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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. [[User:Martin Hogbin|Martin Hogbin]] ([[User talk:Martin Hogbin|talk]]) 13:46, 7 May 2010 (UTC) |
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. [[User:Martin Hogbin|Martin Hogbin]] ([[User talk:Martin Hogbin|talk]]) 13:46, 7 May 2010 (UTC) |
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:Clear and precise formulation! Thank you. – Makes it "to the point", finally showing the ''real problem:'' (delete this if you like)<br />In exactly 2/3 of millions the player will win by switching, and only in 1/3 of millions she will win by staying. For most people it is hard to believe, let alone to understand. But a fact. No doubt. The famous 50:50 paradoxon. But no one can change this fact, neither Morgan.<br />Some smart, but totally fabricated, say perverse interpretation occured: If the host gives ''information about the actual status'' (although he never can change the actual, current status!), e.g. having a preference for opening one particular non selected door, then he can meet his preference in 2/3 of millions only: |
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:In 1/3 (when he has got two goats to choose from), ''i.e. when staying will win!''<br />And in another half of 2/3 (''when he got the car and just one goat,'' and when the goat by chance is behind his preferred door), ''i.e. when switching will win!'' |
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:In these 2/3 of millions, both ''switching and staying in average will have a chance of 1/2 each''.<br />But in 1/3 of millions the car will be behind his preferred door, and by exceptionally opening the "avoided and unusual" door he shows that switching will win anyway with a chance of 1. But the host never can change the current status, though. He just may indicate information ''about the actual constellation.'' Why not opening all three doors at once? But all of that is a dishonorable game, when secret facts about the actual constellation should be revealed. Just suitable for students in probability. Not really affecting the famous 50:50 paradoxon in any way. |
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:This fact being emphasized in every university, and you can find it in multiple internet pages of universities all over the world. This should clearly be stated in the article, and Morgan et al. may not be allowed to confuse. Facts should be presented clearly in WP. Sources? Wherever you have a look. Regards, --[[User:Gerhardvalentin|Gerhardvalentin]] ([[User talk:Gerhardvalentin|talk]]) 22:58, 7 May 2010 (UTC) |
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This is not clear.
I have read the first half this article and I still have learned nothing. I'm quite a smart guy so it is a mystery how most other people will have had this problem explained to them by this article. Some of the sentences are a bit unstructured too. I can fully accept that I got the wrong answer at the beginning, but I think I am not unreasonable for thinking that I should have a better idea of why that is by now. I'm off to find a non-wikipedia source to explain the monty hall problem to me. 121.73.7.84 (talk) 07:26, 22 March 2010 (UTC)
After consulting another site I believe (i may be wrong) that I understand this dilemma, however the explanation on this page is not clearly written and is unnecessarily confusing. This is unhelpful when trying to explain an unfamiliar concept: Quote "The player, having chosen a door, has a 1/3 chance of having the car behind the chosen door and a 2/3 chance that its behind one of the other doors. As the host opening a door to reveal a goat gives the player no new information about what is behind the door he has chosen, the probability of there being a car remains 1/3. Hence the car is then with chance 2/3 behind the remaining unopened door (Wheeler 1991; Schwager 1994). Switching doors thus wins the car with a probability of 2/3, so the player should switch (Wheeler 1991; Mack 1992; Schwager 1994; vos Savant 1996:8; Martin 2002). In order to convert this popular story into a mathematically rigorous solution, one has to argue why the probability that the car is behind door 1 does not change on opening door 2 or 3. This can be answered by an appeal to symmetry: under the complete assumptions made above, nothing is changed in the problem if we renumber the doors arbitrarily, and in particular, if we switch numbers 2 and 3. Therefore, the conditional probability that the car is behind door 1, given the player chose 1 and Monty opened 2, is the same as the conditional probability that the car is behind door 1, given the player chose 1 and Monty opened 3. The average of these two (equal) probabilities is 1/3, hence each of them separately is 1/3, too."
As I understand it, if a contestant chooses a door at random, the remaining doors will both have goats 1/3 of the time and a goat/car mix 2/3 of the time. 1/3 of the time the remaining doors both have goats, so by switching you would lose on this 1/3 of occasions. But on the other 2/3 of occasions the host will pick the door with a goat to avoid the door with the car. Since this is the case 2/3 of the time, by switching to the other door not chosen by the host you will succeed 2/3 of the time.
I'm sure someone will tell me if i've misunderstood the problem 121.73.7.84 (talk) 08:10, 22 March 2010 (UTC)
- You seem to understand it correctly. Just out of curiosity do you find the revised solution section as proposed here easier to understand? -- Rick Block (talk) 14:08, 22 March 2010 (UTC)
- Well done 121.73.7.84! You have managed to fight your way through the ridiculously obfuscated explanation given in this article. I would explain the answer it like this: if you swap, you always get the opposite of your initial choice and you have a 2/3 chance of initially picking a goat, but unfortunately, some editors here have a attachment to a perversely pedantic published paper on the subject. This means that simple explanations, such as the one that I have given are regarded as wrong and must be replaced with complicated explanations such as you have quoted. The simple mathematical puzzle that everyone gets wrong is thus turned into an obscure and complicated probability problem that nobody cares about. You are welcome to join in the discussion on how to improve the article. Martin Hogbin (talk) 18:35, 22 March 2010 (UTC)
- Martin - don't blame Morgan et al. for this. The paragraph quoted is from the existing "Popular Solution" section that was forked from a common "Solution" section essentially at your request. That you "popularists" have turned this section into an unreadable mess has nothing whatsoever to do with Morgan et al. (the perversely pedantic paper you're referring to). I'll repeat my question to 121.73.7.84 - do you find the solution section as proposed here easier to understand? -- Rick Block (talk) 19:17, 22 March 2010 (UTC)
- You cannot blame me for the quoted explanation. My view has always been that we should have some simple, convincing solutions like the one I give above, that do not mention anything about conditions or door numbers or the like first, with full explanations and aids to understanding, then, after that we should discuss which door the host opens, and why it may or may not matter. Martin Hogbin (talk) 19:50, 22 March 2010 (UTC)
- Martin - don't blame Morgan et al. for this. The paragraph quoted is from the existing "Popular Solution" section that was forked from a common "Solution" section essentially at your request. That you "popularists" have turned this section into an unreadable mess has nothing whatsoever to do with Morgan et al. (the perversely pedantic paper you're referring to). I'll repeat my question to 121.73.7.84 - do you find the solution section as proposed here easier to understand? -- Rick Block (talk) 19:17, 22 March 2010 (UTC)
- Rick, the reason the "popular" section has been "turned into an unreadable mess" is the "probabilists" (for lack of a better word, obtained by comparing the "Popular Solution" section to the poorly-named "Probabilistic Solution" section) insist the "popular" solution must be framed in a way that distinguishes it from the "Probabilistic Solution." If the equal-choice version is accepted as the main problem, as all but a few pedantic papers do, then Morgan's p is 1/2 and Moragn's whole point disappears. The "popular solution" can then be written without having to cite references every third word or so, and we can quit inserting weasel words into the article that are there only defend the text against this argument. In other words, frame the main article around what we all should know was intended (no p), and pull Moragn's analysis out into a side case.
- But while I'm here, let me ask you a question. In his book The Drunkard's Walk, Leonard Mlodinow admitted he altered Savant's wording when he asked the MHP this way: "Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, and the host, who knows what's behind the doors, opens another door, which has a goat. He then says to you, 'Do you want to switch your pick to the other door?' Is it to your advantage to switch your choice?" His answer, and he is a citable source, was that the probabilities were 1/3 and 2/3 for the two doors. Do you think that is correct? JeffJor (talk) 15:59, 23 March 2010 (UTC)
- The analysis where p is a variable is already pulled out into a side case. For the 400th time, the issue is NOT whether p is taken to be 1/2 or not, but whether the question is asking about the conditional probability given the player knows which door she has picked and which door the host has opened or the chance of winning by switching averaged across all possible player picks and doors the host might open. These are always different questions even though they have the same answer if p is 1/2. The "unconditional solutions" either (pick one) are not addressing the conditional probability, or are implicitly assuming (without comment or justification - except Selvin's where he later explicitly said what assumptions his unconditional solution is based on) that the conditional answer is the same as the average chance of winning by switching.
- Is your question whether I agree with Mlodinow's answer, or whether I agree it is Mlodinow's answer? For the purposes of this page the first question is irrelevant. I haven't read the source but I'm perfectly willing to believe it says what you're claiming it says. -- Rick Block (talk) 23:56, 23 March 2010 (UTC)
A reasonable person, including those who have published various 'simple' or 'omniconditional' solutions, could assume that 'Suppose you're on a game show...' means that the host will not be indicating to the contestant where the car is located. Accordingly, that person would not see a need to redundantly state 'p = 1/2'.
As the 'simple' or 'omniconditional' solutions are true in all cases, the specific scenario where the contestant chooses door 1 and the host reveals a goat behind door 3 is nothing more than a subset, requiring no additional attention. Glkanter (talk) 04:45, 24 March 2010 (UTC)
- Sorry folks for restarting the same argument yet again. I just wanted to point out that we have had yet another reader who has found the article hard to understand in its current form. If you look back through the archives you see quite a few. There are, no doubt, many more who have not understood the explanation given here but have not bothered to comment. Martin Hogbin (talk) 09:49, 24 March 2010 (UTC)
- I beg to differ with you both, Rick and Nijdam. Whether the issue is "is p taken to be 1/2 or not," "is the question asking about the conditional probability given the player knows which door she has picked," or even "is the quesiton about the chance of winning by switching averaged across all possible player picks" are related to the question I asked. That's why I asked it, because too many people refuse to see the relationship. And being intentionally evasive about it does not help. Is Mlodinow's answer of 1/3 correct, by your understanding of the question I quoted? And to address your other point, Rick, the "unconditional solution" is to a problem that is isomorphic with what you call the "conditional" one, and isomorphism is an acceptable way to express a correct solution. One that can greatly simplify understanding for lay people, which is waht we are trying to accomplish. The article is confusing because it has to be written, due to the nature of your objections, without such isomorphism. It spends far too much time explaining how each thought can be found in literature. And on why that literature is "correct" or "scientific" or "probabilistic" or any of a number of other weasel words inserted just to combat this argument, instead of simply making itself clear even if that means compromising on what you think is necessary rigor. JeffJor (talk) 14:01, 24 March 2010 (UTC)
- After further thought, I decided to be less evasive (although I wasn't trying to be - I wanted a first-impression answer). Mlodinow intentionally dropped the references to the door numbers in his question, quite possibly to avoid this very argument. And he is a citable source. Setting aside for the moment the issues of when, how often, by whom, and just how accurate the various versions of the question are, would it be acceptable to lead off the article with Mlodinow's version - call it maybe the "Simple Problem" - and answer it with a simple solution - an isomorph if you insist - that does not rely on conditional probability or any history? That way the unintuitive result can become established as correct before any need to demonstrate rigor, or explain who thinks who else is wrong or misinterpreted anything. Then go into details about the various versions and history, which is really an enitrely separate issue. JeffJor (talk) 14:31, 24 March 2010 (UTC)
- Jeff - you're the one here focused on what is "correct". I'm perfectly satisfied with discussing what is published in reliable sources. I really couldn't care less what you think is "correct". You really shouldn't care what I think is "correct". As far as leading with Mlodinow's version, it is not the most well known version (which means most of the literature is not referring to this version, i.e. it's as much a variant as a version where the host's preference between goats is explicitly treated as a variable p) so starting with it seems like a curious choice. Once again, I'd request comments on the readability of the solution section proposed here, which is intended to present three types of solutions published by reliable sources without characterizing any of them as "more correct" than any other. -- Rick Block (talk) 14:59, 24 March 2010 (UTC)
- The article should lead off with the original problem formulation that caused the ruckus and not with anybody's favoured version that might be found in literature. Similarly it is a bad idea to pick a particular (re)definition to make one's favourite solution rigorous or without ambiguities. Attempting such things is imho just fooling readers, furthermore much of the problem history and arguments about it cannot be understood without knowing the original statement and its ambiguities. What can do imho however, is to have a more modularized approach in several section. From my perspective Morgan doesn't have to be mentioned in the "popular section", but the popular section needs to explain how it resolves the ambiguity. You can also make a (sourced) argument under which aspect popular and conditional are actually equivalent, but you cannot simply skip that information and use it as an argument for presenting the reader with a different problem formulation.--Kmhkmh (talk) 15:14, 24 March 2010 (UTC)
- The real, the virtual "ploblem" is that Morgan et al., who never did address the "MHP" at all (!) (Monty Hall Problem: "1 door versus 2 doors" - easily recognizable - "33:67 versus 50:50"), but who addressed some osseous "Morgan et al. Ploblem", far aside from the famous paradox (!), and who have been rumbled, debunked and unmasked therefore, still have some insular fans: Some kinda narror-hidebound still feel attracted to some deviantonist relevation: What will the "Morgan et al.-moderator" do? Will he "always open a door"? Or not? Will he "always offer to switch"? Or not? And "if he does, will he flash obscene secrets?", - or not?... - and so on, and so on. Far aside the famous paradox. Probability theory seems to be in demand. Still some math scientists deal with such fallacious issues and cling to similar "conclusions", far aside the MHP. Too bad. World's most eminent scientists have exposed that Morgan was and still is a false prophet in this respect, long ago. They're hushed up in WP. Not distinguishing the MHP from the "Morgan-problem", being an entirely different issue... A pity and a shame for WP. -- Gerhardvalentin (talk) 21:23, 24 March 2010 (UTC)
- Ah yes, you know what the real problem is. We don't need to try to understand or care what multiple sources have to say (Morgan et al. are FAR from the only source saying the question the MHP asks is about conditional probability, and how the host chooses between two goats therefore matters). I know what the real problem is as well. See WP:The Truth. -- Rick Block (talk) 00:26, 25 March 2010 (UTC)
More discussion
- Rick, I have changed the title above. As I have said before, we have should all congratulate ourselves on remaining civil throughout this baffling difference of opinion. Despite years of discussion about every detail of the problem, neither side has managed to convince the other or even understand the view of the other side. Our mediator Andrevan admits that this difference of opinion is hard particularly to deal with.
- I should add there is an equally strong difference of opinion on what the sources say and how best to present this, so an appeal to 'just say what the sources do' does not solve the problem. There are fundamental, intractable and entrenched differences of opinion here. We are all fed up with the lack of progress but our only option is to carry on with discussion, mediation, and other dispute resolution tactics until we arrive at consensus or compromise. Until that time there are likely to be an continued expression of different opinions on this page.
- I believe your section title might be construed by some as the first step in name-calling and incivility and have therefore taken the liberty of changing it. Martin Hogbin (talk) 10:07, 26 March 2010 (UTC)
- Fine - and it's probably a good idea. -- Rick Block (talk) 00:50, 27 March 2010 (UTC)
- Rick, it is hard to understand where to comment on that link, since it is an archived version of this page. But that explanation is the exact kind that user:121.73.7.84 said turns people off, so I suspect he either did read it, or started to didn't want to finish because of the reasons he gave. It starts off confusing the issue by needlessly pointing out that there are "varieties of solutions." That will discourage readers who only want to be given THE solution (whatever that means). It introduces issues (some crossed out) whose importance isn't mentioned, and that are not used in any of the three. That will also confuse people. "Conditional probability" is a math buzzword that will scare them. JeffJor (talk) 19:47, 25 March 2010 (UTC)
- Can we let the user speak for him/herself? As I read his/her comments, the proposed section is exactly the kind of solution he/she is suggesting. Non-mathematical, everyday terminology first, followed by a more mathematically rigorous solution. -- Rick Block (talk) 01:59, 26 March 2010 (UTC)
- And any mention of how "in some variations of the problem the conditional probability may differ" will put them off further - they will wonder how the probability can differ if everything happens the exact same way in the cases you are comparing, but the probability ends up different. I know where the difference you think exists is, but they won't. At that point they will be convinced that they will not be able to understand anything you have written, and stop trying to. Anything they actually understood up to that point will also be dismissed as "I must have misunderstood that, too." I know you feel you need to say all that (the reason you feel that way is because you think it is bad to use only the incorrect solution, despite claiming that you don't care which is correct), but it is unnecessary and confusing at the part of the article where this section would be inserted. All of what you want mentioned, that I said you shouldn't, can be mentioned in the places where the variations they affect are discussed. JeffJor (talk) 19:47, 25 March 2010 (UTC)
- The reason I feel the article needs to say "all that" has nothing to do with what I think is "correct", but to ensure the article represents what reliable sources say about the problem in an NPOV fashion. Presenting only the simple solutions, and deferring the conditional solution to a later "mathematical variants" section is (IMO) POV. This has nothing whatsoever to do with what I think is correct or incorrect. -- Rick Block (talk) 01:59, 26 March 2010 (UTC)
- It also gets away from the fact that the question is not "what is the probability," but "should you switch." Any probabilities you need to evaluate that question are what you have called "Bayesian Probability," not the "Frequentist Probability" Morgan calculates, and that you use in your examples. The difference is not, as you have said before, in the definition of probability. The difference is in the state of knowledge of the random process for which you are calculating a probability. A Bayesean will make assumptions about that process that a Frequentist will not. The contestant has to be a Bayesean, and base her decision on what she beleives the random process to be whether or not it is fully explaned, not on what it actually is on day the game show is filmed. It is only if the contestant beleives that a host bias exists, that Morgan's solution has any impact on the contestant's decision. The contestant has to decide based on what she thinks would happen if she repeated this same game, with the same rules, 900 times. Not on what happens to 900 other players of the game. If the host reveals no information to her about Door #2 (i.e., she doesn't beleive there is a bias), then the probabilties she should use are 1/3 and 2/3. These aren't "averages" as some have claimed, they are "expectations" based on having no information about Door #2. The two are calculated the same, but conceptually different. And there is no contradiction if her Baysean probability is not the same as what a Frequentist, who knows what the random process actually is - gets.
- The better "simple solution," but you porbably will call it OR, is this: The player initially had a 1/3 chance of having picked the car, and the player will lose by switching. In the other 2/3 of the time, the host can open one of the other doors without revealing any information about the last door. In all of those cases, the player wins by switching. Said another way, if the contestant played this game 900 times, then regardless of which door she initially picked and which door the host opened, she would win by switching about 600 times. This says nothing incorrect, that a Morganist would object to. It just pulls a sleight of hand trick to dismiss the possibility that opening Door 3 reveals information about Door 2, which is where that difference I mentioned earlier is. JeffJor (talk) 19:47, 25 March 2010 (UTC)
- If the game was to be repeated 900 times, the player, to be sure of the right answer for her situation, will choose all 900 times door 1. Only 450 times door No 3 will be opened, and will she has the option, as in the original situation, to switch to door No 2. And in 300 of thes 450 times se will get the car if she switches. But this differs conceptually from the 600 of 900 times, if she just chooses some door and decides to switch anyhow. Nijdam (talk) 22:55, 25 March 2010 (UTC)
- Jeff - If you're putting words into the mouths of those who present simple solutions to "fix" those solutions, anyone who knows what OR is would agree that what you're doing is OR. The sources are perfectly capable of speaking for themselves. -- Rick Block (talk) 01:59, 26 March 2010 (UTC)
Question for user:121.73.7.84
I think the question for user:121.73.7.84 has been buried (at least twice) in the thread above. I'll ask again - is the revised solution section as proposed here easier to understand?
If anyone else would like to respond to this question feel free. However, if you want to bitch and moan about how certain references interpret the problem, please do it elsewhere. -- Rick Block (talk) 22:34, 24 March 2010 (UTC)
Crikey, i didn't realise I had stumbled into a hotly debated topic. I am not a mathematician and wikipedia is not a site specifically for mathematicians so the feedback I would offer is to also give an explanation of the problem that does not require understanding of mathematical jargon. A mathematical explanation using correct scientific terms and equations should certainly be included, however, (imo) this tends to switch most people off - hence this suggestion for the dual forms of explanation. In my experience the general public tend to think in verbal-type concepts with everyday words rather than equations and official mathematical terminology.
Best of luck 121.73.7.84 (talk) 07:10, 25 March 2010 (UTC)
- Did you read the proposed solution, linked above? -- Rick Block (talk) 13:09, 25 March 2010 (UTC)
- In case it's not clear, the proposed solution section was a work in progress - please ignore the strkeouts (pretend those phrases aren't there), and comment here. -- Rick Block (talk) 01:30, 26 March 2010 (UTC)
Are you sure you've linked me to the section you intended? I don't think the explanation you have linked to is easily understandable for a lay person. Terms such as "reasonable heuristic", "conditional probability", "by reason of symmetry", "this subset differs from the total set regarding a property influencing the probability", etc, etc., -while perhaps mathematically correct terms- are not remedial enough when attempting to explain an already abstract concept to a general reader. 121.73.7.84 (talk) 12:37, 26 March 2010 (UTC)
- I'm not sure what you're reading. I mean the suggestion in the collapsed section below. -- Rick Block (talk) 14:27, 26 March 2010 (UTC)
Proposed text
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Yes, that is much better 121.73.7.84 (talk) 22:24, 26 March 2010 (UTC)
- I agree that is better, but maybe not the best we could have. One of the problems is that once you have got it, it can become quite hard to see what the problem was originally. Thus we need to aim the first part of the article at someone who has never seen the problem before. Like many things, the answer is obvious when looked at the right way. Martin Hogbin (talk) 09:32, 27 March 2010 (UTC)
@user:121.73.7.84 - When reading the suggested wording above does the third section make you think that the first two (simple) explanations are wrong, or that they're simply other ways to look at the problem? And, although I assume the third (conditional probability) section is mostly a "switch off" section, have you puzzled out what it is saying and if so do you find it helpful (and, if not, do you find it interferes with your understanding of the other two sections)? -- Rick Block (talk) 17:24, 28 March 2010 (UTC)
Which is the third section? 121.73.7.84 (talk) 23:28, 28 March 2010 (UTC)
I've also just noticed the following at the beginning in the FAQ section:
Q: There are clearly two doors, how can it not be 50/50? A: The simplest explanation may be that any player's initial probability of not picking the car is 2/3. Monty's action of revealing a goat behind a door doesn't change that. Therefore, the player should accept the offer to switch. It doubles his likelihood of winning the car.
This is terrible since it IS Monty's actions that are ultimately the key to understanding this problem. The above doesn't "simply" explain the problem at all, and it sends people off on a tangent away from an analysis of Monty's actions which is the key to this problem. The doubling of probability doesn't make sense otherwise. 121.73.7.84 (talk) 23:52, 28 March 2010 (UTC)
- What I mean by the third section is the third subsection. The first subsection is Simplest approach, the second is Enumeration of all cases where the player picks Door 1, the third is The probability of winning by switching given the player picks Door 1 and the host opens Door 3. -- Rick Block (talk) 04:38, 29 March 2010 (UTC)
Question for Martin
My understanding is the only thing Martin is arguing for at this point is to include the existing "Aids to understanding" section between the existing "Popular solution" and the existing "Probabilistic solution" sections (so as not to confuse the reader with the complexities of conditional probability). The confusing verbiage user:121.73.7.84 specifically mentions (above) is in the current "Popular solution" section. Is the "just rearrange things" proposal an example of what might be better than what I've proposed - or are more changes envisioned than just rearranging things? If more changes, can someone create a draft we can all look at (doesn't have to be perfect)? Here's the simple rearrangement. -- Rick Block (talk) 16:37, 27 March 2010 (UTC)
- To explain, there have been many versions of the 'Popular solution' section, some better than others. The best, in my opinion, have made no mention of the fact that it might matter which door the host opens when he has a choice. This is what I would want to start with. I have edited this version to show what I would want to.
Martin's proposal
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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." |
- So, to be clear, are you saying that your modified version would be better than what I'm suggesting in the section above? -- Rick Block (talk) 19:41, 27 March 2010 (UTC)
- Yes. The principle is simple. No confusing complications in the first bit, then explain everything. Martin Hogbin (talk) 20:27, 27 March 2010 (UTC)
In my opinion Rick Block's explanation is simpler for a random lay person to follow. Martin's explanation has more "switch off" elements which could fluster people. This is just my opinion. 121.73.7.84 (talk) 08:35, 28 March 2010 (UTC)
- Thanks for your opinion. Perhaps you could say which bits are the 'switch off' elements. This would be useful to us all. Martin Hogbin (talk) 09:01, 28 March 2010 (UTC)
- As long as we're posing questions, I've posed some above as well. -- Rick Block (talk) 17:25, 28 March 2010 (UTC)
Well, firstly I think starting with the diagram is not an easy way to initially explain this solution. Secondly the following text is hard to follow unless you already know the solution:
Quote: "The above diagram shows that a player who switches always gets the opposite of their original choice, and since the probability of that choice being a goat is twice that of being a car, it is always advantageous to switch. In other words, the probability of originally choosing a goat is 2/3 and the probability of originally choosing the car is 1/3. Once Monty Hall has removed a "goat door," the contestant who chose the door with a goat behind it will necessarily win the car, and the contestant who originally chose the car will necessarily "win" the goat. Because the chances are 2/3 of being a contestant who originally chose a goat, probability will always favor switching choices." 121.73.7.84 (talk) 23:24, 28 March 2010 (UTC)
The solution I gave, while perhaps sounding inelegant was at least simple, quick and -i believe- understandable 121.73.7.84 (talk) 23:26, 28 March 2010 (UTC)
On further reflection, if you just let the diagram explain itself and eliminate the inarticulate and confusing text, quote: "The analysis can be illustrated in terms of the equally likely events that the player has initially chosen the car, goat A, or goat B" - then it is clearer. Saying: "the analysis can be illustrated" or writing "initially chosen the the car, goat A, or goat B" when they mean "has either chosen goat A or Goat B or the car." 121.73.7.84 (talk) 23:39, 28 March 2010 (UTC)
Response from Martin
@121.73.7.84 Thanks for you continued interest in the article. We need more input from people less familiar with the problem to help us formulate the first part of the article in the most understandable way. The point of contention here is whether to include any of the details in the 'Probabilistic solution' (in particular, reference to the possibility that it might matter which door the host opens in some cases) in the first part of the article. We all want to make the explanations as clear as possible.
@Rick As you know the above text is not my wording nor my preferred layout, it was just my quick modification to your suggestion to remove the confusing stuff. I would really like to have several simple solutions at the start of the article, none of which mention the host legal door choice. Martin Hogbin (talk) 09:33, 29 March 2010 (UTC)
My advice is to start as if you're teaching your 5-year old the ABCs and then progress to more elegant, more mathematically technical explanations. 121.73.7.84 (talk) 10:00, 29 March 2010 (UTC)
- @Martin - No, I don't know the text above is not your preferred wording or layout. It was my understanding you've been saying for a while all you really want is to move the "Aids to understanding" section. If this was to be only the first step of a more comprehensive overhaul I must have missed that. If you want to restructure and also rewrite, can you draft a sketch? I mean, it's not like you're averse to writing (with as much as you've written on this and other related pages you could have rewritten the entire article 100 times over by now). I don't know about anyone else, but I'm finding arguing in the abstract to be incredibly unproductive. In concrete terms, I think the version I've drafted above is in a completely different (better) league than what's currently shown as "Martin's suggestion".
- @user:121.73.7.84 - what you're suggesting is the exact structure I'm aiming for in my suggestion above. Some background about the ongoing dispute if you're interested - what Martin and the other "anti-conditionalists" are arguing for is to have a more or less complete article mentioning only the "simple" explanations, and to present any more technical explanation effectively in an appendix, or "variations" or "experts only" section. The argument against this approach is it does not give equal weight to the viewpoint published by multiple sources that most published "simple" explanations (specifically, vos Savant's) are missing something (see selected quotes in a bullet list 3 or 4 replies into this thread). One of Wikipedia's core content policies is to take a neutral point of view (see WP:NPOV). Pushing significant published criticism of the simple solutions into an appendix IMO would be a gross violation of this policy. -- Rick Block (talk) 19:51, 29 March 2010 (UTC)
My proposal explained
At the time I first made my proposal I was reasonably happy with the 'Popular solutions' section and thus just moving a section would have been acceptable to me (although not ideal), however the main purpose of my suggestion is to end this long-standing argument. If we had essentially two separate sections that treat (interpret) the question in two different ways then there would be little remaining argument.
The first section would fully (including 'Aids to understanding') deal with the notable 'puzzle' question in which there is no consideration of the possibility that the legal door chosen by the host might matter. Once the decision to have this section is accepted, the main argument would no longer be relevant and we could all work together, following to WP policies, to improve the article. Contrary to what some may think, I do wish to rewrite this article on my own. The first section might contain several simple solutions from a variety of sources.
The second section could deal with the problem as interpreted by Morgan et al and fully explain its conditional nature, including mention of Morgan's criticism of the simple solutions. Later sections could treat the problem in other ways. Martin Hogbin (talk) 09:37, 30 March 2010 (UTC)
- To what sources would you reference the interpretation that there is no "possibility that the legal door chosen by the host might matter"? To clarify, do you mean by this that the question is taken to be the unconditional question (in which case p does not matter), or that p must be taken as 1/2? And, how is it NPOV to exclude from this section published criticisms (and, again, for the 400,000th time, it's not just Morgan!) of such an interpretation, like say Lucas, Rosenhouse, and Schepler: ... any proposed solution to the MHP failing to pay close attention to Monty’s selection procedure is incomplete?
- Note, specifically, this section of WP:NPOV, which says (in part): Segregation of text or other content into different regions or subsections, based solely on the apparent POV of the content itself, ... may also create an apparent hierarchy of fact: details in the main passage appear "true" and "undisputed", whereas other, segregated material is deemed "controversial", and therefore more likely to be false — an implication that may not be appropriate. A more neutral approach can result from folding debates into the narrative, rather than distilling them into separate sections that ignore each other. -- Rick Block (talk) 14:00, 30 March 2010 (UTC)
Has anyone actually watched the show itself?
I've read through the archives, and seen some fascinating material covered there (even though, as a non-mathematician, some of it flew over my head). One obvious question comes to mind, though - has anyone ever actually gone through past tapes of LMAD to look for instances of this specific game, and found out what the real-world results ended up being (assuming that there's a sufficient sample size)? One would think that if there were enough instances, then the results would speak for themselves. Bcdm (talk) 07:25, 18 April 2010 (UTC)
- Curiously, the show itself is of little relevance to solving what has become known as the Monty Hall problem. On the actual show Monty Hall never actually gave the player the option to swap doors. The scenario generally known as the Monty Hall problem was first proposed by Selvin in a letter to 'The American Statistician' and although it referred to the show it proposed a question, 'Should the player swap?' that was never asked on the actual show. Some years later, this version of events took on a life of its own when Whitaker asked a similar question to the 'Ask Marilyn' column in 'Parade' magazine. Martin Hogbin (talk) 08:24, 18 April 2010 (UTC)
Would it be accurate to say
Since 2/3 doors are the wrong choice, the chances are that you picked the wrong door in the first place, so when shown the other wrong one, the remaining one is more likely to be correct since it's not the one you know is wrong, and not the one you know to be probably wrong. --66.66.187.132 (talk) 03:45, 5 May 2010 (UTC)
- That's one way to look at it. Another way to look at it, given you've picked door 1 and host has opened door 3, is that the host has a 100% chance of opening door 3 if the car is behind door 2 (1/3 chance of this happening) but only a 50% chance of opening door 3 if the car is behind door 1 (1/6 chance of this happening). The situation is similar if the host opens door 2 (1/3 chance the car is behind door 3 and 1/6 chance it's behind door 1). So whichever door the host opens, the chances are twice as good that the car is behind the other unopened door than the one you originally picked - because the host has 2 choices for which door to open if the car is behind the door you pick, but has only one choice if the car is not behind the door you originally picked. -- Rick Block (talk) 04:06, 5 May 2010 (UTC)
Yes, 66.66.187.132, that's all there is to it. The 2/3 chance that your door choice is wrong is not changed by the host's actions. And since the total probability has to add up to 1, the remaining door has only a 1/3 chance of being wrong (more importantly, a 2/3 chance of being the car). Glkanter (talk) 05:55, 5 May 2010 (UTC)
This article is overly complicated, confusing and just plain wrong in more than one place.
The OVERALL odds of winning the game ARE in fact 1/2!
Yes if you play the "switch strategy" the odds of winning are 2/3 and staying will result in 1/3. However the contestant is asked to make to choices: 1) choose one of three doors 2) stay or switch after a goat is revealed
If both of these choices are made randomly than the odds of winning the car are 1/2. One can prove it two ways:
1) averaging strategies: one third plus two thirds is one, one over two is one half.
2) one car, two doors (what are you not getting here?)
The event matrices are good visual descriptions on how switching improves your odds but the article fails to differentiate between the odds of winning with a random selection and acting on a strategy of switching or staying.
Explaining this will leave the reader more satisfied. I would be happy to rewrite the article if that bot will stop reverting my edits. Whatever, it just wouldn't be wikipedia if articles weren't full of wrong. —Preceding unsigned comment added by 101glover (talk • contribs) 09:29, 5 May 2010 (UTC)
- Yes, you are correct if all choices made by the player are random the probability that the player will win is 1/2 but that is not what the Monty Hall problem is all about. The most famous statement of the problem is given at the start of the article:
- Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
- Note the question that I have put in bold. The questioner wants to know if it is better to swap or not. Vos Savant answered correctly that it is better to swap and that you double your odds of winning if you do so. Do you agree with this? Martin Hogbin (talk) 10:57, 5 May 2010 (UTC)
- That's not enough. You have to include the condition that Monty ALWAYS behaves this way. That is, he ALWAYS chooses a door that he knows has a goat and ALWAYS opens that door. If you give Monty the CHOICE of opening the door or not, then there is no unique answer. For example, Monty From Hell: Monty only gives you a choice if you have chosen the door with the prize. Angelic Monty: Monty only gives you a choice if you have chosen a door with a goat.
- Vos Savant was careless when she originally stated the problem, because she neglected to state this condition clearly (actually, at all). Bill Jefferys (talk) 23:25, 5 May 2010 (UTC)
- To justify the answer that the chance of winning by switching is 2/3 knowing which door the player picked and which door the host opened 1) the car must be placed randomly, 2) the host must always open a losing door and make the offer to switch, and 3) the host must choose between two losing doors with equal probability. It can certainly be argued that vos Savant implied all of these, and most people apparently assume these conditions whether they're explicitly included in the problem statement or not.
- However, back to the original objection here, the problem is definitely not talking about the chances if the player makes a random choice of whether to switch or not. Perhaps a more clear way to ask the question (assuming the clarifications about the car placement and host's behavior) is to ask about an expected frequency, i.e. if the show were on 900 times and the players initially picked a random door, how many of the players who pick door 1 and then see the host open door 3 would win if they switch?
- If their initial pick is random, we'd expect about 300 players to initially pick door 1. If the car is placed randomly, of the 300 who picked door 1 we'd expect the car to be behind each door about 100 times. If the host opens door 3, it's because the car is behind door 2 (100 times) or the host has chosen to open door 3 when the car is behind door 1 (50 times). So, there will be about 150 players who choose door 1 and see the host open door 3, and of these 100 win if they switch. The chance of winning the car by switching (given the player picks door 1 and sees the host open door 3) is therefore 100/150 = 2/3. Of course, if all 150 of these players randomly choose whether to switch, about 75 will switch and 75 will stay with their initial choice. Of the 75 who switch about 2/3 (50) will win the car and of the 75 who stay about 1/3 (25) will win the car, so about 75 altogether (i.e. 1/2). This doesn't mean the chances of winning by switching are 1/2 but rather the chances of winning by making a random choice of whether to switch are 1/2.
- See the difference? -- Rick Block (talk) 04:34, 6 May 2010 (UTC)
- As Rick has said, there were some significant assumptions made by vos Savant that she did not male clear in her reply. These have been the subject of later criticism of her response but they are fully covered in this article.
- The point being made by 66.66.187.132 was a different one. He was talking about the overall probability of winning if the player chooses randomly whether to switch or not, which is 1/2 but which is a question that I have never seen asked before and is not generally considered to be the Monty Hall problem. I guess he was misled by use of the term 'overall'. Martin Hogbin (talk) 08:53, 6 May 2010 (UTC)
- Martin - of course you've seen this question before. It's vos Savant's "little green woman" variant (from her 3rd column about the problem). -- Rick Block (talk) 13:16, 6 May 2010 (UTC)
- Yes you are quite right Rick, I had forgotten about that. Martin Hogbin (talk) 14:56, 6 May 2010 (UTC)
- Martin - of course you've seen this question before. It's vos Savant's "little green woman" variant (from her 3rd column about the problem). -- Rick Block (talk) 13:16, 6 May 2010 (UTC)
- A lot of grief could be avoided if the article, instead of quoting vos Savant's flawed statement, instead led off with a correct statement of the problem, for example, as Martin Gardner originally put it (I don't have a copy of his book available but I am morally certain that he would have been careful to give an accurate statement of the problem).
- As it is, leading off with vos Savant's flawed version and then trying to clean it up in the next paragraph or two just sows confusion. Bill Jefferys (talk) 17:14, 6 May 2010 (UTC)
- Interesting suggestion. Thanks for your input. Glkanter (talk) 17:49, 6 May 2010 (UTC)
- I assume by "Savant's flawed version" you actually mean Whitaker's version, in that case the answer is no. We cannot use Gardner because strictly speaking his original problem is not MHP and we cannot replace the ambiguous "original" MHP (Whitaker) by a later non ambiguous one. This is because WP is not in the business of defining the problem but has to report as it is and furthermore the ambiguity is a central part of the problem itself and much of its controversy.--Kmhkmh (talk) 18:41, 6 May 2010 (UTC)
- I disagree. By far, the most notable and well known version of the MHP is as a mathematical puzzle in which all necessary assumptions are made for the solution to be simple with the player having a 2/3 chance of winning if they switch. This was actually the question that vos Savant intended to answer. Unfortunately, she did not make some of her assumptions clear until after she had given her reply. Martin Hogbin (talk) 18:50, 6 May 2010 (UTC)
- Frankly you lost me completely here. Earlier one of your main arguments was that the MHP is not a math problem/puzzle and now you're telling me it is? That aside, I don't quite see, where your statement contradicts what I've wrote above, so where is the actual disagreement?--Kmhkmh (talk) 19:23, 6 May 2010 (UTC)
- I disagree. By far, the most notable and well known version of the MHP is as a mathematical puzzle in which all necessary assumptions are made for the solution to be simple with the player having a 2/3 chance of winning if they switch. This was actually the question that vos Savant intended to answer. Unfortunately, she did not make some of her assumptions clear until after she had given her reply. Martin Hogbin (talk) 18:50, 6 May 2010 (UTC)
- Martin - I can't tell who you're disagreeing with here, Bill or Kmhkmh. I thought you have consistently argued that the most notable and well known version is Whitaker's (as published by vos Savant). In the current structure of the article (problem, solution, explanations, variants, history) the problem section starts with Whitaker's version, mentions the ambiguities, and presents a fully unambiguous version. This understanding of the problem (which I think is the mathematical puzzle you're referring to) is the topic of the solution and explanations sections. Variations are covered in the variants section. I think this structure matches the bulk of the references, many of which use Whitaker's version (modulo minor misquotings) but then present a solution that actually applies only to the fully unambiguous version (usually ignoring the ambiguities).
- To Bill's point - I think the article essentially already does this. Although the "Problem" section starts off with Whitaker's version it proceeds directly into an unambiguous version. For summary in the lead, Whitaker's version is used (since it's more well known, and shorter) but even in the lead the fact that this version of the problem is underconstrained is mentioned.
- And, finally, I completely agree with Kmhkmh. We can't change the problem statement (or solutions) to our liking but need to take what's published about it. We're not writing an original paper about this problem - we're summarizing what the published literature has to say about it. -- Rick Block (talk) 19:31, 6 May 2010 (UTC)
- In case i wasn't quite clear above, I'm fine with the current introduction/lead as it is. I was under the impression Bill suggested to change that and replace it by a non ambiguous problem version. While that approach is fine for an article on MHP published elsewhere, it should not/cannot be handled in WP in such a manner for the reason stated above.--Kmhkmh (talk) 19:42, 6 May 2010 (UTC)
- And, finally, I completely agree with Kmhkmh. We can't change the problem statement (or solutions) to our liking but need to take what's published about it. We're not writing an original paper about this problem - we're summarizing what the published literature has to say about it. -- Rick Block (talk) 19:31, 6 May 2010 (UTC)
You know, if I were to read today's comments a certain way, I could conclude that the article is not only at the level of a Featured Article, but has attained Nirvana, and could not possibly be improved upon by mortal editors. Kudos! Glkanter (talk) 19:50, 6 May 2010 (UTC)
- @Rick and Kmhkmh, this is meant to be an encyclopedia article on the Monty Hall problem not a literature review on the subject. Everything we write should be supported by reliable sources but the sources do not, indeed cannot, tell us how to compose the article. Martin Hogbin (talk) 13:04, 7 May 2010 (UTC)
What is notable
The most notable statement of the problem is undoubtedly that of Whitaker but, as we all know, this statement leaves out much of the information needed to solve the problem and thus needs interpretation. The most notable interpretation of Whitaker's question is undoubtedly that of vos Savant. She, eventually, made clear that she had interpreted the problem as a simple mathematical puzzle, in which the host always offers the swap and the host always opens a legal door randomly to reveal a goat. This simple interpretation turns out to be exactly the same as the question asked and later clarified by Selvin some years earlier. This is the notable problem formulation because it resulted in thousands of letters claiming the answer was 1/2 and not 2/3. No one argued that it might make a difference which door the host opened when he had a choice, or that the player had to make her choice before the host revealed a goat. The only point in question was whether the answer was 1/2 or 2/3. This simple problem formulation, in which it is quite obvious that it makes no difference which legal door the host opens, then found its way into countless puzzle books and web sites and it is by far most the important, interesting, and notable version and it is the one that we should primarily be addressing here.
A second, and far less well known MHP, was later created by Morgan et al who showed that in their, somewhat perverse, interpretation of the problem, it could matter which door the host opened when he had a choice. This is indeed an interesting point for those starting to study conditional probability as it shows how a seemingly unimportant detail can have a significant effect on the a probability, and this version should, of course, be addressed here also.
However, nothing in WP policy, or common sense, tells us that the Morgan interpretation should control the whole article including dictating how we should address and fully explain the solution to the simple problem. That is the most important function of this article and it is up to editors here to decide how to do it. Martin Hogbin (talk) 13:46, 7 May 2010 (UTC)
- Clear and precise formulation! Thank you. – Makes it "to the point", finally showing the real problem: (delete this if you like)
In exactly 2/3 of millions the player will win by switching, and only in 1/3 of millions she will win by staying. For most people it is hard to believe, let alone to understand. But a fact. No doubt. The famous 50:50 paradoxon. But no one can change this fact, neither Morgan.
Some smart, but totally fabricated, say perverse interpretation occured: If the host gives information about the actual status (although he never can change the actual, current status!), e.g. having a preference for opening one particular non selected door, then he can meet his preference in 2/3 of millions only:
- In 1/3 (when he has got two goats to choose from), i.e. when staying will win!
And in another half of 2/3 (when he got the car and just one goat, and when the goat by chance is behind his preferred door), i.e. when switching will win!
- In these 2/3 of millions, both switching and staying in average will have a chance of 1/2 each.
But in 1/3 of millions the car will be behind his preferred door, and by exceptionally opening the "avoided and unusual" door he shows that switching will win anyway with a chance of 1. But the host never can change the current status, though. He just may indicate information about the actual constellation. Why not opening all three doors at once? But all of that is a dishonorable game, when secret facts about the actual constellation should be revealed. Just suitable for students in probability. Not really affecting the famous 50:50 paradoxon in any way.
- This fact being emphasized in every university, and you can find it in multiple internet pages of universities all over the world. This should clearly be stated in the article, and Morgan et al. may not be allowed to confuse. Facts should be presented clearly in WP. Sources? Wherever you have a look. Regards, --Gerhardvalentin (talk) 22:58, 7 May 2010 (UTC)