quit meta-arguing about it, and sit down to write a well-sourced, well-written, accessible formal derivation of the result for the standard problem. Put it in a proposal page and let's discuss it. I'll be happy to help edit it, if you like, and just as happy to withdraw my objections to replacing the current "Bayesian Analysis" section when I see something better. I haven't so far. Talk is cheap.--glopk (talk) 20:10, 9 March 2010 (UTC)

After that the article can indulge in specific (more advanced) viewpoints and approaches. Create a mathematical analysis section and stuff all the various approach and different in there in parallel, but don't just pick your favored (non standard) one instead without providing any context or perspective.--Kmhkmh (talk) 13:57, 9 March 2010 (UTC)

Remove references to "The Monty Hall Trap" by Phil Martin

I would like to remove references to "The Monty Hall Trap" by Phil Martin because I believe his article contains numerous false claims about both probability and the game of bridge. I don't feel that it's right for me to unilaterally remove the references without first discussing this with the person(s) who posted those references. How should I proceed?Secondfoxbat (talk) 07:57, 26 February 2010 (UTC)

Why not explain here what you think the false claims are? Martin Hogbin (talk) 09:02, 26 February 2010 (UTC)

I am currently writing a paper on the problems that I see, so I am reluctant to say too much here. Briefly, Mr. Martin's idea of the Monty hall trap is not one which anyone falls for. His trap, even if true, does not apply to bridge. As far as I can tell, his concept of "biased" data is new with him, but it impugns the idea of declarer drawing an inference from the opening lead. His argument that playing the same deal first as a South declarer, then as a North declarer is based on a fallacy. And there's more.... Secondfoxbat (talk) 01:14, 27 February 2010 (UTC)

It's in the "History" section, so the question should be whether this article has any historical importance. It is one of only a few references before the Parade articles appeared. It also shows the multi-disciplinary appeal of the problem. Rather than delete it, you might abbreviate the paragraph a bit. I believe this reference is far less significant than the Nalebuff one immediately before (which gets only a single sentence as opposed to a paragraph for the Martin one). -- Rick Block (talk) 02:49, 27 February 2010 (UTC)

Thanks for giving me the proper perspective. I agree that the Nalebuff reference is much more significant. I'll shorten the Martin reference so it is shorter.Secondfoxbat (talk) 04:27, 28 February 2010 (UTC)

Someone changed the reference to "The Monty Hall Trap" to give more credit than is due. The best that can be said for Mr. Martin's paper is that it attempted to show a connection between the Monty Hall problem and the game of bridge. Anything more is not factual.Secondfoxbat (talk) 17:50, 5 March 2010 (UTC)

I'm the one who changed it. My intent was to make it more flatly descriptive. "Attempted to show" conveys a critical, rather than neutral, POV (see WP:NPOV). What I changed it to was "presented Selvin's problem as an example of what Martin calls the probability trap of treating non-random information as if it were random, and relates this to concepts in the game of bridge such as ...". Do you agree that this is more or less literally what Martin's article says (as opposed to agree with the content of what Martin is saying)? We need to say what the article says, without adding any "spin" indicating we either agree or disagree with the content. -- Rick Block (talk) 01:28, 6 March 2010 (UTC)

Thanks for pointing out the Wikipedia perspective. I suppose we could agree on a shortened version of your change: "presented Selvin's problem as an example of what Martin calls the probability trap of treating non-random information as if it were random, and relates this to concepts in the game of bridge." If anything more is said about how he relates the probability trap to bridge, it should not be about restricted choice, but about not drawing inferences from the opening lead and not using the Vacant Places probability calculation. Secondfoxbat (talk) 06:47, 7 March 2010 (UTC)

Today I rewrote the article Principle of restricted choice (bridge) without developing a pre-existing reference to the Monty Hall problem; with new explanation in terms of Bayesian updating; mainly explanation of the same point in bridge terms. (The point is multiplication of prior probabilities by likelihoods or prior odds by likelihood ratios.)
Phillip Martin 1989 is silent about any relation of his bridge analysis to the rule of restricted choice although he takes it for granted in a sense and clearly aims to present a more advanced and more tenuous application of the same principles. We may infer that he overblows Monty Hall in order to promote the article and the editors permit the same in order to promote the journal. It's a shame because he and they seriously need to give more space to the bridge analysis which is not simple and not "familiar" (not the umpteenth rehash, as is "Restricted choice" and its Talk page).
(Secondfoxbat says) Briefly, Mr. Martin's idea of the Monty hall trap is not one which anyone falls for. His trap, even if true, does not apply to bridge. Stated more carefully, I believe Sfb means, Mr. Martin thinks that few bridge players fall for the Monty hall trap as hundreds of PhDs and thousands of Parade readers would do one year later. He overrates bridge players, in terms of his value system. --much as Morgan, et al., overrate statisticians or "PhDs" in terms of their values, but not so badly. Restricted choice is old hat in bridge, and it was old hat twenty years ago. --P64 (talk) 05:08, 12 March 2010 (UTC)

Bayes section is REALLY bad: replace by link to Bayes Theorem, Example 2

We can cut the whole section and replace it by a link to:

http://en.wikipedia.org/wiki/Bayes%27_theorem#Example_3:_The_Monty_Hall_problem

Gill110951 (talk) 18:55, 6 March 2010 (UTC)

The point of the section is to present a fully rigorous proof using a specific formalism, which is why it is more or less presented as an appendix. -- Rick Block (talk) 00:08, 7 March 2010 (UTC)
The problem with the current version is, that it might insinuate to less versed readers that a Bayesian philosophy is required to provide a full formal proof based on Bayes' formula, which of course is nonsense. The (same) formal proof with the Bayes formula holds for both a Bayesian and a frequentist viewpoint, i.e. though the philosophical underpinnings or interpretation are different for both viewpoints, they more or less produce the same formal proof and of course Bayes' formula is valid in both.--Kmhkmh (talk) 13:59, 7 March 2010 (UTC)
I also strongly object the section called "Bayesian approach". As I mentioned before, it hardly can be considered a real Bayesian approach, and it only shows a unnecessarily complex way of just using Bayes' formula to calculate the solution. Many textbooks just give the straightforward calculation, and I made a proposal some time ago. Look in Talk:Monty Hall problem/Construction#Mathematical formulation.Nijdam (talk) 15:15, 7 March 2010 (UTC)
KmhKmh's point was made and answered time ago (like most on this board). The "Bayesian Analysis" section makes no claim to being the only correct formal proof. However, it is a formal proof and it is self-contained save for references to basic probability axioms and theorems. It is wrong to say that the "same" proof holds in a frequentist viewpoint, because a self-contained proof based on such a viewpoint would look much different - at a minimum it ought to begin by stating what is the population under consideration. If the article is to have a proof, it must be in one system of formal notation among those available in probability theory, be it Bayesian, frequentist, or measure-theory. The choice of the Bayesian formalism in the (now deleted) section is referenced in the literature specifically for the MHP (Jeff Gill's textbook), and is therefore appropriate for the article. I may add that, in my personal opinion, it is eminently appropriate because the MHP is essentially puzzle of logic.--glopk (talk) 00:55, 9 March 2010 (UTC)
I have reverted Nijdam's change and reinstated the Bayesian analysis section. I notice that it is (at least) the second time that Nijdam replaces that section (which has already passed two F.A. reviews along with the rest of the article) with the same unsourced and poorly formatted "Mathematical analysis". Please avoid defacing the article.--glopk (talk) 01:13, 9 March 2010 (UTC)
Though nijdam's replacement might not be the best choice, it was by no means defacing the article. Criticism of this section and its misleading nature has crept up several times (for the last see above) and it was actually contrary to your claim never properly addressed (at least not from my perspective).
Yes it was: see this archived discussion. All the points you make above (length, complexity, bias toward a Bayesian interpretation) were set forth by Nijdam and refuted therein.--glopk (talk) 07:05, 9 March 2010 (UTC)
No they were not and I'm not Nijdam nor do I argue his solution is better. My point is that the way the bayesian approach is framed in the article as "the formal treatment with Bayes' theorem" is misleading. That problem was not addressed and still exists. As far as your discussion with Nijdam and conservapedia there is concerned. You could also argue vice versa that people with a (bayesian) mission might be better suited for conservapedia than wikipedia. And Nijdam's is right as far as accessibility is concerned. Note this is not an argument against a bayesian analysis in the article, but how it is framed and relates to the rest of the article. The problem is that the section to some degree still suggest a bayesian point of view is required for a formal proof using Bayes' theorem, in particular since other sections where referring to it for the formal proof based on Bayes and that is grossly misleading. --Kmhkmh (talk) 14:01, 9 March 2010 (UTC)
The section is referred once and by one other section only - please re-check your count. Nowhere its text says or implies that "Bayes Theorem" and "Bayesian Probability Theory" are the same thing, pray say what exactly is misleading about it. The section is not intrusive: it is placed at the end similar to the way formal proof are placed in appendices in scholarly articles - as WP page templates unfortunately do not allow for appendices. Again, these point have already been argued ad nauseam and settled.--glopk (talk) 07:05, 9 March 2010 (UTC)
The misleading part is the context in which this occurs and how it framed (see below) and not with a particular claim in the section.--Kmhkmh (talk) 14:01, 9 March 2010 (UTC)
Also i can't really see why the bayesian is supposed to be particularly appropriate, while the vast majority of publications seems to use a measure theoretic approach. Personally I don't mind having Jeff Gill's approach as part of the article, put it needs to be put in relation and in context with the rest of the sources, which imho it currently is not.
The article does not even try to cite all authoritative references on the probabilistic treatment of the MHP - there are way too many, and it'd take an analogous of the Erdos number to measure how "in context" they are with each other. Besides, most formal treatments assume an informed reader (at least as informed as to have sat through the first three lectures of a class on PT). Jeff Gill's textbook's one is among the few that are completely worked out, elementary and self-contained. Are you proposing another one, or just criticizing?--glopk (talk) 07:05, 9 March 2010 (UTC)
I'm criticizing they it presented within the article not that it is used per se. Again in doubt a measure theoretic approach would be more appropriate and more representative with the majority of sources. My issue is not with having Jeff Gill in the article, my issue is with having him as the only formal analysis with Bayes and the article referring to it as "the" formal treatment. That is misleading. That's meant with in context, it is framed in an unappropriate manner. As par as relation is concerned, if the article gives only one formal proof, it is good idea to pick the most common one, i.e. a measure theoretic one. If you are looking explicit source for detailed measure theoretic approach, you could use Henze, but I guess there are many other as well (most likely including the much debated Morgan, didn't read his paper personally)---Kmhkmh (talk) 13:45, 9 March 2010 (UTC)
I think the article and the editing may need some serious rethinking and possible restructuring. All involved should take a step back and seriously reconsidering what they are doing. Aside from the nauseating vos Savant vs Morgan issue, there seems potential threat of getting a(n equally pointless) bayesian versus frequentist debate as well in connection with a turf war between editors about who provides what formal proof. Also we have another editor, who's pushing his own (pre)publications and personal research interest into the article, which looks rather questionable to me as well. All that needs to stop.--Kmhkmh (talk) 03:25, 9 March 2010 (UTC)
I don't see anyone other than you and Nijdam suggesting that there is an ongoing "bayesian vs frequentist debate" in the editing of the article. But I do see one between those who want to preserve the quality of an informative article and those who prefer to keep pushing their three lines of ASCII math.--glopk (talk) 07:05, 9 March 2010 (UTC)
In that I suggest you do some serious rereading. Also note I was talking of a Potential--Kmhkmh (talk) 13:45, 9 March 2010 (UTC)

[outindent] OK. Let me clarify my position once again. I am not into any mission to push the use of a Bayesian formalism over and above any other correct one. Rather, I simply expressed and justified above my preference for it. Following that preference I re-edited the section as it currently stands, starting from one that was already in the article at the time of the first FA review. I also then proposed to title it simply "Formal analysis", but the current title was preferred in the discussion, I forget why - and still think the title is wrong. However, I do have a strong preference for having in the article:

A - One formal proof: no adding more beasts to the already crowded zoo, please?

B - Well-sourced: published in an English-language publication with > 10K copies printed, and available in a good university library or (better) Google Books or similarly accessible online full-text source.

C - Well-edited: formatting of the formulae at least as pleasant as the current one (which, btw, has been carefully crafted to be well readable on a variety of screens formats and sizes).

D - Accessible to a reader with only elementary notions of probability: do not presume of the reader more than you would of a student that has attended the first one-two lectures of an undergraduate-level course in Probability Theory or Statistics. Jeff Gill's book proposes the MHP in the exercises section at the end of the first chapter, and that's a level of readership I consider appropriate.

E - Complete and self-contained save at most for the fundamental axioms and theorems: do not force the causal reader through a series of jumps, either put a complete proof or nothing.

The "Bayesian section" as it current stands satisfies all the above. The replacements that have variously been proposed so far satisfied none. I applaud everyone's good efforts to improve the section, or indeed the whole article. But please let your changes make it better than the current bar, rather than an equation-jam copied from a napkin at the corner's cafe.--glopk (talk) 16:55, 9 March 2010 (UTC)

current dispute

This page now includes mention of formal Mediation. There is nothing formal on this page, however. Maybe somewhere in the archives?

One participant asks whether another has completed the reading assignments, so to speak. The links to archives of this page and the companion mathematics page show about ten archives created during the last three months :-(

Can anyone suggest how to approach this daunting lot? --P64 (talk) 18:50, 8 March 2010 (UTC)

The formal mediation request, which identifies the issues at a high level, is at Wikipedia:Requests for mediation/Monty Hall problem. Ongoing mediation is taking place on the associated talk page (which already has one archive). There are summary statements from most of the involved parties (pre-mediation) in this thread which might be a good starting point. -- Rick Block (talk) 19:50, 8 March 2010 (UTC)
... and the 5 references I refer to above are listed (with relevant quotes) in this thread. The Morgan et al., Gillman, and Grinstead and Snell references are all already in the article, with full citations. -- Rick Block (talk) 20:14, 8 March 2010 (UTC)
FA status is awarded by other editors of English wikipedia, right? not by the Mathematics or Statistics or TV project? Anyway, do the reviewers commend specific achievements by the article? It's relevant to any editorial dispute, I suppose, because those commendations should have some weight.
By the way, I doubt that there is a valid distinction between original research and original pedagogy or original illustration. I understand that this acute observation may be consigned to eternal by-the-waydom at wikipedia. --P64 (talk) 20:50, 12 March 2010 (UTC)

Gill 2009a, 2009b, 2010

Good articles, 2009a and 2009b. [...]

Rick Block, Thanks for the prompt replies above and below.
Now I have visited a library where I am able to read all three articles. I will remove comments about them to my user space because they are inappropriate here. (done) --P64 (talk) 01:20, 9 March 2010 (UTC)

What is the status of these articles? In particular, what does "prepublication" mean? Is the first a forthcoming peer-reviewed publication? a University of Leiden working paper? --P64 (talk) 20:08, 8 March 2010 (UTC)

The author of these papers, Richard D. Gill, is editing here as user:Gill110951 and added these references. -- Rick Block (talk) 20:21, 8 March 2010 (UTC)

Show or Hide?

Hide by default, Show at reader option

Editor Michael Hardy (show-or-hide) provides code with suggested writing and layout style whereby a passage that should be an appendix or long footnote may be incorporated in the body of an article but hidden from display by default and opened at the reader's optional click.

Most of section "Bayesian analysis" may be hidden by default, for example.

Show by default, Hide at reader option remains to be explored. --P64 (talk) 17:19, 9 March 2010 (UTC)

Per MOS:COLLAPSE: Scrolling lists and boxes that toggle text display between hide and show are acceptable for use, but should not be used in article prose. So, the question would be whether math details should be considered "prose" or not. I'd say probably yes. In particular the "Bayesian analysis" section is already positionally an appendix, so I don't see any particular need to try to abbreviate it. -- Rick Block (talk) 19:16, 9 March 2010 (UTC)

Mathematical formulation

Glopk is right in his own way, but so am I, as I refer to the cited literature. In their formulation the probability concerned is in fact the conditional given door 1 has inmitially been chosen. That's also what I mentioned. If Glopk wants to make a remark (referenced of course) about it, fine with me, but for the rest stay out of my text.Nijdam (talk) 11:20, 11 March 2010 (UTC)

If you are looking for measure theoretic approach that explicitly uses a 3 steps experiment design, you can use Henze as a source (a popular German probability primer, can be read via google books: [2]). Personally I prefer Henze's/glopk's approach, it may look a bit more cumbersome than the shortened approach used in some of the other sources, but it models the problem "closest" to reality and is not relying on any "smart" shortcuts to reduce the probability space. For detailed formal analysis that seems to be the best approach. Model the problem as it occurs in "reality" (3 steps) and show that it yields the same result as any smart formal or informal arguments (such as reduction of the probability space to a 2 step design). If you feel the 3 step modeling is not necessary please argue that here before editing. Also note that there's no such thing as "your text" in an WP article. What matters here is that we get the best (or at least a good) description for readers and it does not matter who provides what line.--Kmhkmh (talk) 11:56, 11 March 2010 (UTC)
If more discussiants think it better to give a formulation with the choice included, it is fine with me. We only have to add "X=1" with X being the door of choice in each formula. But the literature I mentioned, and there is more, mostly depart from the situation in which door No 1 is chosen. The formulation is not meant for the fully skilled mathematician, who otherwise may critizise it, nor is it meant to show how wonderful mathematical formulas may be. Nijdam (talk) 14:15, 11 March 2010 (UTC)
BTW: if someone (i.e. Glopk) wants to make major changes, ask him to argue that here before editing.Nijdam (talk) 14:17, 11 March 2010 (UTC)
I looked Henze's formulation up, and I'm not very happy with it. He assumes the player chooses randomly, which is an unnecessary assumption. Nijdam (talk) 14:38, 11 March 2010 (UTC)
A couple of things here.
  • You started the major editing not glopk.
  • It is true that you don't have to require a random choice of the player , however you could could make similar argument for the placing of the car and then we are starting to get into trouble. Instead of switching the randomness between car placing and player pick (and reducing the probability space). I'd rather go with "natural choice" to model each step, i.e. you have model the placing of the car and the player pick as separate steps.
  • also the mathematical formulation is meant for "skilled matematican", since the less formal approach is already given before. The point of the "mathematical" formulation is exactly to provide a complete detailed formal model for those, who are familiar with te terminology.
--Kmhkmh (talk) 15:19, 11 March 2010 (UTC)
I agree with all the points made by Kmhkmh above. Modeling the user's choice with as a variable on its own neatly sidesteps all imprecise/cumbersome verbiage, a la say, Door 1, or renumbering the doors if necessary. Equivalently, it removes the need to advocate an abstract property of the problem, namely symmetry. It thus makes for a cleaner and elegant formal specification of the host's strategy. Further, such modeling is supported in the literature: in addition to Henze, I note that the previous Bayesian notation was equivalently using three indices in specifying the host's strategy. Last, it can be further worked out to obtain the "unconditional" solution by marginalizing w.r.t. the player's selection, i.e. treating S as a nuisance variable. On the editing side, I also note that (a) Nijdam's formulation is quite cumbersome and ugly - really, how many equal = signs can you put in a row before your eyes start spinning? (b) My proposed formulation uses notation that is textbook-standard, and accessible to a beginner's student of probability theory, just as the old Bayesian analysis section did.
For all the above arguments, I am reverting the last change of Nijdam and adding a reference to Henze.--glopk (talk) 16:18, 11 March 2010 (UTC)
I don't have a particular preference either way about 2 or 3 variables, however I strongly prefer (insist) that the approach match the cited references (Glopk's current version does, but neither Nijdam's or Glopk's earlier versions do) and would prefer English sources (so not Henze). More FYI than anything else - K&W start with the player having chosen Door 1. Morgan et al. don't use Bayes Theorem at all, but rather the definition of conditional probability with an explicitly enumerated sample space. Mueser and Granberg say they're extending Morgan et al., so by inference they're not using Bayes Theorem either. Lucas, Rosenhouse, and Schepler (not currently cited in the article) include a Bayes Theorem expansion also assuming the player has chosen Door 1. As I recollect, there's a book by Ruma Falk with a Bayes expansion but I don't remember if she starts with the player having selected door 1 or not (her paper that is cited in the article only has a Bayes expansion of the 3 prisoners problem). Gillman uses Bayes Theorem assuming the player has chosen Door 1, but then solves for the probability of winning using the ratio of the probability the host opens door 3 given the car is behind door 1 to the probability the host opens door 3 given the car is behind door 2 (i.e. he never computes the denominator of the Bayes expansion - he only observes that it is the same regardless of whether the car is behind door 1 or door 2). Rosenthal's approach (based on what he calls the "proportionality principle" which follows from Bayes) is essentially a version of Gillman's approach. -- Rick Block (talk) 20:02, 11 March 2010 (UTC)
All valid points. Time for a trip to the library. I suggest we keep the current (Glopk's 3-variable) with the Henze reference temporarily, until I find an English one - bet it'll take me all of 10 minutes ;-) --glopk (talk) 20:13, 11 March 2010 (UTC)
Note the issue whether the player has already picked door 1 is separate from modelling issue. The modelling issue about modelling all 3 steps that actually occur versus modelling only 2 steps and by combining 2 occuring steps into 1 ("as it intuitively makes sense"). In each model however the player picks door 1 anyway. As far as the sources are concerned - I agree that English sources are preferable, but non English ones are acceptable (provided we have enough authors around to verify them).--Kmhkmh (talk) 21:38, 11 March 2010 (UTC)

This is not clear.

I have read the first half this article and I still have learned nothing. I'm quite a smart guy so it is a mystery how most other people will have had this problem explained to them by this article. Some of the sentences are a bit unstructured too. I can fully accept that I got the wrong answer at the beginning, but I think I am not unreasonable for thinking that I should have a better idea of why that is by now. I'm off to find a non-wikipedia source to explain the monty hall problem to me. 121.73.7.84 (talk) 07:26, 22 March 2010 (UTC)

After consulting another site I believe (i may be wrong) that I understand this dilemma, however the explanation on this page is not clearly written and is unnecessarily confusing. This is unhelpful when trying to explain an unfamiliar concept: Quote "The player, having chosen a door, has a 1/3 chance of having the car behind the chosen door and a 2/3 chance that its behind one of the other doors. As the host opening a door to reveal a goat gives the player no new information about what is behind the door he has chosen, the probability of there being a car remains 1/3. Hence the car is then with chance 2/3 behind the remaining unopened door (Wheeler 1991; Schwager 1994). Switching doors thus wins the car with a probability of 2/3, so the player should switch (Wheeler 1991; Mack 1992; Schwager 1994; vos Savant 1996:8; Martin 2002). In order to convert this popular story into a mathematically rigorous solution, one has to argue why the probability that the car is behind door 1 does not change on opening door 2 or 3. This can be answered by an appeal to symmetry: under the complete assumptions made above, nothing is changed in the problem if we renumber the doors arbitrarily, and in particular, if we switch numbers 2 and 3. Therefore, the conditional probability that the car is behind door 1, given the player chose 1 and Monty opened 2, is the same as the conditional probability that the car is behind door 1, given the player chose 1 and Monty opened 3. The average of these two (equal) probabilities is 1/3, hence each of them separately is 1/3, too."

As I understand it, if a contestant chooses a door at random, the remaining doors will both have goats 1/3 of the time and a goat/car mix 2/3 of the time. 1/3 of the time the remaining doors both have goats, so by switching you would lose on this 1/3 of occasions. But on the other 2/3 of occasions the host will pick the door with a goat to avoid the door with the car. Since this is the case 2/3 of the time, by switching to the other door not chosen by the host you will succeed 2/3 of the time.

I'm sure someone will tell me if i've misunderstood the problem 121.73.7.84 (talk) 08:10, 22 March 2010 (UTC)

You seem to understand it correctly. Just out of curiosity do you find the revised solution section as proposed here easier to understand? -- Rick Block (talk) 14:08, 22 March 2010 (UTC)
Well done 121.73.7.84! You have managed to fight your way through the ridiculously obfuscated explanation given in this article. I would explain the answer it like this: if you swap, you always get the opposite of your initial choice and you have a 2/3 chance of initially picking a goat, but unfortunately, some editors here have a attachment to a perversely pedantic published paper on the subject. This means that simple explanations, such as the one that I have given are regarded as wrong and must be replaced with complicated explanations such as you have quoted. The simple mathematical puzzle that everyone gets wrong is thus turned into an obscure and complicated probability problem that nobody cares about. You are welcome to join in the discussion on how to improve the article. Martin Hogbin (talk) 18:35, 22 March 2010 (UTC)
Martin - don't blame Morgan et al. for this. The paragraph quoted is from the existing "Popular Solution" section that was forked from a common "Solution" section essentially at your request. That you "popularists" have turned this section into an unreadable mess has nothing whatsoever to do with Morgan et al. (the perversely pedantic paper you're referring to). I'll repeat my question to 121.73.7.84 - do you find the solution section as proposed here easier to understand? -- Rick Block (talk) 19:17, 22 March 2010 (UTC)
You cannot blame me for the quoted explanation. My view has always been that we should have some simple, convincing solutions like the one I give above, that do not mention anything about conditions or door numbers or the like first, with full explanations and aids to understanding, then, after that we should discuss which door the host opens, and why it may or may not matter. Martin Hogbin (talk) 19:50, 22 March 2010 (UTC)
Rick, the reason the "popular" section has been "turned into an unreadable mess" is the "probabilists" (for lack of a better word, obtained by comparing the "Popular Solution" section to the poorly-named "Probabilistic Solution" section) insist the "popular" solution must be framed in a way that distinguishes it from the "Probabilistic Solution." If the equal-choice version is accepted as the main problem, as all but a few pedantic papers do, then Morgan's p is 1/2 and Moragn's whole point disappears. The "popular solution" can then be written without having to cite references every third word or so, and we can quit inserting weasel words into the article that are there only defend the text against this argument. In other words, frame the main article around what we all should know was intended (no p), and pull Moragn's analysis out into a side case.
But while I'm here, let me ask you a question. In his book The Drunkard's Walk, Leonard Mlodinow admitted he altered Savant's wording when he asked the MHP this way: "Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, and the host, who knows what's behind the doors, opens another door, which has a goat. He then says to you, 'Do you want to switch your pick to the other door?' Is it to your advantage to switch your choice?" His answer, and he is a citable source, was that the probabilities were 1/3 and 2/3 for the two doors. Do you think that is correct? JeffJor (talk) 15:59, 23 March 2010 (UTC)
In the light of the ongoing discussion here, this seems to me a rather misplaced, not to say stupid, question.Nijdam (talk) 09:07, 24 March 2010 (UTC)
The analysis where p is a variable is already pulled out into a side case. For the 400th time, the issue is NOT whether p is taken to be 1/2 or not, but whether the question is asking about the conditional probability given the player knows which door she has picked and which door the host has opened or the chance of winning by switching averaged across all possible player picks and doors the host might open. These are always different questions even though they have the same answer if p is 1/2. The "unconditional solutions" either (pick one) are not addressing the conditional probability, or are implicitly assuming (without comment or justification - except Selvin's where he later explicitly said what assumptions his unconditional solution is based on) that the conditional answer is the same as the average chance of winning by switching.
Is your question whether I agree with Mlodinow's answer, or whether I agree it is Mlodinow's answer? For the purposes of this page the first question is irrelevant. I haven't read the source but I'm perfectly willing to believe it says what you're claiming it says. -- Rick Block (talk) 23:56, 23 March 2010 (UTC)

A reasonable person, including those who have published various 'simple' or 'omniconditional' solutions, could assume that 'Suppose you're on a game show...' means that the host will not be indicating to the contestant where the car is located. Accordingly, that person would not see a need to redundantly state 'p = 1/2'.

As the 'simple' or 'omniconditional' solutions are true in all cases, the specific scenario where the contestant chooses door 1 and the host reveals a goat behind door 3 is nothing more than a subset, requiring no additional attention. Glkanter (talk) 04:45, 24 March 2010 (UTC)

Sorry folks for restarting the same argument yet again. I just wanted to point out that we have had yet another reader who has found the article hard to understand in its current form. If you look back through the archives you see quite a few. There are, no doubt, many more who have not understood the explanation given here but have not bothered to comment. Martin Hogbin (talk) 09:49, 24 March 2010 (UTC)
I beg to differ with you both, Rick and Nijdam. Whether the issue is "is p taken to be 1/2 or not," "is the question asking about the conditional probability given the player knows which door she has picked," or even "is the quesiton about the chance of winning by switching averaged across all possible player picks" are related to the question I asked. That's why I asked it, because too many people refuse to see the relationship. And being intentionally evasive about it does not help. Is Mlodinow's answer of 1/3 correct, by your understanding of the question I quoted? And to address your other point, Rick, the "unconditional solution" is to a problem that is isomorphic with what you call the "conditional" one, and isomorphism is an acceptable way to express a correct solution. One that can greatly simplify understanding for lay people, which is waht we are trying to accomplish. The article is confusing because it has to be written, due to the nature of your objections, without such isomorphism. It spends far too much time explaining how each thought can be found in literature. And on why that literature is "correct" or "scientific" or "probabilistic" or any of a number of other weasel words inserted just to combat this argument, instead of simply making itself clear even if that means compromising on what you think is necessary rigor. JeffJor (talk) 14:01, 24 March 2010 (UTC)
After further thought, I decided to be less evasive (although I wasn't trying to be - I wanted a first-impression answer). Mlodinow intentionally dropped the references to the door numbers in his question, quite possibly to avoid this very argument. And he is a citable source. Setting aside for the moment the issues of when, how often, by whom, and just how accurate the various versions of the question are, would it be acceptable to lead off the article with Mlodinow's version - call it maybe the "Simple Problem" - and answer it with a simple solution - an isomorph if you insist - that does not rely on conditional probability or any history? That way the unintuitive result can become established as correct before any need to demonstrate rigor, or explain who thinks who else is wrong or misinterpreted anything. Then go into details about the various versions and history, which is really an enitrely separate issue. JeffJor (talk) 14:31, 24 March 2010 (UTC)
Jeff - you're the one here focused on what is "correct". I'm perfectly satisfied with discussing what is published in reliable sources. I really couldn't care less what you think is "correct". You really shouldn't care what I think is "correct". As far as leading with Mlodinow's version, it is not the most well known version (which means most of the literature is not referring to this version, i.e. it's as much a variant as a version where the host's preference between goats is explicitly treated as a variable p) so starting with it seems like a curious choice. Once again, I'd request comments on the readability of the solution section proposed here, which is intended to present three types of solutions published by reliable sources without characterizing any of them as "more correct" than any other. -- Rick Block (talk) 14:59, 24 March 2010 (UTC)
The article should lead off with the original problem formulation that caused the ruckus and not with anybody's favoured version that might be found in literature. Similarly it is a bad idea to pick a particular (re)definition to make one's favourite solution rigorous or without ambiguities. Attempting such things is imho just fooling readers, furthermore much of the problem history and arguments about it cannot be understood without knowing the original statement and its ambiguities. What can do imho however, is to have a more modularized approach in several section. From my perspective Morgan doesn't have to be mentioned in the "popular section", but the popular section needs to explain how it resolves the ambiguity. You can also make a (sourced) argument under which aspect popular and conditional are actually equivalent, but you cannot simply skip that information and use it as an argument for presenting the reader with a different problem formulation.--Kmhkmh (talk) 15:14, 24 March 2010 (UTC)
The real, the virtual "ploblem" is that Morgan et al., who never did address the "MHP" at all (!) (Monty Hall Problem: "1 door versus 2 doors" - easily recognizable - "33:67 versus 50:50"), but who addressed some osseous "Morgan et al. Ploblem", far aside from the famous paradox (!), and who have been rumbled, debunked and unmasked therefore, still have some insular fans: Some kinda narror-hidebound still feel attracted to some deviantonist relevation: What will the "Morgan et al.-moderator" do? Will he "always open a door"? Or not? Will he "always offer to switch"? Or not? And "if he does, will he flash obscene secrets?", - or not?... - and so on, and so on. Far aside the famous paradox. Probability theory seems to be in demand. Still some math scientists deal with such fallacious issues and cling to similar "conclusions", far aside the MHP. Too bad. World's most eminent scientists have exposed that Morgan was and still is a false prophet in this respect, long ago. They're hushed up in WP. Not distinguishing the MHP from the "Morgan-problem", being an entirely different issue... A pity and a shame for WP. -- Gerhardvalentin (talk) 21:23, 24 March 2010 (UTC)
Ah yes, you know what the real problem is. We don't need to try to understand or care what multiple sources have to say (Morgan et al. are FAR from the only source saying the question the MHP asks is about conditional probability, and how the host chooses between two goats therefore matters). I know what the real problem is as well. See WP:The Truth. -- Rick Block (talk) 00:26, 25 March 2010 (UTC)
The main reason for people to be critical of Morgan, is they otherwise have to admit, they don't understand the MHP. Nijdam (talk) 17:13, 25 March 2010 (UTC)
Nijdam, your argument, above, would be more convincing if you were to reference the original source, for example, like this. Glkanter (talk) 00:58, 26 March 2010 (UTC)

More discussion

Rick, I have changed the title above. As I have said before, we have should all congratulate ourselves on remaining civil throughout this baffling difference of opinion. Despite years of discussion about every detail of the problem, neither side has managed to convince the other or even understand the view of the other side. Our mediator Andrevan admits that this difference of opinion is hard particularly to deal with.
I should add there is an equally strong difference of opinion on what the sources say and how best to present this, so an appeal to 'just say what the sources do' does not solve the problem. There are fundamental, intractable and entrenched differences of opinion here. We are all fed up with the lack of progress but our only option is to carry on with discussion, mediation, and other dispute resolution tactics until we arrive at consensus or compromise. Until that time there are likely to be an continued expression of different opinions on this page.
I believe your section title might be construed by some as the first step in name-calling and incivility and have therefore taken the liberty of changing it. Martin Hogbin (talk) 10:07, 26 March 2010 (UTC)
Fine - and it's probably a good idea. -- Rick Block (talk) 00:50, 27 March 2010 (UTC)
Rick, it is hard to understand where to comment on that link, since it is an archived version of this page. But that explanation is the exact kind that user:121.73.7.84 said turns people off, so I suspect he either did read it, or started to didn't want to finish because of the reasons he gave. It starts off confusing the issue by needlessly pointing out that there are "varieties of solutions." That will discourage readers who only want to be given THE solution (whatever that means). It introduces issues (some crossed out) whose importance isn't mentioned, and that are not used in any of the three. That will also confuse people. "Conditional probability" is a math buzzword that will scare them. JeffJor (talk) 19:47, 25 March 2010 (UTC)
Can we let the user speak for him/herself? As I read his/her comments, the proposed section is exactly the kind of solution he/she is suggesting. Non-mathematical, everyday terminology first, followed by a more mathematically rigorous solution. -- Rick Block (talk) 01:59, 26 March 2010 (UTC)
And any mention of how "in some variations of the problem the conditional probability may differ" will put them off further - they will wonder how the probability can differ if everything happens the exact same way in the cases you are comparing, but the probability ends up different. I know where the difference you think exists is, but they won't. At that point they will be convinced that they will not be able to understand anything you have written, and stop trying to. Anything they actually understood up to that point will also be dismissed as "I must have misunderstood that, too." I know you feel you need to say all that (the reason you feel that way is because you think it is bad to use only the incorrect solution, despite claiming that you don't care which is correct), but it is unnecessary and confusing at the part of the article where this section would be inserted. All of what you want mentioned, that I said you shouldn't, can be mentioned in the places where the variations they affect are discussed. JeffJor (talk) 19:47, 25 March 2010 (UTC)
The reason I feel the article needs to say "all that" has nothing to do with what I think is "correct", but to ensure the article represents what reliable sources say about the problem in an NPOV fashion. Presenting only the simple solutions, and deferring the conditional solution to a later "mathematical variants" section is (IMO) POV. This has nothing whatsoever to do with what I think is correct or incorrect. -- Rick Block (talk) 01:59, 26 March 2010 (UTC)
It also gets away from the fact that the question is not "what is the probability," but "should you switch." Any probabilities you need to evaluate that question are what you have called "Bayesian Probability," not the "Frequentist Probability" Morgan calculates, and that you use in your examples. The difference is not, as you have said before, in the definition of probability. The difference is in the state of knowledge of the random process for which you are calculating a probability. A Bayesean will make assumptions about that process that a Frequentist will not. The contestant has to be a Bayesean, and base her decision on what she beleives the random process to be whether or not it is fully explaned, not on what it actually is on day the game show is filmed. It is only if the contestant beleives that a host bias exists, that Morgan's solution has any impact on the contestant's decision. The contestant has to decide based on what she thinks would happen if she repeated this same game, with the same rules, 900 times. Not on what happens to 900 other players of the game. If the host reveals no information to her about Door #2 (i.e., she doesn't beleive there is a bias), then the probabilties she should use are 1/3 and 2/3. These aren't "averages" as some have claimed, they are "expectations" based on having no information about Door #2. The two are calculated the same, but conceptually different. And there is no contradiction if her Baysean probability is not the same as what a Frequentist, who knows what the random process actually is - gets.
The better "simple solution," but you porbably will call it OR, is this: The player initially had a 1/3 chance of having picked the car, and the player will lose by switching. In the other 2/3 of the time, the host can open one of the other doors without revealing any information about the last door. In all of those cases, the player wins by switching. Said another way, if the contestant played this game 900 times, then regardless of which door she initially picked and which door the host opened, she would win by switching about 600 times. This says nothing incorrect, that a Morganist would object to. It just pulls a sleight of hand trick to dismiss the possibility that opening Door 3 reveals information about Door 2, which is where that difference I mentioned earlier is. JeffJor (talk) 19:47, 25 March 2010 (UTC)
If the game was to be repeated 900 times, the player, to be sure of the right answer for her situation, will choose all 900 times door 1. Only 450 times door No 3 will be opened, and will she has the option, as in the original situation, to switch to door No 2. And in 300 of thes 450 times se will get the car if she switches. But this differs conceptually from the 600 of 900 times, if she just chooses some door and decides to switch anyhow. Nijdam (talk) 22:55, 25 March 2010 (UTC)
Jeff - If you're putting words into the mouths of those who present simple solutions to "fix" those solutions, anyone who knows what OR is would agree that what you're doing is OR. The sources are perfectly capable of speaking for themselves. -- Rick Block (talk) 01:59, 26 March 2010 (UTC)

Question for user:121.73.7.84

I think the question for user:121.73.7.84 has been buried (at least twice) in the thread above. I'll ask again - is the revised solution section as proposed here easier to understand?

If anyone else would like to respond to this question feel free. However, if you want to bitch and moan about how certain references interpret the problem, please do it elsewhere. -- Rick Block (talk) 22:34, 24 March 2010 (UTC)


Crikey, i didn't realise I had stumbled into a hotly debated topic. I am not a mathematician and wikipedia is not a site specifically for mathematicians so the feedback I would offer is to also give an explanation of the problem that does not require understanding of mathematical jargon. A mathematical explanation using correct scientific terms and equations should certainly be included, however, (imo) this tends to switch most people off - hence this suggestion for the dual forms of explanation. In my experience the general public tend to think in verbal-type concepts with everyday words rather than equations and official mathematical terminology.

Best of luck 121.73.7.84 (talk) 07:10, 25 March 2010 (UTC)

Did you read the proposed solution, linked above? -- Rick Block (talk) 13:09, 25 March 2010 (UTC)
In case it's not clear, the proposed solution section was a work in progress - please ignore the strkeouts (pretend those phrases aren't there), and comment here. -- Rick Block (talk) 01:30, 26 March 2010 (UTC)

Are you sure you've linked me to the section you intended? I don't think the explanation you have linked to is easily understandable for a lay person. Terms such as "reasonable heuristic", "conditional probability", "by reason of symmetry", "this subset differs from the total set regarding a property influencing the probability", etc, etc., -while perhaps mathematically correct terms- are not remedial enough when attempting to explain an already abstract concept to a general reader. 121.73.7.84 (talk) 12:37, 26 March 2010 (UTC)

I'm not sure what you're reading. I mean the suggestion in the collapsed section below. -- Rick Block (talk) 14:27, 26 March 2010 (UTC)
Proposed text
Solution
Different sources present solutions to the problem using a variety of approaches.
Simplest approach
The player initially has a 1/3 chance of picking the car. The host always opens a door revealing a goat, so if the player doesn't switch the player has a 1/3 chance of winning the car. Similarly, the player has a 2/3 chance of initially picking a goat and if the player switches after the host has revealed the other goat the player has a 2/3 chance of winning the car. (some appropriate reference, perhaps Grinstead and Snell)
What this solution is saying is that if 900 contestants all switch, regardless of which door they initially pick and which door the host opens about 600 would win the car.
Enumeration of all cases where the player picks Door 1
If the player has picked, say, Door 1, there are three equally likely cases.
Door 1 Door 2 Door 3 result if switching
Car Goat Goat Goat
Goat Car Goat Car
Goat Goat Car Car
A player who switches ends up with a goat in only one of these cases but ends up with the car in two, so the probability of winning the car by switching is 2/3. (some appropriate reference, perhaps vos Savant)
What this solution is saying is that if 900 contestants are on the show and roughly 1/3 pick Door 1 and they all switch, of these 300 players about 200 would win the car.
The probability of winning by switching given the player picks Door 1 and the host opens Door 3
Tree showing the probability of every possible outcome if the player initially picks Door 1
This is a more complicated type of solution involving conditional probability. The difference between this approach and the previous one can be expressed as whether the player must decide to switch before the host opens a door or is allowed to decide after seeing which door the host opens (Gillman 1992).
The probabilities in all cases where the player has initially picked Door 1 can be determined by referring to the figure below or to an equivalent decision tree as shown to the right (Chun 1991; Grinstead and Snell 2006:137-138 presents an expanded tree showing all initial player picks). Given the player has picked Door 1 but before the host opens a door, the player has a 1/3 chance of having selected the car. Referring to either the figure or the tree, in the cases the host then opens Door 3, switching wins with probability 1/3 if the car is behind Door 2 but loses only with probability 1/6 if the car is behind Door 1. The sum of these probabilities is 1/2, meaning the host opens Door 3 only 1/2 of the time. The conditional probability of winning by switching for players who pick Door 1 and see the host open Door 3 is computed by dividing the total probability of winning in the case the host opens Door 3 (1/3) by the probability of all cases where the host opens Door 3 (1/2), therefore this probability is (1/3)/(1/2)=2/3.
???   between   - Given the player has picked door 1, in the cases the host then opens door 3, switching wins with "probability" of 3/3 if the car is behind door 2, and looses "only?" with "probability" of 3/3 if the car is behind door 1. - (delete this if you like)  -- Gerhardvalentin (talk) 20:07, 28 March 2010 (UTC)
Although this is the same answer as the simpler solutions for the unambiguous problem statement as presented above, in some variations of the problem the conditional probability may differ from the average probability and the probability given only that the player initially picks Door 1, see Variants below. Some proponents of solutions using conditional probability consider the simpler solutions to be incomplete, since the simpler solutions do not explicitly use the constraint in the problem statement that the host must choose which door to open randomly if both hide goats (multiple references, e.g. Morgan et al., Gillman, ...).
What this type of solution is saying is that if 900 contestants are on the show and roughly 1/3 pick Door 1, of these 300 players about 150 will see the host open Door 3. If they all switch, about 100 would win the car.
A formal proof that the conditional probability of winning by switching is 2/3 is presented below, see Bayesian analysis.


Car hidden behind Door 3 Car hidden behind Door 1 Car hidden behind Door 2
Player initially picks Door 1
Player has picked Door 1 and the car is behind Door 3 Player has picked Door 1 and the car is behind it Player has picked Door 1 and the car is behind Door 2
Host must open Door 2 Host randomly opens either goat door Host must open Door 3
Host must open Door 2 if the player picks Door 1 and the car is behind Door 3 Host opens Door 2 half the time if the player picks Door 1 and the car is behind it Host opens Door 3 half the time if the player picks Door 1 and the car is behind it Host must open Door 3 if the player picks Door 1 and the car is behind Door 2
Probability 1/3 Probability 1/6 Probability 1/6 Probability 1/3
Switching wins Switching loses Switching loses Switching wins
If the host has opened Door 3, these cases have not happened If the host has opened Door 3, switching wins twice as often as staying

Yes, that is much better 121.73.7.84 (talk) 22:24, 26 March 2010 (UTC)

I agree that is better, but maybe not the best we could have. One of the problems is that once you have got it, it can become quite hard to see what the problem was originally. Thus we need to aim the first part of the article at someone who has never seen the problem before. Like many things, the answer is obvious when looked at the right way. Martin Hogbin (talk) 09:32, 27 March 2010 (UTC)

@user:121.73.7.84 - When reading the suggested wording above does the third section make you think that the first two (simple) explanations are wrong, or that they're simply other ways to look at the problem? And, although I assume the third (conditional probability) section is mostly a "switch off" section, have you puzzled out what it is saying and if so do you find it helpful (and, if not, do you find it interferes with your understanding of the other two sections)? -- Rick Block (talk) 17:24, 28 March 2010 (UTC)


Which is the third section? 121.73.7.84 (talk) 23:28, 28 March 2010 (UTC)


I've also just noticed the following at the beginning in the FAQ section:

Q: There are clearly two doors, how can it not be 50/50? A: The simplest explanation may be that any player's initial probability of not picking the car is 2/3. Monty's action of revealing a goat behind a door doesn't change that. Therefore, the player should accept the offer to switch. It doubles his likelihood of winning the car.

This is terrible since it IS Monty's actions that are ultimately the key to understanding this problem. The above doesn't "simply" explain the problem at all, and it sends people off on a tangent away from an analysis of Monty's actions which is the key to this problem. The doubling of probability doesn't make sense otherwise. 121.73.7.84 (talk) 23:52, 28 March 2010 (UTC)

What I mean by the third section is the third subsection. The first subsection is Simplest approach, the second is Enumeration of all cases where the player picks Door 1, the third is The probability of winning by switching given the player picks Door 1 and the host opens Door 3. -- Rick Block (talk) 04:38, 29 March 2010 (UTC)

Question for Martin

My understanding is the only thing Martin is arguing for at this point is to include the existing "Aids to understanding" section between the existing "Popular solution" and the existing "Probabilistic solution" sections (so as not to confuse the reader with the complexities of conditional probability). The confusing verbiage user:121.73.7.84 specifically mentions (above) is in the current "Popular solution" section. Is the "just rearrange things" proposal an example of what might be better than what I've proposed - or are more changes envisioned than just rearranging things? If more changes, can someone create a draft we can all look at (doesn't have to be perfect)? Here's the simple rearrangement. -- Rick Block (talk) 16:37, 27 March 2010 (UTC)

To explain, there have been many versions of the 'Popular solution' section, some better than others. The best, in my opinion, have made no mention of the fact that it might matter which door the host opens when he has a choice. This is what I would want to start with. I have edited this version to show what I would want to.
Martin's proposal

Popular solution

[I have deleted this as it serves no useful purpose Martin Hogbin (talk) 17:19, 27 March 2010 (UTC)]

The analysis can be illustrated in terms of the equally likely events that the player has initially chosen the car, goat A, or goat B (Economist 1999):

1.
Monty-CurlyPicksCar.svg
Host reveals
either goat
Pfeil.png

Pfeil.png
Monty-DoubleSwitchfromCar.svg
Player picks car
(probability 1/3)
Switching loses.
2.
Monty-CurlyPicksGoatA.svg Host must
reveal Goat B

Pfeil.png
Monty-SwitchfromGoatA.svg
Player picks Goat A
(probability 1/3)
Switching wins.
3.
Monty-CurlyPicksGoatB.svg Host must
reveal Goat A

Pfeil.png
Monty-SwitchfromGoatB.svg
Player picks Goat B
(probability 1/3)
Switching wins.
The player has an equal chance of initially selecting the car, Goat A, or Goat B. Switching results in a win 2/3 of the time.

The above diagram shows that a player who switches always gets the opposite of their original choice, and since the probability of that choice being a goat is twice that of being a car, it is always advantageous to switch. In other words, the probability of originally choosing a goat is 2/3 and the probability of originally choosing the car is 1/3. Once Monty Hall has removed a "goat door," the contestant who chose the door with a goat behind it will necessarily win the car, and the contestant who originally chose the car will necessarily "win" the goat. Because the chances are 2/3 of being a contestant who originally chose a goat, probability will always favor switching choices.

Player's pick has a 1/3 chance while the other two doors have a 2/3 chance.
Player's pick has a 1/3 chance, other two doors a 2/3 chance split 2/3 for the still unopened one and 0 for the one the host opened

Another way to understand the solution is to consider the two original unchosen doors together. Instead of one door being opened and shown to be a losing door, an equivalent action is to combine the two unchosen doors into one since the player cannot choose the opened door (Adams 1990; Devlin 2003; Williams 2004; Stibel et al., 2008).

As Cecil Adams puts it (Adams 1990), "Monty is saying in effect: you can keep your one door or you can have the other two doors." The player therefore has the choice of either sticking with the original choice of door, or choosing the sum of the contents of the two other doors, as the 2/3 chance of hiding the car hasn't been changed by the opening of one of these doors.

As Keith Devlin says (Devlin 2003), "By opening his door, Monty is saying to the contestant 'There are two doors you did not choose, and the probability that the prize is behind one of them is 2/3. I'll help you by using my knowledge of where the prize is to open one of those two doors to show you that it does not hide the prize. You can now take advantage of this additional information. Your choice of door A has a chance of 1 in 3 of being the winner. I have not changed that. But by eliminating door C, I have shown you that the probability that door B hides the prize is 2 in 3.'"

Aids to understanding Why the probability is not 1/2

The contestant has a 1 in 3 chance of selecting the car door in the first round. Then, from the set of two unselected doors, Monty Hall non-randomly removes a door that he knows is a goat door. If the contestant originally chose the car door (1/3 of the time) then the remaining door will contain a goat. If the contestant chose a goat door (the other 2/3 of the time) then the remaining door will contain the car.

The critical fact is that Monty does not randomly choose a door - he always chooses a door that he knows contains a goat after the contestant has made their choice. This means that Monty's choice does not affect the original probability that the car is behind the contestant's door. When the contestant is asked if the contestant wants to switch, there is still a 1 in 3 chance that the original choice contains a car and a 2 in 3 chance that the original choice contains a goat. But now, Monty has removed one of the other doors and the door he removed cannot have the car, so the 2 in 3 chance of the contestant's door containing a goat is the same as a 2 in 3 chance of the remaining door having the car.

This is different from a scenario where Monty is choosing his door at random and there is a possibility he will reveal the car. In this instance the revelation of a goat would mean that the chance of the contestant's original choice being the car would go up to 1 in 2. This difference can be demonstrated by contrasting the original problem with a variation that appeared in vos Savant's column in November 2006. In this version, Monty Hall forgets which door hides the car. He opens one of the doors at random and is relieved when a goat is revealed. Asked whether the contestant should switch, vos Savant correctly replied, "If the host is clueless, it makes no difference whether you stay or switch. If he knows, switch" (vos Savant, 2006).

Another way of looking at the situation is to consider that if the contestant chooses to switch then they are effectively getting to see what is behind 2 of the 3 doors, and will win if either one of them has the car. In this situation one of the unchosen doors will have the car 2/3 of the time and the other will have a goat 100% of the time. The fact that Monty Hall shows one of the doors has a goat before the contestant makes the switch is irrelevant, because one of the doors will always have a goat and Monty has chosen it deliberately. The contestant still gets to look behind 2 doors and win if either has the car, it is just confirmed that one of doors will have a goat first.

Increasing the number of doors

It may be easier to appreciate the solution by considering the same problem with 1,000,000 doors instead of just three (vos Savant 1990). In this case there are 999,999 doors with goats behind them and one door with a prize. The player picks a door. The game host then opens 999,998 of the other doors revealing 999,998 goats—imagine the host starting with the first door and going down a line of 1,000,000 doors, opening each one, skipping over only the player's door and one other door. The host then offers the player the chance to switch to the only other unopened door. On average, in 999,999 out of 1,000,000 times the other door will contain the prize, as 999,999 out of 1,000,000 times the player first picked a door with a goat. A rational player should switch. Intuitively speaking, the player should ask how likely is it, that given a million doors, he or she managed to pick the right one. The example can be used to show how the likelihood of success by switching is equal to (1 minus the likelihood of picking correctly the first time) for any given number of doors. It is important to remember, however, that this is based on the assumption that the host knows where the prize is and must not open a door that contains that prize, randomly selecting which other door to leave closed if the contestant manages to select the prize door initially.

This example can also be used to illustrate the opposite situation in which the host does not know where the prize is and opens doors randomly. There is a 999,999/1,000,000 probability that the contestant selects wrong initially, and the prize is behind one of the other doors. If the host goes about randomly opening doors not knowing where the prize is, the probability is likely that the host will reveal the prize before two doors are left (the contestant's choice and one other) to switch between. This is analogous to the game play on another game show, Deal or No Deal; In that game, the contestant chooses a numbered briefcase and then randomly opens the other cases one at a time.

Stibel et al. (2008) propose working memory demand is taxed during the Monty Hall problem and that this forces people to "collapse" their choices into two equally probable options. They report that when increasing the number of options to over 7 choices (7 doors) people tend to switch more often; however most still incorrectly judge the probability of success at 50/50.

Simulation

Simulation of 30 outcomes of the Monty Hall problem

A simple way to demonstrate that a switching strategy really does win two out of three times on the average is to simulate the game with playing cards (Gardner 1959b; vos Savant 1996:8). Three cards from an ordinary deck are used to represent the three doors; one 'special' card such as the Ace of Spades should represent the door with the car, and ordinary cards, such as the two red twos, represent the goat doors.

The simulation, using the following procedure, can be repeated several times to simulate multiple rounds of the game. One card is dealt face-down at random to the 'player', to represent the door the player picks initially. Then, looking at the remaining two cards, at least one of which must be a red two, the 'host' discards a red two. If the card remaining in the host's hand is the Ace of Spades, this is recorded as a round where the player would have won by switching; if the host is holding a red two, the round is recorded as one where staying would have won.

By the law of large numbers, this experiment is likely to approximate the probability of winning, and running the experiment over enough rounds should not only verify that the player does win by switching two times out of three, but show why. After one card has been dealt to the player, it is already determined whether switching will win the round for the player; and two times out of three the Ace of Spades is in the host's hand.

If this is not convincing, the simulation can be done with the entire deck, dealing one card to the player and keeping the other 51 (Gardner 1959b; Adams 1990). In this variant the Ace of Spades goes to the host 51 times out of 52, and stays with the host no matter how many non-Ace cards are discarded.

Another simulation, suggested by vos Savant, employs the "host" hiding a penny, representing the car, under one of three cups, representing the doors; or hiding a pea under one of three shells.

Probabilistic solution

Morgan et al. (1991) state that many popular solutions are incomplete, because they do not explicitly address their interpretation of Whitaker's original question (Seymann), which is the specific case of a player who has picked Door 1 and has then seen the host open Door 3. These solutions correctly show that the probability of winning for all players who switch is 2/3, but without certain assumptions this does not necessarily mean the probability of winning by switching is 2/3 given which door the player has chosen and which door the host opens. This probability is a conditional probability (Morgan et al. 1991; Gillman 1992; Grinstead and Snell 2006:137; Gill 2009b). The difference is whether the analysis is of the average probability over all possible combinations of initial player choice and door the host opens, or of only one specific case—for example the case where the player picks Door 1 and the host opens Door 3. Another way to express the difference is whether the player must decide to switch before the host opens a door or is allowed to decide after seeing which door the host opens (Gillman 1992). Although these two probabilities are both 2/3 for the unambiguous problem statement presented above, the conditional probability may differ from the overall probability and either or both may not be able to be determined depending on the exact formulation of the problem (Gill 2009b).

Tree showing the probability of every possible outcome if the player initially picks Door 1

The conditional probability of winning by switching given which door the host opens can be determined referring to the expanded figure below, or to an equivalent decision tree as shown to the right (Chun 1991; Grinstead and Snell 2006:137-138), or formally derived as in the mathematical formulation section below. For example, if the host opens Door 3 and the player switches, the player wins with overall probability 1/3 if the car is behind Door 2 and loses with overall probability 1/6 if the car is behind Door 1—the possibilities involving the host opening Door 2 do not apply. To convert these to conditional probabilities they are divided by their sum, so the conditional probability of winning by switching given the player picks Door 1 and the host opens Door 3 is (1/3)/(1/3 + 1/6), which is 2/3. This analysis depends on the constraint in the explicit problem statement that the host chooses randomly which door to open after the player has initially selected the car.


Car hidden behind Door 3 Car hidden behind Door 1 Car hidden behind Door 2
Player initially picks Door 1
Player has picked Door 1 and the car is behind Door 3 Player has picked Door 1 and the car is behind it Player has picked Door 1 and the car is behind Door 2
Host must open Door 2 Host randomly opens either goat door Host must open Door 3
Host must open Door 2 if the player picks Door 1 and the car is behind Door 3 Host opens Door 2 half the time if the player picks Door 1 and the car is behind it Host opens Door 3 half the time if the player picks Door 1 and the car is behind it Host must open Door 3 if the player picks Door 1 and the car is behind Door 2
Probability 1/3 Probability 1/6 Probability 1/6 Probability 1/3
Switching wins Switching loses Switching loses Switching wins
If the host has opened Door 2, switching wins twice as often as staying If the host has opened Door 3, switching wins twice as often as staying

Mathematical formulation The above solution may be formally proven using Bayes' theorem, similar to Gill, 2002, Henze, 1997 and many others. Different authors use different formal notations, but the one below may be regarded as typical. Consider the discrete random variables:

: the number of the door hiding the Car,
: the number of the door Selected by the player, and
: the number of the door opened by the Host.

As the host's placement of the car is random, all values of C are equally likely. The initial (unconditional) probability of C is then

, for every value of C.

Further, as the initial choice of the player is independent of the placement of the car, variables C and S are independent. Hence the conditional probability of C given S is

, for every value of C and S.

The host's behavior is reflected by the values of the conditional probability of H given C and S:

  if H = S, (the host cannot open the door picked by the player)
  if H = C, (the host cannot open a door with a car behind it)
  if S = C, (the two doors with no car are equally likely to be opened)
  if H C and S C, (there is only one door available to open)

The player can then use Bayes' rule to compute the probability of finding the car behind any door, after the initial selection and the host's opening of one. This is the conditional probability of C given H and S:

,

where the denominator is computed as the marginal probability

.

Thus, if the player initially selects Door 1, and the host opens Door 3, the probability of winning by switching is

Sources of confusion When first presented with the Monty Hall problem an overwhelming majority of people assume that each door has an equal probability and conclude that switching does not matter (Mueser and Granberg, 1999). Out of 228 subjects in one study, only 13% chose to switch (Granberg and Brown, 1995:713). In her book The Power of Logical Thinking, vos Savant (1996:15) quotes cognitive psychologist Massimo Piattelli-Palmarini as saying "... no other statistical puzzle comes so close to fooling all the people all the time" and "that even Nobel physicists systematically give the wrong answer, and that they insist on it, and they are ready to berate in print those who propose the right answer."

Most statements of the problem, notably the one in Parade Magazine, do not match the rules of the actual game show (Krauss and Wang, 2003:9), and do not fully specify the host's behavior or that the car's location is randomly selected (Granberg and Brown, 1995:712). Krauss and Wang (2003:10) conjecture that people make the standard assumptions even if they are not explicitly stated. Although these issues are mathematically significant, even when controlling for these factors nearly all people still think each of the two unopened doors has an equal probability and conclude switching does not matter (Mueser and Granberg, 1999). This "equal probability" assumption is a deeply rooted intuition (Falk 1992:202). People strongly tend to think probability is evenly distributed across as many unknowns as are present, whether it is or not (Fox and Levav, 2004:637).

A competing deeply rooted intuition at work in the Monty Hall problem is the belief that exposing information that is already known does not affect probabilities (Falk 1992:207). This intuition is the basis of solutions to the problem that assert the host's action of opening a door does not change the player's initial 1/3 chance of selecting the car. For the fully explicit problem this intuition leads to the correct numerical answer, 2/3 chance of winning the car by switching, but leads to the same solution for slightly modified problems where this answer is not correct (Falk 1992:207).

According to Morgan et al. (1991) "The distinction between the conditional and unconditional situations here seems to confound many." That is, they, and some others, interpret the usual wording of the problem statement as asking about the conditional probability of winning given which door is opened by the host, as opposed to the overall or unconditional probability. These are mathematically different questions and can have different answers depending on how the host chooses which door to open when the player's initial choice is the car (Morgan et al., 1991; Gillman 1992). For example, if the host opens Door 3 whenever possible then the probability of winning by switching for players initially choosing Door 1 is 2/3 overall, but only 1/2 if the host opens Door 3. In its usual form the problem statement does not specify this detail of the host's behavior, nor make clear whether a conditional or an unconditional answer is required, making the answer that switching wins the car with probability 2/3 equally vague. Many commonly presented solutions address the unconditional probability, ignoring which door was chosen by the player and which door opened by the host; Morgan et al. call these "false solutions" (1991). Others, such as Behrends (2008), conclude that "One must consider the matter with care to see that both analyses are correct."

So, to be clear, are you saying that your modified version would be better than what I'm suggesting in the section above? -- Rick Block (talk) 19:41, 27 March 2010 (UTC)
Yes. The principle is simple. No confusing complications in the first bit, then explain everything. Martin Hogbin (talk) 20:27, 27 March 2010 (UTC)

In my opinion Rick Block's explanation is simpler for a random lay person to follow. Martin's explanation has more "switch off" elements which could fluster people. This is just my opinion. 121.73.7.84 (talk) 08:35, 28 March 2010 (UTC)

Thanks for your opinion. Perhaps you could say which bits are the 'switch off' elements. This would be useful to us all. Martin Hogbin (talk) 09:01, 28 March 2010 (UTC)
As long as we're posing questions, I've posed some above as well. -- Rick Block (talk) 17:25, 28 March 2010 (UTC)


Well, firstly I think starting with the diagram is not an easy way to initially explain this solution. Secondly the following text is hard to follow unless you already know the solution:

Quote: "The above diagram shows that a player who switches always gets the opposite of their original choice, and since the probability of that choice being a goat is twice that of being a car, it is always advantageous to switch. In other words, the probability of originally choosing a goat is 2/3 and the probability of originally choosing the car is 1/3. Once Monty Hall has removed a "goat door," the contestant who chose the door with a goat behind it will necessarily win the car, and the contestant who originally chose the car will necessarily "win" the goat. Because the chances are 2/3 of being a contestant who originally chose a goat, probability will always favor switching choices." 121.73.7.84 (talk) 23:24, 28 March 2010 (UTC)

The solution I gave, while perhaps sounding inelegant was at least simple, quick and -i believe- understandable 121.73.7.84 (talk) 23:26, 28 March 2010 (UTC)

On further reflection, if you just let the diagram explain itself and eliminate the inarticulate and confusing text, quote: "The analysis can be illustrated in terms of the equally likely events that the player has initially chosen the car, goat A, or goat B" - then it is clearer. Saying: "the analysis can be illustrated" or writing "initially chosen the the car, goat A, or goat B" when they mean "has either chosen goat A or Goat B or the car." 121.73.7.84 (talk) 23:39, 28 March 2010 (UTC)

Response from Martin

@121.73.7.84 Thanks for you continued interest in the article. We need more input from people less familiar with the problem to help us formulate the first part of the article in the most understandable way. The point of contention here is whether to include any of the details in the 'Probabilistic solution' (in particular, reference to the possibility that it might matter which door the host opens in some cases) in the first part of the article. We all want to make the explanations as clear as possible.

@Rick As you know the above text is not my wording nor my preferred layout, it was just my quick modification to your suggestion to remove the confusing stuff. I would really like to have several simple solutions at the start of the article, none of which mention the host legal door choice. Martin Hogbin (talk) 09:33, 29 March 2010 (UTC)


My advice is to start as if you're teaching your 5-year old the ABCs and then progress to more elegant, more mathematically technical explanations. 121.73.7.84 (talk) 10:00, 29 March 2010 (UTC)

@Martin - No, I don't know the text above is not your preferred wording or layout. It was my understanding you've been saying for a while all you really want is to move the "Aids to understanding" section. If this was to be only the first step of a more comprehensive overhaul I must have missed that. If you want to restructure and also rewrite, can you draft a sketch? I mean, it's not like you're averse to writing (with as much as you've written on this and other related pages you could have rewritten the entire article 100 times over by now). I don't know about anyone else, but I'm finding arguing in the abstract to be incredibly unproductive. In concrete terms, I think the version I've drafted above is in a completely different (better) league than what's currently shown as "Martin's suggestion".
@user:121.73.7.84 - what you're suggesting is the exact structure I'm aiming for in my suggestion above. Some background about the ongoing dispute if you're interested - what Martin and the other "anti-conditionalists" are arguing for is to have a more or less complete article mentioning only the "simple" explanations, and to present any more technical explanation effectively in an appendix, or "variations" or "experts only" section. The argument against this approach is it does not give equal weight to the viewpoint published by multiple sources that most published "simple" explanations (specifically, vos Savant's) are missing something (see selected quotes in a bullet list 3 or 4 replies into this thread). One of Wikipedia's core content policies is to take a neutral point of view (see WP:NPOV). Pushing significant published criticism of the simple solutions into an appendix IMO would be a gross violation of this policy. -- Rick Block (talk) 19:51, 29 March 2010 (UTC)

My proposal explained

At the time I first made my proposal I was reasonably happy with the 'Popular solutions' section and thus just moving a section would have been acceptable to me (although not ideal), however the main purpose of my suggestion is to end this long-standing argument. If we had essentially two separate sections that treat (interpret) the question in two different ways then there would be little remaining argument.

The first section would fully (including 'Aids to understanding') deal with the notable 'puzzle' question in which there is no consideration of the possibility that the legal door chosen by the host might matter. Once the decision to have this section is accepted, the main argument would no longer be relevant and we could all work together, following to WP policies, to improve the article. Contrary to what some may think, I do wish to rewrite this article on my own. The first section might contain several simple solutions from a variety of sources.

The second section could deal with the problem as interpreted by Morgan et al and fully explain its conditional nature, including mention of Morgan's criticism of the simple solutions. Later sections could treat the problem in other ways. Martin Hogbin (talk) 09:37, 30 March 2010 (UTC)

To what sources would you reference the interpretation that there is no "possibility that the legal door chosen by the host might matter"? To clarify, do you mean by this that the question is taken to be the unconditional question (in which case p does not matter), or that p must be taken as 1/2? And, how is it NPOV to exclude from this section published criticisms (and, again, for the 400,000th time, it's not just Morgan!) of such an interpretation, like say Lucas, Rosenhouse, and Schepler: ... any proposed solution to the MHP failing to pay close attention to Monty’s selection procedure is incomplete?
Note, specifically, this section of WP:NPOV, which says (in part): Segregation of text or other content into different regions or subsections, based solely on the apparent POV of the content itself, ... may also create an apparent hierarchy of fact: details in the main passage appear "true" and "undisputed", whereas other, segregated material is deemed "controversial", and therefore more likely to be false — an implication that may not be appropriate. A more neutral approach can result from folding debates into the narrative, rather than distilling them into separate sections that ignore each other. -- Rick Block (talk) 14:00, 30 March 2010 (UTC)

Has anyone actually watched the show itself?

I've read through the archives, and seen some fascinating material covered there (even though, as a non-mathematician, some of it flew over my head). One obvious question comes to mind, though - has anyone ever actually gone through past tapes of LMAD to look for instances of this specific game, and found out what the real-world results ended up being (assuming that there's a sufficient sample size)? One would think that if there were enough instances, then the results would speak for themselves. Bcdm (talk) 07:25, 18 April 2010 (UTC)

Curiously, the show itself is of little relevance to solving what has become known as the Monty Hall problem. On the actual show Monty Hall never actually gave the player the option to swap doors. The scenario generally known as the Monty Hall problem was first proposed by Selvin in a letter to 'The American Statistician' and although it referred to the show it proposed a question, 'Should the player swap?' that was never asked on the actual show. Some years later, this version of events took on a life of its own when Whitaker asked a similar question to the 'Ask Marilyn' column in 'Parade' magazine. Martin Hogbin (talk) 08:24, 18 April 2010 (UTC)

Would it be accurate to say

Since 2/3 doors are the wrong choice, the chances are that you picked the wrong door in the first place, so when shown the other wrong one, the remaining one is more likely to be correct since it's not the one you know is wrong, and not the one you know to be probably wrong. --66.66.187.132 (talk) 03:45, 5 May 2010 (UTC)

That's one way to look at it. Another way to look at it, given you've picked door 1 and host has opened door 3, is that the host has a 100% chance of opening door 3 if the car is behind door 2 (1/3 chance of this happening) but only a 50% chance of opening door 3 if the car is behind door 1 (1/6 chance of this happening). The situation is similar if the host opens door 2 (1/3 chance the car is behind door 3 and 1/6 chance it's behind door 1). So whichever door the host opens, the chances are twice as good that the car is behind the other unopened door than the one you originally picked - because the host has 2 choices for which door to open if the car is behind the door you pick, but has only one choice if the car is not behind the door you originally picked. -- Rick Block (talk) 04:06, 5 May 2010 (UTC)

Yes, 66.66.187.132, that's all there is to it. The 2/3 chance that your door choice is wrong is not changed by the host's actions. And since the total probability has to add up to 1, the remaining door has only a 1/3 chance of being wrong (more importantly, a 2/3 chance of being the car). Glkanter (talk) 05:55, 5 May 2010 (UTC)

This article is overly complicated, confusing and just plain wrong in more than one place.

The OVERALL odds of winning the game ARE in fact 1/2!

Yes if you play the "switch strategy" the odds of winning are 2/3 and staying will result in 1/3. However the contestant is asked to make to choices: 1) choose one of three doors 2) stay or switch after a goat is revealed

If both of these choices are made randomly than the odds of winning the car are 1/2. One can prove it two ways:

1) averaging strategies: one third plus two thirds is one, one over two is one half.

2) one car, two doors (what are you not getting here?)

The event matrices are good visual descriptions on how switching improves your odds but the article fails to differentiate between the odds of winning with a random selection and acting on a strategy of switching or staying.

Explaining this will leave the reader more satisfied. I would be happy to rewrite the article if that bot will stop reverting my edits. Whatever, it just wouldn't be wikipedia if articles weren't full of wrong. —Preceding unsigned comment added by 101glover (talkcontribs) 09:29, 5 May 2010 (UTC)

Yes, you are correct if all choices made by the player are random the probability that the player will win is 1/2 but that is not what the Monty Hall problem is all about. The most famous statement of the problem is given at the start of the article:
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
Note the question that I have put in bold. The questioner wants to know if it is better to swap or not. Vos Savant answered correctly that it is better to swap and that you double your odds of winning if you do so. Do you agree with this? Martin Hogbin (talk) 10:57, 5 May 2010 (UTC)
That's not enough. You have to include the condition that Monty ALWAYS behaves this way. That is, he ALWAYS chooses a door that he knows has a goat and ALWAYS opens that door. If you give Monty the CHOICE of opening the door or not, then there is no unique answer. For example, Monty From Hell: Monty only gives you a choice if you have chosen the door with the prize. Angelic Monty: Monty only gives you a choice if you have chosen a door with a goat.
Vos Savant was careless when she originally stated the problem, because she neglected to state this condition clearly (actually, at all). Bill Jefferys (talk) 23:25, 5 May 2010 (UTC)
To justify the answer that the chance of winning by switching is 2/3 knowing which door the player picked and which door the host opened 1) the car must be placed randomly, 2) the host must always open a losing door and make the offer to switch, and 3) the host must choose between two losing doors with equal probability. It can certainly be argued that vos Savant implied all of these, and most people apparently assume these conditions whether they're explicitly included in the problem statement or not.
However, back to the original objection here, the problem is definitely not talking about the chances if the player makes a random choice of whether to switch or not. Perhaps a more clear way to ask the question (assuming the clarifications about the car placement and host's behavior) is to ask about an expected frequency, i.e. if the show were on 900 times and the players initially picked a random door, how many of the players who pick door 1 and then see the host open door 3 would win if they switch?
If their initial pick is random, we'd expect about 300 players to initially pick door 1. If the car is placed randomly, of the 300 who picked door 1 we'd expect the car to be behind each door about 100 times. If the host opens door 3, it's because the car is behind door 2 (100 times) or the host has chosen to open door 3 when the car is behind door 1 (50 times). So, there will be about 150 players who choose door 1 and see the host open door 3, and of these 100 win if they switch. The chance of winning the car by switching (given the player picks door 1 and sees the host open door 3) is therefore 100/150 = 2/3. Of course, if all 150 of these players randomly choose whether to switch, about 75 will switch and 75 will stay with their initial choice. Of the 75 who switch about 2/3 (50) will win the car and of the 75 who stay about 1/3 (25) will win the car, so about 75 altogether (i.e. 1/2). This doesn't mean the chances of winning by switching are 1/2 but rather the chances of winning by making a random choice of whether to switch are 1/2.
See the difference? -- Rick Block (talk) 04:34, 6 May 2010 (UTC)
As Rick has said, there were some significant assumptions made by vos Savant that she did not male clear in her reply. These have been the subject of later criticism of her response but they are fully covered in this article.
The point being made by 66.66.187.132 was a different one. He was talking about the overall probability of winning if the player chooses randomly whether to switch or not, which is 1/2 but which is a question that I have never seen asked before and is not generally considered to be the Monty Hall problem. I guess he was misled by use of the term 'overall'. Martin Hogbin (talk) 08:53, 6 May 2010 (UTC)
Martin - of course you've seen this question before. It's vos Savant's "little green woman" variant (from her 3rd column about the problem). -- Rick Block (talk) 13:16, 6 May 2010 (UTC)
Yes you are quite right Rick, I had forgotten about that. Martin Hogbin (talk) 14:56, 6 May 2010 (UTC)
A lot of grief could be avoided if the article, instead of quoting vos Savant's flawed statement, instead led off with a correct statement of the problem, for example, as Martin Gardner originally put it (I don't have a copy of his book available but I am morally certain that he would have been careful to give an accurate statement of the problem).
As it is, leading off with vos Savant's flawed version and then trying to clean it up in the next paragraph or two just sows confusion. Bill Jefferys (talk) 17:14, 6 May 2010 (UTC)
Interesting suggestion. Thanks for your input. Glkanter (talk) 17:49, 6 May 2010 (UTC)
I assume by "Savant's flawed version" you actually mean Whitaker's version, in that case the answer is no. We cannot use Gardner because strictly speaking his original problem is not MHP and we cannot replace the ambiguous "original" MHP (Whitaker) by a later non ambiguous one. This is because WP is not in the business of defining the problem but has to report as it is and furthermore the ambiguity is a central part of the problem itself and much of its controversy.--Kmhkmh (talk) 18:41, 6 May 2010 (UTC)
I disagree. By far, the most notable and well known version of the MHP is as a mathematical puzzle in which all necessary assumptions are made for the solution to be simple with the player having a 2/3 chance of winning if they switch. This was actually the question that vos Savant intended to answer. Unfortunately, she did not make some of her assumptions clear until after she had given her reply. Martin Hogbin (talk) 18:50, 6 May 2010 (UTC)
Frankly you lost me completely here. Earlier one of your main arguments was that the MHP is not a math problem/puzzle and now you're telling me it is? That aside, I don't quite see, where your statement contradicts what I've wrote above, so where is the actual disagreement?--Kmhkmh (talk) 19:23, 6 May 2010 (UTC)
Martin - I can't tell who you're disagreeing with here, Bill or Kmhkmh. I thought you have consistently argued that the most notable and well known version is Whitaker's (as published by vos Savant). In the current structure of the article (problem, solution, explanations, variants, history) the problem section starts with Whitaker's version, mentions the ambiguities, and presents a fully unambiguous version. This understanding of the problem (which I think is the mathematical puzzle you're referring to) is the topic of the solution and explanations sections. Variations are covered in the variants section. I think this structure matches the bulk of the references, many of which use Whitaker's version (modulo minor misquotings) but then present a solution that actually applies only to the fully unambiguous version (usually ignoring the ambiguities).
To Bill's point - I think the article essentially already does this. Although the "Problem" section starts off with Whitaker's version it proceeds directly into an unambiguous version. For summary in the lead, Whitaker's version is used (since it's more well known, and shorter) but even in the lead the fact that this version of the problem is underconstrained is mentioned.
And, finally, I completely agree with Kmhkmh. We can't change the problem statement (or solutions) to our liking but need to take what's published about it. We're not writing an original paper about this problem - we're summarizing what the published literature has to say about it. -- Rick Block (talk) 19:31, 6 May 2010 (UTC)
In case i wasn't quite clear above, I'm fine with the current introduction/lead as it is. I was under the impression Bill suggested to change that and replace it by a non ambiguous problem version. While that approach is fine for an article on MHP published elsewhere, it should not/cannot be handled in WP in such a manner for the reason stated above.--Kmhkmh (talk) 19:42, 6 May 2010 (UTC)

You know, if I were to read today's comments a certain way, I could conclude that the article is not only at the level of a Featured Article, but has attained Nirvana, and could not possibly be improved upon by mortal editors. Kudos! Glkanter (talk) 19:50, 6 May 2010 (UTC)

@Rick and Kmhkmh, this is meant to be an encyclopedia article on the Monty Hall problem not a literature review on the subject. Everything we write should be supported by reliable sources but the sources do not, indeed cannot, tell us how to compose the article. Martin Hogbin (talk) 13:04, 7 May 2010 (UTC)

What is notable

The most notable statement of the problem is undoubtedly that of Whitaker but, as we all know, this statement leaves out much of the information needed to solve the problem and thus needs interpretation. The most notable interpretation of Whitaker's question is undoubtedly that of vos Savant. She, eventually, made clear that she had interpreted the problem as a simple mathematical puzzle, in which the host always offers the swap and the host always opens a legal door randomly to reveal a goat. This simple interpretation turns out to be exactly the same as the question asked and later clarified by Selvin some years earlier. This is the notable problem formulation because it resulted in thousands of letters claiming the answer was 1/2 and not 2/3. No one argued that it might make a difference which door the host opened when he had a choice, or that the player had to make her choice before the host revealed a goat. The only point in question was whether the answer was 1/2 or 2/3. This simple problem formulation, in which it is quite obvious that it makes no difference which legal door the host opens, then found its way into countless puzzle books and web sites and it is by far most the important, interesting, and notable version and it is the one that we should primarily be addressing here.

A second, and far less well known MHP, was later created by Morgan et al who showed that in their, somewhat perverse, interpretation of the problem, it could matter which door the host opened when he had a choice. This is indeed an interesting point for those starting to study conditional probability as it shows how a seemingly unimportant detail can have a significant effect on the a probability, and this version should, of course, be addressed here also.

However, nothing in WP policy, or common sense, tells us that the Morgan interpretation should control the whole article including dictating how we should address and fully explain the solution to the simple problem. That is the most important function of this article and it is up to editors here to decide how to do it. Martin Hogbin (talk) 13:46, 7 May 2010 (UTC)

Clear and precise formulation! Thank you. – Makes it "to the point", finally showing the real problem: (delete this if you like)
In exactly 2/3 of millions the player will win by switching, and only in 1/3 of millions she will win by staying. For most people it is hard to believe, let alone to understand. But a fact. No doubt. The famous 50:50 paradoxon. But no one can change this fact, neither Morgan.
Some smart, but totally fabricated, say perverse interpretation occured: If the host gives information about the actual status (although he never can change the actual, current status!), e.g. having a preference for opening one particular non selected door, then he can meet his preference in 2/3 of millions only:
In 1/3 (when he has got two goats to choose from), i.e. when staying will win!
And in another half of 2/3 (when he got the car and just one goat, and when the goat by chance is behind his preferred door), i.e. when switching will win!
In these 2/3 of millions, both switching and staying in average will have a chance of 1/2 each.
But in 1/3 of millions the car will be behind his preferred door, and by exceptionally opening the "avoided and unusual" door he shows that switching will win anyway with a chance of 1. But the host never can change the current status, though. He just may indicate information about the actual constellation. Why not opening all three doors at once? But all of that is a dishonorable game, when secret facts about the actual constellation should be revealed. Just suitable for students in probability. Not really affecting the famous 50:50 paradoxon in any way.
This fact being emphasized in every university, and you can find it in multiple internet pages of universities all over the world. This should clearly be stated in the article, and Morgan et al. may not be allowed to confuse. Facts should be presented clearly in WP. Sources? Wherever you have a look. Regards, --Gerhardvalentin (talk) 22:58, 7 May 2010 (UTC)
I would say that my problem with the introduction to the article is that the discussion assumes that the reader has information that hasn't already been stated to him. For example, it says:
However, such possible behaviors had little or nothing to do with the controversy that arose (vos Savant 1990), and the intended behavior was clearly implied by the author (Seymann 1991).
What is the "intended behavior"? The reader of the article does not know at this point, and will only read about it if she reads further. I would suggest something like this:
However, such possible behaviors had little or nothing to do with the controversy that arose (vos Savant 1990), and the intended behavior -- that Monty must open a door that he knows has a goat...(add whatever you need) -- was clearly implied by the author (Seymann 1991).
Making it clear in the initial section what the "intended behavior" is supposed to have been would go a long way towards alleviating my discomfort with the article as it stands.
Martin has correctly diagnosed the real point. People answer the "notable problem" incorrectly, even if it is stated unambiguously. The article needs to make clear what the "notable problem" is in the introduction. As it stands, we have an introduction to the history of the popular perception of the problem and the controversy that the Parade article engendered, but in what seems to me to be a confusing way, since the introduction never says what the "notable problem" is! Bill Jefferys (talk) 00:26, 8 May 2010 (UTC)
Bill, you are not the first person to make that point and you will most likely not be the last. I would like to change this article to give a complete, clear and, convincing solution to the basic MHP first and then proceed to the potential complications but there is a minority of editors who have resolutely resisted any change. Martin Hogbin (talk) 21:06, 9 May 2010 (UTC)

@Martin: From WP:NPOV: Neutral point of view (NPOV) is a fundamental Wikimedia principle and a cornerstone of Wikipedia. All Wikipedia articles and other encyclopedic content must be written from a neutral point of view, representing fairly, proportionately, and as far as possible without bias, all significant views that have been published by reliable sources. This is non-negotiable and expected of all articles and all editors.

Whether the "most notable interpretation of Whitaker's question is undoubtedly that of vos Savant" is the wrong question to be asking. The right question is what is the proportion of reliable sources that agree with her interpretation.

Bill hasn't been directly involved in this dispute, but for his benefit the issue here is whether vos Savant's solution, which mathematically solves for the overall probability of winning by switching given the player has picked door 1 (as opposed to the conditional probability of winning given the player has, for example, picked door 1 and has seen the host open door 3) should be the primary way the solution is presented. The solutions Martin favors make no attempt to justify that the probability they compute is the same as the conditional probability, and omit any mention of the critical assumption that makes these two probabilities necessarily equal (i.e. that the host choose between two goats randomly with equal probability) - as if the question asks about the overall probability rather than the conditional probability. Of course you CAN argue that these two probabilities must be the same (for example, if the problem is assumed to be symmetric - which it is if and only if the host chooses between two goats equally), but none of the sources presenting these kinds of solutions do this, so the article can't on their behalf. Furthermore, Martin desires to relegate any mention of a conditional solution to a "variants" section.

My argument is that this presentation would not represent "fairly, proportionately, and as far as possible without bias, all significant views that have been published by reliable sources". In my reading of the literature there are at least as many sources that explicitly address the conditional probability as ignore it, and in addition a significant number that explicitly criticize solutions that do not address the conditional probability. Martin's claim that the interpretation of Morgan et al. (this reference) is less well known, even perverse (?!), could not IMO be further from the truth.

We're already in formal mediation about this, see Wikipedia talk:Requests for mediation/Monty Hall problem. Further argument outside of the mediation process seems relatively pointless. -- Rick Block (talk) 04:19, 8 May 2010 (UTC)

We are nominally in formal mediation but nothing has happened for ages now. Andrevan has made clear that he is happy for discussion to continue outside the mediation. Indeed it is better for us to discuss here rather than on the mediation page.
Neutral POV is fine, but that does not make an article into a literature survey. The question is one of notability. Google the MHP and you will see almost nothing about the door that the host chooses. An Open University site even directs readers here, saying, 'The very informative article on the Monty Hall problem in the Wikipedia online encyclopaedia has several other arguments that might persuade you [my emphasis]. The whole and only reason that the MHP is notable is that people think that the answer is 1/2 not 2/3. This is not my POV it is what overwhelming evidence tells us. Even Gill, an academic studying the MHP was not aware of the Morgan paper. It is, by any measure, a sideline or an academic diversion from the real problem, which is a simple mathematical puzzle, which everybody gets wrong. Martin Hogbin (talk) 10:36, 8 May 2010 (UTC)
Martin - the question is not notability (or popularity), it's prevalence of a particular viewpoint among reliable sources. The policy is set up this way specifically so that the predominant published viewpoint (or viewpoints) is what is presented here regardless of popularity. I absolutely agree the MHP is notable because people get it wrong, but I think you're wrong about what problem most people think they're solving when they end up with 1/2 vs. 2/3. 1/2 is the "obvious" (but wrong) answer for the conditional problem (player has picked a specific door and can see which door the host has opened). 2/3 is the obvious (and right) answer for the unconditional problem (overall chance of winning by switching regardless of which door the host opens). By not distinguishing these as different problems what vos Savant and the others presenting "simple" solutions (never explicitly connected to any conditional case) are saying is effectively that the problem is the unconditional problem and (by implication) the answer to the conditional problem (whether it's 1/2 or anything else) is fundamentally irrelevant. Another (more generous) way to look at it is these solutions solve the conditional problem indirectly by solving a slightly different problem that has the same answer (but only given a specific unstated assumption - i.e. symmetry, or equivalently, that the host chooses between two goats equally). What Morgan et al. and the other sources you don't like are saying is that the problem is explicitly about the conditional probability so you must solve it conditionally, which you can do directly.
Are you denying that Morgan et al. is only one of many reliable sources that take this viewpoint? I've previously offered at least 5. How many would it take to convince you that this is not a "sideline or an academic diversion from the real problem" but rather a widely held significant view? -- Rick Block (talk) 17:17, 8 May 2010 (UTC
Hardly anyone finds the anwer 1/2 obvious, for either the conditional or unconditional case (in fact I doubt that most people distinguish the two cases). The idea that this might be the case seems to be unique to you. We cannot base this article on your own theory.
I am well aware that a number of statistics text books and papers mention the conditional nature of the problem, that is because the conditional aspect of the problem is of value in teaching elementary conditional probability but it is of little interest to most people, and for that matter probability experts. Also I accept that, if you want to be pedantic, even when the host chooses a legal door randomly, the problem is one of conditional probability. I even have no problem in including this in the article. However the fact is that the host's choice of legal door makes no difference to the probability of winning in this case and the vast majority of sources (online and written) do not consider this fact important. Most sources treat the MHP as a simple puzzle in which the host door choice is unimportant and that is what we must do. Martin Hogbin (talk) 21:42, 8 May 2010 (UTC)
What I mean by the 1/2 answer being "obvious" is that it is the answer most people immediately see as "correct" (since there are two closed doors and one open door). The problem for which this answer is "obvious" is the conditional problem, i.e. the situation faced by a player who has (for example) selected door 1 and has seen the host open door 3. This idea is not unique to me - it's also what Krauss & Wang say ("... most participants take the opening of Door 3 for granted, and base their reasoning on this fact. In a pretest, ... we saw 35 out of the remaining 36 participants (97%) indeed drew an open Door 3"). So, again, how many sources does it take to convince you that the viewpoint that the problem is inherently conditional is a mainstream POV that, per WP:NPOV policy MUST be represented fairly, proportionately, and as far as possible without bias? -- Rick Block (talk) 22:32, 8 May 2010 (UTC)
The connection between the problem being conditional and the difficulty people have in solving it correctly is your own pet theory. K&W certainly do not make this connection in their paper. Their main point is that excessive information, in particular door numbers, only serves to confuse people. Martin Hogbin (talk) 20:57, 9 May 2010 (UTC)
Would you please answer the question? How many sources does it take to convince you that the viewpoint that the problem is inherently conditional is a mainstream POV that, per WP:NPOV policy MUST be represented fairly, proportionately, and as far as possible without bias? -- Rick Block (talk) 21:50, 9 May 2010 (UTC)
I have already agreed that the problem is one of conditional probability, just like every other probability problem. The point that we are arguing about is how to best give a solution to the problem in this article and, in particular, whether the condition that the host opens a specific door should be considered important in our initial description of the problem. The vast majority of sources, online and written, do not consider this a significant condition and do not consider it in their solutions. Selvin, when he originally posed the problem, specified that the host would choose randomly when he had a choice of goat doors, with the clear intention of making the host's choice of legal door irrelevant to the problem. Vos Savant continued this tradition by making the same assumption in her solution. Most sources that addresses the problem proceed on this basis. There are a number of sources, clearly aimed at those studying conditional probability, that consider the case where the host may not choose a legal door randomly and thus the specific door opened by the host becomes an important condition of the problem. This case, however, is of minority academic interest, and is one which should influence the structure of our article.
I am not aware of any source before 1991 that specifically considers the door that the host opens when he has a choice to be significant. Please tell me if you know of one. Martin Hogbin (talk) 11:29, 10 May 2010 (UTC)
At this point I frankly couldn't care less whether you agree the problem is conditional or not - the question is about the prevalence of views expressed in reliable sources. What we're arguing about is whether it is appropriate to marginalize the POV expressed by Morgan et al. and others that the solution should be explicitly conditional, and the corresponding criticism of solutions that do not explicitly address any conditional case. You're claiming this POV conflicts with that of "the vast majority of sources". My question is how many sources do I need to cite to convince you that this is NOT a marginal POV, but a POV that is at least as common among reliable sources as (say) vos Savants? The date of publication is not significant to this question, although Selvin's second letter does consider how the host opens a door when he has a choice to be significant. So, once more, would you please answer the question? How many sources does it take to convince you that the viewpoint that the problem is inherently conditional is a mainstream POV that, per WP:NPOV policy MUST be represented fairly, proportionately, and as far as possible without bias?-- Rick Block (talk) 14:19, 10 May 2010 (UTC)
Firstly, Selvin's second letter defines the host to choose randomly because Selvin knows that this could be significant, if he had not defined the host to act that way. In other word he makes toe hosts random legal choice clear in order to avoid the very issue we are arguing about.
To answer your question of how many sources you need to quote to show that Morgan's approach is as common as vos Savant's, maybe a hundred would do it.
Finally, although it is my personal POV that Morgan's approach is a marginal academic diversion, I am not trying to push this in the article. As I have often said, as an attempt at compromise, I am perfectly willing to give Morgan's solution equal space to the simple solutions in the article. I simply want to ensure that the simple solutions are fully, completely, and convincingly explained first - the way that all good text books do things. It is you who is pushing your POV that the Morgan paper somehow has the power to direct the structure of the article for us. Martin Hogbin (talk) 16:24, 10 May 2010 (UTC)
Defining the host's choice between two losers as random doesn't avoid the issue we're arguing about - it is the issue we're arguing about. The host's choice in this case doesn't matter and doesn't need to be specified if the question is interpreted unconditionally (the average chance of winning by switching is 2/3 no matter how the host chooses between two losers) - it only matters if the question is interpreted conditionally (the conditional chance of winning is 2/3 if and only if the host chooses randomly between two losers). So, if you agree it matters, you're agreeing the question is the conditional question (and, vice versa).
Yes, that is exactly why Selvin did define the host to choose a legal door randomly, to avoid all this argument. Martin Hogbin (talk) 19:15, 10 May 2010 (UTC)
Perhaps we should ask the mediator about this, but are you seriously suggesting it would take 100 references to convince you that the POV that the question being asked is the conditional question is a common POV? This seems like a completely unreasonable stance to me. I'd be willing to find as many references that take this POV as you can find that take vos Savant's POV.
Yes, there are far more sources which do not consider it important which legal door the host opens. Try a Google search on the MHP. You seemed to want to do a source count although we both probably agree that this is not the way to decide. Martin Hogbin (talk) 19:15, 10 May 2010 (UTC)
I'm not saying the Morgan paper should direct the structure of the article. What I'm saying is that the article must not present the simple solution as if it is the main, true, correct answer agreed by all sources - since it clearly is not - but rather the article should present both types of solutions as valid. The solution section I've proposed (per this thread) attempts to do this (and it is a compromise since my POV is that the problem is conditional and the unconditional solutions don't quite address it). Your counter suggestion continues to be to present a fully fledged unconditional solution (matching your POV) deferring any mention of a conditional solution or any criticism of the unconditional solutions to an "experts only", academic diversion section. You're insisting your POV dominate the article, not me. -- Rick Block (talk) 18:22, 10 May 2010 (UTC)
Rick, where have I ever suggested that we, 'present the simple solution as if it is the main, true, correct answer agreed by all sources'? Let me explain one more time. We first present the simple solution, with no comment that it is either complete or incomplete. This is exactly what the sources that present such solutions do. It is also what most good textbooks do, they start with a simple explanation. Then, we present Morgan's argument that the simple solution is incomplete. We state their argument in full together with theit solution and supporting sources. At this time we should also present sources which criticise Morgan. Martin Hogbin (talk) 19:15, 10 May 2010 (UTC)
Since the current version effectively does what you say, i.e. "first present the simple solution, with no comment that it is either complete or incomplete" and then "present Morgan's argument that the simple solution is incomplete", and you're still arguing about this I can only infer you must mean something different. What you've said before is you want to move "aids to understanding" before "Morgan's argument" (with criticisms of this argument) which has the form of a more or less complete article, followed by a controversial objection. This form conveys that the first section is the generally accepted (main, true, correct) answer and what comes later is not - i.e. it structurally makes your POV dominate the article. -- Rick Block (talk) 23:24, 10 May 2010 (UTC)

@Bill - regardless of the above, we can work on the lead. Perhaps the paragraph following the quoted problem description could say:

As the player cannot be certain which of the two remaining unopened doors is the winning door, most people assume that each of these doors has an equal probability and conclude that switching does not matter. In the usual interpretation of the problem the host always opens a door, always makes the offer to switch, and randomly chooses which door to open if both hide goats. Under these conditions, the player should switch—doing so doubles the probability of winning the car, from 1/3 to 2/3.

Would that address your concern? -- Rick Block (talk) 04:19, 8 May 2010 (UTC)

@Rick - Yes, that would address my concern. Thanks! Bill Jefferys (talk) 21:23, 8 May 2010 (UTC)

...and this simple solution distributes equally over any and all door selection and door revealed pairings.

I read Selvin's original table solution as indicating that all combinations are equally likely.

I get the feeling that this is what is being argued. Glkanter (talk) 23:31, 8 May 2010 (UTC)

No, what is being argued is whether the the problem is about the unconditional probability (i.e. the overall chance of winning by switching), or the conditional probability (the chance of winning by switching if you've, for example, picked door 1 and have seen the host open door 3). What Selvin says about this is "The basis to my solution is that Monty Hall knows which box contains the keys and when he can open either of two boxes without exposing the keys, he chooses between them at random." For the unconditional problem, how the host chooses between two incorrect choices is irrelevant - the answer is 2/3 chance of winning by switching no matter how the host picks between two goats. That Selvin considered the host picking between two incorrect choices part of the basis of his solution clearly implies he was thinking the problem was asking about the conditional probability. Moreover, his alternate solution computes the conditional probability. -- Rick Block (talk) 23:56, 8 May 2010 (UTC)
I agree, "The basis to my solution is that Monty Hall knows which box contains the keys and when he can open either of two boxes without exposing the keys, he chooses between them at random..." and this simple solution distributes equally over any and all door selection and door revealed pairings. Glkanter (talk) 00:55, 9 May 2010 (UTC)
If the host chooses a door hiding a goat uniformly at random, the conditional and unconditional answers are exactly equal. This fact is quite clear without the use of conditional probability. Martin Hogbin (talk) 18:29, 9 May 2010 (UTC)
How many sources present an unconditional answer making no mention of whether it must be the same as the conditional answer?
It is not possible to answer this question. Hundreds of sources present an answer that is the same as both the unconditional answer and the symmetrical conditional answer, namely 2/3. These sources, quite sensibly, do not bother to distinguish between these cases. Martin Hogbin (talk) 15:53, 10 May 2010 (UTC)
How many sources present a conditional answer?
I have not counted them. Maybe a dozen or so. I am sure that you can tell me. Martin Hogbin (talk) 15:53, 10 May 2010 (UTC)
How many sources criticize sources only presenting an unconditional answer as not exactly answering the question? -- Rick Block (talk) 21:09, 9 May 2010 (UTC)
Again, you will know better than I will. It is clear that these are all based on one source, since no source did this before 1991. Martin Hogbin (talk) 15:53, 10 May 2010 (UTC)
Now one for you Rick How many sources criticise the one source that started all this nonsense? Martin Hogbin (talk) 15:53, 10 May 2010 (UTC)
I think you have this exactly backwards. Vos Savant is the primary source presenting an unconditional solution completely ignoring the conditional answer. There are many (I doubt that it's hundreds) of reliable sources (mostly non-academic, which doesn't necessarily mean they're not reliable) that follow HER lead. I think there are at least as many sources that present a conditional solution - and these tend to be academic and peer reviewed sources which (per WP:SOURCES) "are usually the most reliable sources where available". The number of sources criticizing sources presenting an unconditional solution is perhaps a dozen or so (PLENTY enough to call it a significant POV), and indeed Morgan et al. was the first but most of the rest are actually independent. For example, there's clearly no way Gillman is based on Morgan et al. As far as how many sources criticize Morgan et al., I think the answer may be two - Seymann's comment and Rosenhouse's book - although both of these are essentially criticizing the tone, not the mathematics. I think you are missing the fact that vos Savant's POV being the most popular among lay people (which, of course it is, since she was writing for an extremely widely distributed medium) does not in any way mean it is the most prevalent among published sources. The question is not how many people hold a particular POV, but how many sources do - and, if sources disagree, the relative reliability of those sources comes into play as well. -- Rick Block (talk) 17:09, 10 May 2010 (UTC)
Seymann was not criticising the tone of Morgan's paper he was suggesting that Morgan had answered the wrong question or at the very least failed to realise that the question could have been read in more than one way. That must seriously lower Morgan's credibility as a peer reviewed source. Martin Hogbin (talk) 19:22, 10 May 2010 (UTC)

Wikipedia's Standards Are Higher Than The American Statistician's

Rick keeps asking Martin 'How many sources...?'

But it's a bogus question.

In Morgan's paper, they appear ignorant of Selvin's letters. By their 'standard', 1 source, even the 'original' is not enough. They just act as if it didn't exist. Even though the letters were published in the exact same journal! Who died and made them boss?

So, like a junkyard dog with nothing else to 'protect' (no more 'host bias' argument, no more 'disputed State of Knowledge, no more 'simple solution is wrong') Rick accuses anyone of proposing changes that do not follow his PERSONAL INTERPRETATION of the published material as violating the sacrosanct Wikipedia NPOV pillar.

Did anyone see that most recent edit to the article? The diff shows, I believe, Line 28 after the edit. Perhaps the worst written English sentence I have ever read. Glkanter (talk) 12:43, 10 May 2010 (UTC)

Um, what are you talking about - this diff? And, I'm asking for sources, not personal opinions. -- Rick Block (talk) 14:36, 10 May 2010 (UTC)
No, not any edit you made, Rick. Some other editor, just yesterday maybe, added a period to a sentence. The diff showed the next paragraph, or at least a sentence. It's just horrible. Sorry for the confusion. Glkanter (talk) 15:39, 11 May 2010 (UTC)
  • Just as a break for mediator from this article will help him become refocussed and reenergised, so too would a break for the parties to this dispute be beneficial. Unless of course you feel that the discussions that are today taking place on this talk page are beneficial, in which case please do proceed. AGK 19:30, 10 May 2010 (UTC)
We had a clear consensus of editors last November. That's when Rick resorted to his 'NPOV' complaint.
To suggest anyone needs to calm down after weeks of silence is preposterous. Glkanter
No, we did not have a "clear consensus". Even with a totally skewed methodology counting "for change" vs. "keep as is", the count (which does not determine consensus) was what 8-5?. And what I've actually said is consensus cannot overrule NPOV - so even if there were a clear consensus (and, again, there wasn't) the article cannot be written in a way that it fails to represent "fairly, proportionately, and as far as possible without bias, all significant views that have been published by reliable sources". -- Rick Block (talk) 21:15, 10 May 2010 (UTC)
The NPOV argument is somewhat bogus as each side considers the other to be pushing their POV. Also, as Glkanter points out, we have just taken a long break from our discussion, in the hopes that the mediation might achieve something. 21:32, 10 May 2010 (UTC)
Well, let's see. "Your side" has been complaining about bias for the Morgan et al. POV since about this version (the version following the last FAR) which presents a simple solution, and then transitions to a conditional solution with two paragraphs in between that say
The reasoning above applies to all players at the start of the game without regard to which door the host opens, specifically before the host opens a particular door and gives the player the option to switch doors (Morgan et al. 1991). This means if a large number of players randomly choose whether to stay or switch, then approximately 1/3 of those choosing to stay with the initial selection and 2/3 of those choosing to switch would win the car. This result has been verified experimentally using computer and other simulation techniques (see Simulation below).
A subtly different question is which strategy is best for an individual player after being shown a particular open door. Answering this question requires determining the conditional probability of winning by switching, given which door the host opens. This probability may differ from the overall probability of winning depending on the exact formulation of the problem (see Sources of confusion, below).
The article now looks like this, with completely separate sections for simple solutions and conditional solutions, with nothing at all in the simple solution section contradicting it or in any way saying it is deficient (hmmm, much like what you claim above is your goal) but yet you're STILL complaining about "pro-Morgan" bias. Separate, but equal, solution sections are apparently not good enough for you and you want more. Glkanter has never exactly said what he wants, but apparently wants to delete all references to Morgan et al. You say you're OK with simply moving any mention of conditional probability to a later section of the article (burying it, in newspaper terminology). This is the point at which I have been pushing back on NPOV grounds. If the two of you are not simply pushing your POV you're doing a damn good imitation of it. -- Rick Block (talk) 22:16, 10 May 2010 (UTC)
This is not a newspaper article it is a encyclopedia article and using terms like 'burying' to describe putting more complex solutions after simple solutions is completely inappropriate. I can only restate that in many mathematical and technical text books it is common to start with a simplified (and therefore incomplete) version of the subject and then proceed to a address more complex issues. In fact study of most subjects is only possible using this approach. Much school mechanics is based on massless inextensible strings operating over frictionless pulleys, all non-existent entities. Whole books at an advanced level are written on Newtonian mechanics even though, in the light of relativity and quantum mechanics, this is known to be wrong. Even if I liked the Morgan paper and agreed with you that the problem must only be treated conditionally I would still recommend starting with the simple version of the problem, followed immediately with sufficient discussion to ensure that this had been fully understood before proceeding to the conditional case, simply because of the astounding difficulty most people have in understanding and accepting the correct solution. Martin Hogbin (talk) 22:39, 10 May 2010 (UTC)
Martin - how is this not exactly the structure of the solution section I've proposed? Are you so accustomed to arguing with me you think need to argue with whatever I suggest without even reading it? Again, the suggestion is in this thread. Except for the fact that you're still arguing with me it sounds like we're agreeing. -- Rick Block (talk) 23:44, 10 May 2010 (UTC)
Not quite. Your rewritten solution makes the distinction between various (effectively equivalent in my opinion) cases in which: any door is chosen by the player, door 1 is chosen by the player, door 1 is chosen by the player, and door 3 is opened by the host. My suggestion was to keep the same wording that we currently have but move one section to a different place.
The crux of my suggestion is to keep all complicating issues (which doors are opened, whether the problem is conditional or not) right out of the way until the simple solution has been fully explained and discussed. This does mean initially giving solutions that some might consider to be incomplete to start with, but his is how most text books work. Martin Hogbin (talk) 10:27, 11 May 2010 (UTC)
But doesn't "fully explaining" the simple solution involve explaining exactly which of these situations the solution directly addresses and then (potentially) why it can be considered to also addresses the others? Per K&W, most people think about the specific case where the player's picked door 1 and the host has opened door 3, and have a very hard time switching to the mental model required for the simple solutions to make sense. I'm making the model behind each solution explicit, so however you're thinking of the problem you'll find a matching solution. Not only does this make the presentation of the solutions more NPOV, but (IMO) it helps our readers by providing a solution matching the mental model in which they're trying to solve the problem. -- Rick Block (talk) 13:34, 11 May 2010 (UTC)
I am in favour of anything which helps our readers to understand why the answer is 2/3 and not 1/2 but your suggestion does not, in my opinion, do this. It has three sections, with rather complicated headings, and it is not clear whether these are intended to relate to the same puzzle or different ones. Also the first two sections have confusing disclaimers at the end. I understand why you do this, you want to ensure that what you say is strictly correct but I would say that we should forget all about pedantry in the first section and completely ignore the distinction between the conditional and unconditional cases and the door numbers chosen by the host and the player. We all know that these make no difference to the answer in the standard case and there are plenty of reliable sources that take this approach.
One the reader has got to grips with the main problem, and if they are interested, they can read about Morgans criticism of the simple solutions, read about Morgan's solution and read about conditional probability and its effect on the solution if the host is considered to choose non-uniformly. Martin Hogbin (talk) 14:46, 11 May 2010 (UTC)
Martin - what K&W are saying (and I agree) is that the "simple" solution is very hard to understand if you start with a mental model of 2 closed doors and an open door. You seem to be saying that the article should more or less force the reader to see the problem unconditionally (as if this is the "right" way to understand the problem), and then present a conditional solution later. I'm suggesting we present solutions that match however the reader is thinking about the problem (including looking at 2 doors and an open door) - without forcing a shift between mental models. Not only do I suspect this would be easier to comprehend, but in addition this structure doesn't make the article inherently endorse one view over another (makes it more NPOV). The intent is the solutions all pertain to the same problem - if this is not clear we should work on clarifying it.
Rick, can we please forget about the distinction between the conditional and unconditional case just for the moment. What I am saying is that I want to start with explanations that are as simple as possible for people to understand and, whatever these explanations are, I do not want to confuse the reader further at that stage with statements that this or that solution is conditional/unconditional or that a given solution is incomplete or does not answer the question. To put it bluntly I would be happy to present an unconditional solution as a solution to the conditional problem. Hardly anybody makes the distinction initially. If the K&W paper helps us to present a convincing solution that is fine with me. Martin Hogbin (talk) 22:01, 11 May 2010 (UTC)
Although I'm more or less awaiting the result of the mediation, I now and then look at the "progress" made here. @Martin: To put it bluntly, I will not accept a, what you call, unconditional solution to the conditional problem. Nijdam (talk) 10:13, 12 May 2010 (UTC)
"...I will not accept..." Very sad. Glkanter (talk) 11:57, 12 May 2010 (UTC)
My bad. I was not previously aware of this little known Wikipedia codicil: WP:NIJDAMISTHEULTIMATEWIKIPEDIADECISIONMAKER. Glkanter (talk) 15:50, 12 May 2010 (UTC)
How is the "simplest approach" in my suggested solution section not exactly this? Do you think it's not wordy enough to be convincing? What, exactly, would you add? Similarly the enumeration of cases where the player has picked door 1. This is exactly the same as vos Savant's approach. Neither says anything about the distinction between conditional and unconditional (why do you think I'm stuck on this?). These solutions are both complete and (I'd bet) easier to understand than the mess that's currently in the "Popular solution" section of the article. Is what you don't like only that they're immediately followed by a conditional solution? This is how NPOV works. -- Rick Block (talk) 06:50, 12 May 2010 (UTC)
The disclaimers you're saying are confusing are the numeric samples (right?), for example the first one which says "What this solution is saying is that if 900 contestants all switch, regardless of which door they initially pick and which door the host opens about 600 would win the car.". These are intended to clarify what the associated solution is actually saying - to help match the mental model. At least some people have an easier time thinking about how many times something might happen than probabilities (even though these are fundamentally exactly the same) - i.e. if I roll a die it should be a 2 or a 5 roughly two times out of every six rolls (as opposed to "with probability 1/3"). Do you personally find these confusing are you saying you think others might? -- Rick Block (talk) 19:21, 11 May 2010 (UTC)
Once again you seem stuck on the conditional/unconditional issue. Pretty well nobody who sees the problem for the first time is aware of the potential difference between the player choosing before the host has opened a door and the player choosing afterwords, thus there is no need to match mental models as you put it. The fact that we seem to make a distinction between a player winning 200 times out of 300 and a player having a probability of winning of 2/3 serves only to confuse. Martin Hogbin (talk) 22:01, 11 May 2010 (UTC)
Martin - The bottom line is there are multiple reliable sources that present only a conditional solution (I've found 14 so far - these are published papers and books - and haven't spent much time looking), and multiple sources that explicitly criticize "simple" solutions. Wikipedia policy says the views of these sources must be represented "fairly, proportionately, and as far as possible without bias" (regardless of whether you agree with them). I'd honestly prefer if we could agree that presenting both approaches in an integrated fashion is actually more helpful to our readers, but whether you agree or not I think ultimately NPOV will drive us to this structure. IMO, what you're suggesting is not NPOV. If you don't think what I'm suggesting is NPOV or easy enough to understand, please suggest another alternative. -- Rick Block (talk) 06:50, 12 May 2010 (UTC)
Rick and Nijdam, you have both completely missed my point. I will explain in a section below. Martin Hogbin (talk) 12:46, 12 May 2010 (UTC)
  • Glkanter: From the completely unproductive nature of this conversation even since my comment, it seems that my suggestion was far from preposterous. The positions in this debate are irreconcilable, at least without outside assistance from a mediator or some other facilitating party; and so further interaction is at best useless and at worst conducive to bickering. YMMV. AGK 22:21, 10 May 2010 (UTC)
Very simply, Wikipedia seems to have no mechanism for resolving this long-running (5 years?) dispute. Dragging out this 'mediation' may have lessened the activity, but has done nothing to move this closer to a resolution of the underlying issues. Glkanter (talk) 22:38, 10 May 2010 (UTC)
AKG, I think we once came very close to a solution, which was my suggestion to simply swap the order of two sections in the article, as suggested in remarks above. This would not completely resolve all the problems but it would have meant that we could probably all work much better together to improve the article. This was stalled by one, or maybe two, editors. Martin Hogbin (talk) 22:58, 10 May 2010 (UTC)
And, we've come equally close (actually closer, since nearly final wording is available) with the merged solution section that I've proposed, that has been stalled by you and Glkanter. -- Rick Block (talk) 23:31, 10 May 2010 (UTC)