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Hate to spoil the conversation but thought y'all might be interested in this. hydnjo ( talk) 04:16, 30 January 2010 (UTC)
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:
They reference vos Savant and Selvin, but not Morgan et al. -- Rick Block ( talk) 06:04, 30 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)
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)
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)
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.
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)
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:
Door 1 | Door 2 | Door 3 | result if switching |
---|---|---|---|
Car | Goat | Goat | Goat |
Goat | Car | Goat | Car |
Goat | Goat | Car | Car |
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)
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.
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)
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)
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 | ? |
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 | ? |
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 |
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 |
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 |
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 |
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 |
Thread moved to /Arguments#Table showing the host opening an unnamed door. -- Rick Block ( talk) 17:54, 7 February 2010 (UTC)
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)
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)
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)
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)
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:
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.
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.
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.
Here we discuss what sources say about why people cannot answer the simple problem correctly. The tests on pigeons might fit in here
Intentional complications, generalisations etc as currently.
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)
Please try to keep these at a minimum.
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)
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 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.
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)
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)
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)
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)
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)
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)
[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)
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)
Good articles, 2009a and 2009b. [...]
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)
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)
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)
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)
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)
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)
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)
Proposed text
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Yes, that is much better 121.73.7.84 ( talk) 22:24, 26 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)
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)
Martin's proposal
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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): 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. ![]() ![]() 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 ![]() 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). ![]() 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.
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:
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
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
The host's behavior is reflected by the values of the conditional probability of H given C and S:
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." |
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)
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)
@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)
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)
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)
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)
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 ( talk • contribs) 09:29, 5 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)
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)
@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)
@Bill - regardless of the above, we can work on the lead. Perhaps the paragraph following the quoted problem description could say:
Would that address your concern? -- Rick Block ( talk) 04:19, 8 May 2010 (UTC)
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)
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)
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Hate to spoil the conversation but thought y'all might be interested in this. hydnjo ( talk) 04:16, 30 January 2010 (UTC)
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:
They reference vos Savant and Selvin, but not Morgan et al. -- Rick Block ( talk) 06:04, 30 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)
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)
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)
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.
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)
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:
Door 1 | Door 2 | Door 3 | result if switching |
---|---|---|---|
Car | Goat | Goat | Goat |
Goat | Car | Goat | Car |
Goat | Goat | Car | Car |
Car hidden behind Door 3 | Car hidden behind Door 1 | Car hidden behind Door 2 | |
---|---|---|---|
Player initially picks Door 1 | |||
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Host must open Door 2 | Host randomly opens either goat door | Host must open Door 3 | |
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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)
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.
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)
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)
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 | ? |
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 | ? |
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 |
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 |
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 |
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 |
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 |
Thread moved to /Arguments#Table showing the host opening an unnamed door. -- Rick Block ( talk) 17:54, 7 February 2010 (UTC)
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)
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)
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)
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)
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:
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.
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.
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.
Here we discuss what sources say about why people cannot answer the simple problem correctly. The tests on pigeons might fit in here
Intentional complications, generalisations etc as currently.
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)
Please try to keep these at a minimum.
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)
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 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.
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)
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)
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)
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)
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)
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)
[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)
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)
Good articles, 2009a and 2009b. [...]
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)
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)
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)
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)
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)
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)
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)
Proposed text
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Yes, that is much better 121.73.7.84 ( talk) 22:24, 26 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)
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)
Martin's proposal
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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):
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. ![]() ![]() 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 ![]() 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). ![]() 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.
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:
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
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
The host's behavior is reflected by the values of the conditional probability of H given C and S:
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." |
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)
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)
@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)
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)
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)
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)
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 ( talk • contribs) 09:29, 5 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)
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)
@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)
@Bill - regardless of the above, we can work on the lead. Perhaps the paragraph following the quoted problem description could say:
Would that address your concern? -- Rick Block ( talk) 04:19, 8 May 2010 (UTC)
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)
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)