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::I think it best not to edit the article at all, so I've changed the wording to reflect that. [[User:Sunray|Sunray]] ([[User talk:Sunray|talk]]) 21:12, 7 October 2010 (UTC) |
::I think it best not to edit the article at all, so I've changed the wording to reflect that. [[User:Sunray|Sunray]] ([[User talk:Sunray|talk]]) 21:12, 7 October 2010 (UTC) |
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I regret to inform you that I have a 'policy' of not agreeing to self-censorship on Wikipedia. I suggest you have the page protection restored. I'll wait a reasonable amount of time. [[User:Glkanter|Glkanter]] ([[User talk:Glkanter|talk]]) 21:18, 7 October 2010 (UTC) |
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Framework for discussion proposal
I've been thinking about this problem for a while, and would like to point out, there could be many different problems, and original question is ambigious. Following are only examples how we could understand original question. First I'd like to summarize them in english, and then in math:
PROBLEM A ("unconditional"): Which people win more often, when given selection of some gate, choose one of them, and after that Monty opens some other gate? Those who always switch, or those who always stick?
PROBLEM B ("conditional"): Let's think about the following situation: when playing the game like the one above, player selected gate 1. Monty opened gate 3. Now let's ask: Which people win more often, when given such a situation, those who switch, or those who stick?
PROBLEM C: What is the best strategy in some game theoretic sense, for winning most often.
PROBLEM D: Given the situation in PROBLEM B, what is the best strategy in some game theoretic sense for winning most often.
Now please let me formalise problems A and B. Firstly I'd like to present the whole sample space for PROBLEM A. Sample space for PROBLEM B is part of this sample space:
No | Car | Me | Monty | I stick | I switch |
---|---|---|---|---|---|
1 | 1 | 1 | 2 | I win | I lose |
2 | 1 | 1 | 3 | I win | I lose |
3 | 1 | 2 | 3 | I lose | I win |
4 | 1 | 3 | 2 | I lose | I win |
5 | 2 | 1 | 3 | I lose | I win |
6 | 2 | 2 | 1 | I win | I lose |
7 | 2 | 2 | 3 | I win | I lose |
8 | 2 | 3 | 1 | I lose | I win |
9 | 3 | 1 | 2 | I lose | I win |
10 | 3 | 2 | 1 | I lose | I win |
11 | 3 | 3 | 1 | I win | I lose |
12 | 3 | 3 | 2 | I win | I lose |
After introducing the table, please let me introduce the following notation:
[1,3,5] means the set {<1,1,2>,<1,2,3>,<2,1,3>}.
Let me define the following sets:
Om=[1,2,3,4,5,6,7,8,9,10,11,12] --- the sample space
S1=[1,2,3,4] --- it means "the car is in gate 1"
S2=[5,6,7,8] --- it means "the car is in gate 2"
S3=[9,10,11,12] --- it means "the car is in gate 3"
J1=[1,2,5,9] --- it means "I've selected gate 1"
J2=[3,6,7,10] --- it means "I've selected gate 2"
J3=[4,8,11,12] --- it means "I've selected gate 3"
W1=[1,2,6,7,11,12] --- it means "I win using strategy 'I stick'"
W2=[3,4,5,8,9,10] --- it means "I win using strategy 'I switch'"
B=[2,5] --- situation when I've selected gate 1 and Monty've selected gate 3
a=P([2])/P([1,2]) --- probability that Monty selects gate 3 when in situation [1,2]
Now let me specify what do we know:
1. For all i, for all j P(Si product Jj)=P(Si)*P(Jj) --- because our selection of gate is independent of where the car is. ("axiom")
2. P(W1)+P(W2)=1 ("theorem")
3. P([2])=a*P([1,2])=a*P(S1 product J1)=a*P(S1)*P(J1) ("theorem")
4. P([5])=P(J1 product S2)=P(J1)*P(S2) ("theorem")
Now we are ready to ask questions:
PROBLEM A: P(W1)=?
P(W1)=P((S1 product J1) sum (S2 product J2) sum (S3 product J3)). Because of beeing mutually exclusive... Because of independence...
P(W1)=P(S1)*P(J1)+P(S2)*P(J2)+P(S3)*P(J3)
PROBLEM B:
P(W1 product B)/P(B)=?
P(W1 product B)/P(B)=P([2])/P([2,5])=P([2])/(P([2])+P([5]))=a*P(S1)/(a*P(S1)+P(S2))=1/(1+ P(S2) / (a*P(S1)) )
Now let's discuss some additional assumptions we might make (not making them now):
Assumption 1: P(J1)=P(J2)=P(J3)=1/3
Assumption 2: P(S1)=P(S2)
Assumption 3: a=1/2
Assumption 4: a=1
If we make assumption 1 only:
PROBLEM A:
P(W1)=1/3*P(S1)+1/3*P(S2)+1/3*P(S3)=1/3
If we make assumption 2 only:
PROBLEM B:
P(W1 product B)/P(B)=1/(1+ 1/a)=a/(a+1)
If we make assumptions 2 and 3 only:
PROBLEM B:
P(W1 product B)/P(B)=1/(1+2)=1/3
If we make assumption 2 and 4 only:
PROBLEM B:
P(W1 product B)/P(B)=1/(1+1)=1/2
Now let me discuss about PROBLEMS C and D. These problems are DIFFERENT problems than PROBLEM A and PROBLEM B. To solve them we need to think about knowledge of strategy of an opponent. Does Monty know your strategy? Could he manipulate car placement etc. Do you know Monty's strategy? Is it important if you know his strategy, or he knows yours? Some trivial examples: if you know probabilities P(S1), P(S2), P(S3) you select the most probable gate, and stick with it. Trivial example: P(S2)=P(S3)=0, and you know it, you can use strategy: "always select gate 1". Sorry, I know nothing about game theory, therefore I can't tell what is correct solution as far as game theory is concerned (although I'm still learning ;-). Also I hope I haven't made some mistake (please review if you have enough time ;-)
Now after presenting whole problem (not really, I haven't touched PROBLEM C and PROBLEM D), please let me explain what do I think. I wonder if it is possible to use this presentation somehow when creating wiki page. I think the most important is to present how people might understand MHP. Full solution might be given, but I wonder if it is possible to use english instead of math, but still be strict enough. I think this presentation is more useful for people who would like to understand MHP in Discuss pages, not in wiki page. So my opinion is:
1. tell people that original problem is ambigious and tell them what are possible ways to understand it, but not necesarilly use math to explain it, we might explain it in english if possible, but it MUST be strict enough, for (smart) people not to be able to misunderstand each problem.
2. tell them solutions to at least PROBLEM A and PROBLEM B, maybe with some sort of explanation, but giving necessary assumptions.
123unoduetre (talk) 20:49, 22 September 2010 (UTC)
Some suggestions for a compromise
As you know I do not want to encumber the simple solutions with talk of conditional probability or anything else that might complicate the explanation.
Although it is my personal opinion that the simple solutions are completely valid, it is universally agreed that the simple solutions are valid for the unconditional problem. My only objection to stating this is that to do so brings in complications that might confuse the reader. Is it possible that we describe the unconditional problem in our simple solution section but in a subtle way that will not draw attention to the distinction between the conditional and unconditional problems at that stage?
For example, rather that say 'players whose strategy is to swap' we might just say 'players who swap'. This can still be argued to be technically correct but avoids bringing up the complication issue of strategy vs decision-made-at-the-time. Maybe the 'conditionalists' should write the simple section in language that they are happy with then the 'unconditionalist' could attempt to tone down the language whilst still keeping it technically correct. Later in the article, of course, the distinction could be discussed fully. That would include stating that many sources do give simple solutions to the problem 'as stated'. Martin Hogbin (talk) 17:54, 19 September 2010 (UTC)
- Excuse me, but I do not uderstand who was the target of this comment. If it was about "Something to include" section, I don't understand what you are saying. We certainly do NOT discuss conditional probability.123unoduetre (talk) 18:00, 19 September 2010 (UTC)
- We have a long standing dispute about sources that say the problem is asking about the conditional probability P(win by switching|player picks door 1 and host opens door 3) as opposed to the unconditional probability P(win by switching) or the "semi-conditional" P(win by switching|player picks door 1). I can only surmise Martin is thinking what I'm suggesting here is somehow related to this discussion - perhaps because the analysis I'm suggesting is technically P(win by switching|player picks door 1). I'm suggesting this analysis NOT because of this dispute, but because it is what I think most sources presenting this sort of analysis say. It is clearly what vos Savant says. -- Rick Block (talk) 19:38, 19 September 2010 (UTC)
- 123unoduetre, I started a new section because this was, exactly as Rick has said, a suggestion that concerns a very long running dispute about conditional/unconditional probability that is currently the subject of formal mediation. The problem is that it is very hard for the current editors to discuss what might be best in the article without seeing it from their particular POV. My suggestion was for a compromise that might help end the dispute so that the existing editors could all work together and consider new suggestions, such as yours, from the same perspective, that of improving the article for the general reader. Martin Hogbin (talk) 21:11, 19 September 2010 (UTC)
- It would indeed help us a lot further if a "conditionalist" would draft with as much sympathy as he or she can muster the simple solution (or the first part of it), and if an "unconditionalist" would draft with as much sympathy as he or she can muster the conditional solution (or the first part of it). And both parties would allow the others to propose rewordings which remove hidden assumptions or hidden bias. (I am disqualified since I am neither. And sadly no one else yet has felt inclined to support the "economist" point of view.) Gill110951 (talk) 12:40, 20 September 2010 (UTC)
I think it is also useful to realise that the conditionalist's solution makes more assumptions in order to gain stronger conclusions. @Nijdam recently challenged me to explain what I thought was wrong with the conditionalist's solution. I answered him here [1], on my talk page. Actually I don't think there is anything wrong with it, though I don't like the way it is sometimes sold (whether on wikipedia or in "reliable sources"). My approach as a professional writer of reliable sources in probability and statistics is always to try to figure out and understand "The Truth" first, in the hopes of finding a way of making it more easily to apprehend, despite wikipedia policy that "The Truth" is irrelevant. In the case of MHP I come to the conclusion that The MHP Truth is multi-facetted (and this is exactly what makes it such a great topic!). Now people who don't follow my arguments may complain that I am pushing Own Research. That's fine, still, it's my opinion that in the field of mathematics it does pay off to search for The Truth, and if it turns out that The Truth is easily and immediately recognisable, then it would be sad that wikipedia would have to wait 10 years for it to appear on wikipedia pages. Fortunately, if The Truth is indeed so easily and immediately recognisable then people will come up with other fundamental wikipedia policies which justify its inclusion. It's like the situation when different fundamental rights of man are in conflict with one another (eg free speech versus a ban on racial or religious or gender discrimination). I do think we're making real progress on MHP thanks to all the dedicated editors here! Gill110951 (talk) 12:32, 20 September 2010 (UTC)
- I agree that it is essential to understand all the issues relating to the MHP to write a good article on it. A cut-and-paste approach to sources does not work. Neither is it the way that WP is intended to work. Sources are meant to support the information given by editors to show that it is correct. They do not tell us how to write the article.
- I am not trying to push either the simple or conditional solutions, I am just trying to make the simple solutions as unconfusing as possible for the average reader. Talk of 'players whose strategy is...', 'solutions that do not address the stated problem' etc do nothing to help the reader understand why the answer is 2/3 rather than 1/2. on the other hand we may be able to make the simple solutions refer to the unconditional problem more subtly, with talk of 'players who switch' or the like. The lead section, which nobody has criticised, makes quite a good job of this. Martin Hogbin (talk) 22:33, 20 September 2010 (UTC)
- Rick: Influenced by Monty's action? Indeed there is some point that you are missing. As I read Ruma Falk, she clearly is saying that the opening of one door by the host cannot change the odds of the door selected by the guest, unless the host is biased to open his favored door, "and you know about this bias", and she says that – without such a given and known host's bias, the odds of the door selected by the guest remain unchanged, because we’ve learned nothing to allow us to revise the odds.
Are there reliable contradicting sources that disagree with Falk?
You say there is an influence. We should pay attention to the fact that – without expressively mentioning of such a given and known host's bias – the opening of a door by the host never can be any relevant condition to change the odds. It has no influence.
"Using" of the door opened by the host as a so called "new condition" without mentioning the host's given and known bias, evidently emerges just as an utilitarian and practical method to provide an alleged "new condition" for the purpose of teaching conditional probability maths, but all of that without really addressing the MHP itself. We should pay regard to the fact that any alleged "influence" can only result from the given and known bias of the host to open his preferred door. Otherwise not. We should mention this fact. Gerhardvalentin (talk) 22:44, 20 September 2010 (UTC)
- Rick: Influenced by Monty's action? Indeed there is some point that you are missing. As I read Ruma Falk, she clearly is saying that the opening of one door by the host cannot change the odds of the door selected by the guest, unless the host is biased to open his favored door, "and you know about this bias", and she says that – without such a given and known host's bias, the odds of the door selected by the guest remain unchanged, because we’ve learned nothing to allow us to revise the odds.
- You're reading something into what Falk is saying that she doesn't actually say (and I'd bet she wouldn't even agree with). What she says is that if the host is biased and you know about it, then it is not true that (in vos Savant's words) "we've learned nothing to allow us to revise the odds". As logical propositions, the structure here is IF A AND B, THEN NOT C, where A is "host is biased", B is "you know about it" and C is "we learn nothing". What you're claiming she's saying is IF NOT B, THEN C, which is neither what she actually says or logically implied by what she says. -- Rick Block (talk) 14:01, 21 September 2010 (UTC)
- There is no other way to gather what Falk clearly is saying: she clearly indicates just this unique reason, and no other reason, because there is no other reason. Without a given and known bias of the host, by opening of one door, you have learned nothing to revise the odds of the door selected by the guest. We should be careful and not turn around the meaning and and the significance of the statement. Gerhardvalentin (talk) 14:16, 21 September 2010 (UTC)
- Falk understands probability in the subjectivist way. Rick understands it in the frequentist way. Different notions of probability make different assumptions reasonable or unreasonable, motivate different approaches to MHP. For Falk, probability is in her mind, her information. For Rick it is in the real world. Both ways of using probability are reasonable. Since they are different kinds of probability, assumptions concerning them have different meaning, and conclusions drawn with them have different meaning. The answer may be "switch", "2/3" but the meaning of "2/3" is completely different. Gill110951 (talk) 18:16, 21 September 2010 (UTC)
- There is no other way to gather what Falk clearly is saying: she clearly indicates just this unique reason, and no other reason, because there is no other reason. Without a given and known bias of the host, by opening of one door, you have learned nothing to revise the odds of the door selected by the guest. We should be careful and not turn around the meaning and and the significance of the statement. Gerhardvalentin (talk) 14:16, 21 September 2010 (UTC)
- You're reading something into what Falk is saying that she doesn't actually say (and I'd bet she wouldn't even agree with). What she says is that if the host is biased and you know about it, then it is not true that (in vos Savant's words) "we've learned nothing to allow us to revise the odds". As logical propositions, the structure here is IF A AND B, THEN NOT C, where A is "host is biased", B is "you know about it" and C is "we learn nothing". What you're claiming she's saying is IF NOT B, THEN C, which is neither what she actually says or logically implied by what she says. -- Rick Block (talk) 14:01, 21 September 2010 (UTC)
Frequentist
- I disagree with you here Gill. The frequentist approach gives the same answer as the Bayesian approach, provided that the correct experiment is repeated. Martin Hogbin (talk) 21:45, 21 September 2010 (UTC)
- Nonsense. Experiment and choice of priors are independent for the subjectivist Bayesian (as well as for the objectivists who believe in MAXENT, as MAXENT is not a necessary constraint on the choice of priors). There is no such thing as a "repetition of correct experiments" for the subjectivist. There are only explicit statements of what the priors and the likelihoods are. glopk (talk) 14:12, 22 September 2010 (UTC)
- I have not mention subjectivists. I said that a frequentist approach would give the same result as a Bayesian approach if the correct experiment is repeated. If you disagree, perhaps you could give me an example where the Bayesian and frequentist approaches give different answers to the MHP. Martin Hogbin (talk) 17:32, 22 September 2010 (UTC)
- (1) My answer covered both subjectivist and objectivist Bayesian (problems with your reading glasses?). (2) Example: use the prior P(C) = 0 for all c.glopk (talk) 22:00, 22 September 2010 (UTC)
- I thought you were going to give me an example of a case where the Bayesian and frequentist answers (probability of winning by switching) were different. Martin Hogbin (talk) 22:39, 22 September 2010 (UTC)
- I thought you knew that frequentists have no notion of "priors" (hence the frequentist response is unaffected), whereas the Bayesian answer with prior P(C) = 0 is undefined (0/0 naively, see Borel–Kolmogorov paradox): different enough? glopk (talk) 23:03, 22 September 2010 (UTC)
- Sorry, I should have said realistic version of the MHP. I am not interested in talking about a version with no prize. Martin Hogbin (talk) 23:33, 22 September 2010 (UTC)
- Ooooh, now you want "realistic", and your criterion for being realistic is that there is a prize (in the player's opinion, since we are talking priors here). OK, I'll make it easy and really really close to the standard version. Please conjure a physically feasible repeated experiment whose relative frequencies of winning by switching converge to a value equal to the Bayesian posterior for the same event, given the following prior: P(C=c) = { 1/3 - 2x iff c=1; 1/3 + x iff c = 0,2 }, where x = 1/exp(N!) and N is the Avogadro number, and the "realistic" convergence threshold is less than x. glopk (talk) 05:09, 23 September 2010 (UTC)
- BY realistic I mean some version of the MHP. There has never been a version suggested with no prize. The experiment does not need to be physically feasible. In principle, any experiment can be repeated as many times as we wish. Martin Hogbin (talk) 08:56, 23 September 2010 (UTC)
- As usual, you didn't even try to read my response, so I have to spell it out for you. I gave you above an example with prizes which is vanishingly close to the standard version of the MHP. Yet, a frequentist analysis for it has to conjure an experiment that is infeasible to such a degree as to make the whole notion of "experiment" redundant: if you tried random sampling for the prior I gave you above, you'd have to collect data for much longer than the age of the universe before you'd get a statistically significant set. So this is the mirror image of the "no prize" example I gave you above: here it's the Bayesian guy that gets the answer in a straightforward manner, whereas the frequentist guy is left in a hopeless situation. I ask again: different enough? glopk (talk) 17:57, 23 September 2010 (UTC)
- BY realistic I mean some version of the MHP. There has never been a version suggested with no prize. The experiment does not need to be physically feasible. In principle, any experiment can be repeated as many times as we wish. Martin Hogbin (talk) 08:56, 23 September 2010 (UTC)
- Ooooh, now you want "realistic", and your criterion for being realistic is that there is a prize (in the player's opinion, since we are talking priors here). OK, I'll make it easy and really really close to the standard version. Please conjure a physically feasible repeated experiment whose relative frequencies of winning by switching converge to a value equal to the Bayesian posterior for the same event, given the following prior: P(C=c) = { 1/3 - 2x iff c=1; 1/3 + x iff c = 0,2 }, where x = 1/exp(N!) and N is the Avogadro number, and the "realistic" convergence threshold is less than x. glopk (talk) 05:09, 23 September 2010 (UTC)
- Sorry, I should have said realistic version of the MHP. I am not interested in talking about a version with no prize. Martin Hogbin (talk) 23:33, 22 September 2010 (UTC)
- I thought you knew that frequentists have no notion of "priors" (hence the frequentist response is unaffected), whereas the Bayesian answer with prior P(C) = 0 is undefined (0/0 naively, see Borel–Kolmogorov paradox): different enough? glopk (talk) 23:03, 22 September 2010 (UTC)
- I thought you were going to give me an example of a case where the Bayesian and frequentist answers (probability of winning by switching) were different. Martin Hogbin (talk) 22:39, 22 September 2010 (UTC)
- (1) My answer covered both subjectivist and objectivist Bayesian (problems with your reading glasses?). (2) Example: use the prior P(C) = 0 for all c.glopk (talk) 22:00, 22 September 2010 (UTC)
- I have not mention subjectivists. I said that a frequentist approach would give the same result as a Bayesian approach if the correct experiment is repeated. If you disagree, perhaps you could give me an example where the Bayesian and frequentist approaches give different answers to the MHP. Martin Hogbin (talk) 17:32, 22 September 2010 (UTC)
- Nonsense. Experiment and choice of priors are independent for the subjectivist Bayesian (as well as for the objectivists who believe in MAXENT, as MAXENT is not a necessary constraint on the choice of priors). There is no such thing as a "repetition of correct experiments" for the subjectivist. There are only explicit statements of what the priors and the likelihoods are. glopk (talk) 14:12, 22 September 2010 (UTC)
- I don't know what you mean here, Martin. Of course all probabilities, whether ontological / objective or epistemological / subjective (whether supposed to be properties of objects in the real world like time, mass, length, or supposed to be a reflection of particular person's knowledge), can be thought about by imagining appropriate repetitions. Both kinds satisfy the same axioms. However, in the real world, motivating assumptions about probabilities does involve settling also this question. If you want to motivate an assumption that some probability is 1/3, you have to say 1/3 of what kinds of repetitions. It means 1/3 of the times, but what "times" are we are talking about? The "times" could consist of viewing a movie of a possible parallel world, the parallel worlds of our imagination corresponding to our beliefs about the real world based on our information about it. Conversely, "objective" probabilities can be thought of as the subjective probabilities of an imaginary subject who knows a great deal more than you or I. The philosophers have been going round in circles for 300 years and will no doubt continue to do so for another 300 years. I think ultimately it is a matter of taste (politics, religion, culture) whether you take one or the the point of view as primary, since you can always embed the other view inside your own. But it is very important to get out onto the table, what repetitions people have in mind when they talk about a probability. What stays the same, what varies; how and why does it vary. Gill110951 (talk) 14:19, 22 September 2010 (UTC)
- Gill, your last two sentences were exactly the point that I was making. With the appropriate repetitions in the frequentist result is the same as the Bayesian result. Martin Hogbin (talk) 17:32, 22 September 2010 (UTC)
- I don't know what you mean here, Martin. Of course all probabilities, whether ontological / objective or epistemological / subjective (whether supposed to be properties of objects in the real world like time, mass, length, or supposed to be a reflection of particular person's knowledge), can be thought about by imagining appropriate repetitions. Both kinds satisfy the same axioms. However, in the real world, motivating assumptions about probabilities does involve settling also this question. If you want to motivate an assumption that some probability is 1/3, you have to say 1/3 of what kinds of repetitions. It means 1/3 of the times, but what "times" are we are talking about? The "times" could consist of viewing a movie of a possible parallel world, the parallel worlds of our imagination corresponding to our beliefs about the real world based on our information about it. Conversely, "objective" probabilities can be thought of as the subjective probabilities of an imaginary subject who knows a great deal more than you or I. The philosophers have been going round in circles for 300 years and will no doubt continue to do so for another 300 years. I think ultimately it is a matter of taste (politics, religion, culture) whether you take one or the the point of view as primary, since you can always embed the other view inside your own. But it is very important to get out onto the table, what repetitions people have in mind when they talk about a probability. What stays the same, what varies; how and why does it vary. Gill110951 (talk) 14:19, 22 September 2010 (UTC)
Frequentist - Mediation Link Placeholder
The tree derived from Carlton's simple solution leads me to conclude that the 1/3 v 2/3 results are identical for the simple and the conditional solutions. Any number (the 1/3 v 2/3 before a door has been revealed, aka Carlton's simple solution) multiplied by 1 (the 100% likelihood that the host will reveal a goat) remains unchanged (the 1/3 v 2/3 after a door has been opened to reveal a goat, aka the non-traditional conditional solution). Glkanter (talk) 14:28, 22 September 2010 (UTC)
- More nonsense. Of course it is possible for the conditional solution to yield a different value for the probability of winning by switching - Whitaker's statement of the problem is compatible with any assignment of the prior distribution P(C) for the location of the car, and with any assignment for the host's strategy P(H | C, S) for selecting the door to open. With P(C) uniform (= 1/3) only the optimal decision need be the same between the simple and conditional solutions, since the posterior P(C | H, S) >= 1/2 regardless of the host's strategy, but the probability of winning by adhering to that decision can still be any number in [1/2, 1]. It's just that in the conditional formulation more "stuff" is made explicit that the simple solutions conveniently sweep under the carpet, and therefore the conditional formulations allow more flexibility for reasoning about the assumptions while remaining consistent with them. glopk (talk) 16:02, 22 September 2010 (UTC)
- It's not 'nonsense'. It just neatly disproves your arguments. Can you demonstrate how that decision tree is flawed in some way? Or that it doesn't address the MHP as put forth by Selvin, vos Savant and Whitaker? But not Morgan's 'slightly different problems'. The article isn't about those. Besides, are you taking issue with the simple solutions, or with the *presentation* of the simple solutions by the reliable sources? Glkanter (talk) 18:53, 22 September 2010 (UTC)
- I am taking issue with the nonsensical statement above: "the 1/3 v 2/3 results are identical for the simple and the conditional solutions". The conditional solution can yield any value in [1/2, 1] for the probability of winning by switching with a uniform prior for the initial placement of the car, depending on the host's strategy P(H|C,S). It's only in the K&W formulation of the MHP (specifically, assuming no host bias), that the simple and conditional solution yield numerically equal values for the probability of switching. I don't have to demonstrate any flaws in any OR "decision tree" pictures to see that: it's math. glopk (talk) 21:55, 22 September 2010 (UTC)
- It's not 'nonsense'. It just neatly disproves your arguments. Can you demonstrate how that decision tree is flawed in some way? Or that it doesn't address the MHP as put forth by Selvin, vos Savant and Whitaker? But not Morgan's 'slightly different problems'. The article isn't about those. Besides, are you taking issue with the simple solutions, or with the *presentation* of the simple solutions by the reliable sources? Glkanter (talk) 18:53, 22 September 2010 (UTC)
Yes, "it's math". And "logic". I'd like to see you prove the decision tree wrong for the MHP about a game show. Based on the premises as derived from Selvin, vos Savant and Whitaker.Glkanter (talk) 23:40, 22 September 2010 (UTC)
Sure, incorporate K&W. They're not an 'original source', so I left them out. But use the whole thing from K&W. No requirement for doors 1 & 3 ("Imagine that you chose Door 1 and the host opens Door 3, which has a goat." is what they say for clarity *after* listing the premises), everything is 'random' or 'uniformly at random', no need for a solution to repeat the premises given. So, tell me where the decision tree derived from Carlton's simple solution fails. Glkanter (talk) 00:09, 23 September 2010 (UTC)
Plus, just because they're provided, not every piece of information *must* be used. For example, Selvin says in his 2nd letter that the host chooses between two goats uniformly at random. Or something to that affect. K&W pick this up as a premise. Carlton's simple solution shows that that information is not required to solve the puzzle. (In fact, Rick Block's old 'broken clock' analogy applies here. That conditional solution you're so fond of ONLY works when the host chooses equally. Carlton's, and other simple solutions show that constraint isn't necessary.) That's not a flaw in the solution. That's elegance. Glkanter (talk) 01:55, 23 September 2010 (UTC)
- I thought I could resist commenting here, but this is just too much. Like many things, you have this exactly backwards. Assuming we're interested in the probability of winning by switching AFTER the host has opened a door, i.e. the counterintuitive 2/3 probability a player has when deciding whether to switch looking at two closed doors and one open door showing a goat, the conditional solution always works since this is precisely what the conditional solution addresses. The simple solution only "works" (works meaning results in the correct numeric value for the probability) if the host chooses equally. The simple solution is the broken clock here. It's always (correctly) telling you the probability of winning if you decide to switch BEFORE the host opens a door. But if you want to know the probability AFTER the host has opened a door (when you're looking at two closed doors and one open door), using the simple solution doesn't tell you the right answer unless the answer happens to be 2/3 (which it can be but doesn't have to be - exactly like a broken clock correctly tells you the time, but only if the time is what the clock says). -- Rick Block (talk) 02:59, 23 September 2010 (UTC)
Well Rick, I've said for a long time that you and I see (nearly) all things 180 degrees differently. Maybe you could explain how that decision tree derived from Carlton's simple solution is flawed as a solution to the MHP? But no 'it fails for slightly different problems', please. But I'm especially interested in how it fails to address the 'after' problem. It shows a door has been opened to reveal a goat. 100% of the time. Exactly as K&W state the premise. No more. No less. Glkanter (talk) 03:26, 23 September 2010 (UTC)
- I'm not going to play this game. See you in mediation. I suggest you bring your sources. -- Rick Block (talk) 03:49, 23 September 2010 (UTC)
You should have followed your original instinct Rick, and not got involved in this thread. You called me out in a big, tough paragraph based on your personal opinions. Which is OK, most everybody has been offering personal opinions in this thread. I asked you to validate your 'broken clock' claims regarding the decision tree, and you accuse me of 'playing games', you decline, and say something about 'sources'. OK. Carlton, K&W, Selvin, vos Savant, Whitaker and basic math are the sources for that tree. Which sources was your comment supported by? Glkanter (talk) 04:10, 23 September 2010 (UTC)
Let's go for all the marbles. Maybe Nijdam would address the decision tree in regards to his longstanding claim that the 'before' 1/3 & 2/3 are not the same as the 'after' 1/3 & 2/3. This shows pretty clearly that they are the same. Glkanter (talk) 04:29, 23 September 2010 (UTC)
- Rick has already correctly answered your comments to my notes above. I have nothing to add, see ya in mediation. glopk (talk) 04:43, 23 September 2010 (UTC)
Gamesmanship. Hard to hold onto that presumption of 'Good Faith' from all editors. Glkanter (talk) 04:56, 23 September 2010 (UTC)
- Much easier than my having to hold onto a presumption of good faith AND competence for a certain pair of editors. glopk (talk) 06:15, 23 September 2010 (UTC)
Consistency
- Why do you choose to take P(C) as uniform but P(H|C,S) to be non-uniform? Martin Hogbin (talk) 17:38, 22 September 2010 (UTC)
- Because I can: I only need one counter-example to falsify Glkanter's general statement that "the 1/3 v 2/3 results are identical for the simple and the conditional solutions". And for the umpteenth time I have to repeat to you the simple mathematical fact that the two solutions don't have to yield equal numerical values for the probability of winning by switching. One must assume uniform car placement and lack of host bias to get equal numerical values. Most simple solutions in the sources are incomplete because they do not explicitly state these necessary assumptions, whether they purport to solve the decision problem of switching after the host's question, or whether they compute the marginal probability of winning for a population of switchers. glopk (talk) 21:55, 22 September 2010 (UTC)
- Yes, I know one must assume the unstated distributions to get and answer to the MHP but one should be consistent in one's assumptions. No information is given in Whitaker's question about either the initial car placement or the host's goat-door choice. There are two self-consistent ways to treat the problem: take both distributions to be non-uniform (and unknown), in which case the problem is insoluble, or take both distributions to be uniform, in which case the answer is exactly 2/3. There is no justification for doing anything else. Morgan et al have now agreed with this. Martin Hogbin (talk) 22:46, 22 September 2010 (UTC)
- Wrong again. (1) Exactly because no information is given about those distribution's in Whitaker's statement, there are infinite self-consistent ways to treat the problem: one may assume any mathematical form for them, and still remain consistent with the statement. (2) "Consistent" is not the same as "justified": a baroque choice for the priors may be hard to justify (did Monty feel baroque that night?), but still yield a solution consistent with the problem statement. (3) Please don't mix an argument about math with one about sources, unless you'd rather hide behind an authority. (4) Perhaps the Argument pages is more appropriate for this discussion, or we may accept Rick's plea to continue on the mediation page. glopk (talk) 23:50, 22 September 2010 (UTC)
- If course, if you want to contrive an interesting problems for statistics students, you are free to treat the unknown distributions differently but there is no mathematical, philosophical, or real-world justification for doing so.
- Wrong again. (1) Exactly because no information is given about those distribution's in Whitaker's statement, there are infinite self-consistent ways to treat the problem: one may assume any mathematical form for them, and still remain consistent with the statement. (2) "Consistent" is not the same as "justified": a baroque choice for the priors may be hard to justify (did Monty feel baroque that night?), but still yield a solution consistent with the problem statement. (3) Please don't mix an argument about math with one about sources, unless you'd rather hide behind an authority. (4) Perhaps the Argument pages is more appropriate for this discussion, or we may accept Rick's plea to continue on the mediation page. glopk (talk) 23:50, 22 September 2010 (UTC)
- Yes, I know one must assume the unstated distributions to get and answer to the MHP but one should be consistent in one's assumptions. No information is given in Whitaker's question about either the initial car placement or the host's goat-door choice. There are two self-consistent ways to treat the problem: take both distributions to be non-uniform (and unknown), in which case the problem is insoluble, or take both distributions to be uniform, in which case the answer is exactly 2/3. There is no justification for doing anything else. Morgan et al have now agreed with this. Martin Hogbin (talk) 22:46, 22 September 2010 (UTC)
- Because I can: I only need one counter-example to falsify Glkanter's general statement that "the 1/3 v 2/3 results are identical for the simple and the conditional solutions". And for the umpteenth time I have to repeat to you the simple mathematical fact that the two solutions don't have to yield equal numerical values for the probability of winning by switching. One must assume uniform car placement and lack of host bias to get equal numerical values. Most simple solutions in the sources are incomplete because they do not explicitly state these necessary assumptions, whether they purport to solve the decision problem of switching after the host's question, or whether they compute the marginal probability of winning for a population of switchers. glopk (talk) 21:55, 22 September 2010 (UTC)
- Why do you choose to take P(C) as uniform but P(H|C,S) to be non-uniform? Martin Hogbin (talk) 17:38, 22 September 2010 (UTC)
- Regarding outside discussions during mediation, I am happy to accept the mediators' advice on the matter. Martin Hogbin (talk) 08:50, 23 September 2010 (UTC)
Why don't we use The Truth here to guide our discussions? Different sources think that Whitaker asks different questions. Different sources find it more or less reasonable to make different supplementary assumptions. (Sometimes because, IMHO, they have different understandings of the word "probability", but that's a side issue; and sometimes because they think of the guest as an active player, not as a passive player). We all agree that we know that the host always opens a door revealing a goat and can do this because he knows where the car is hidden.
Do we all agree that the following (uno due tre) are Mathematically Indisputable True Facts? Do we all agree that each makes in succession a stronger assumption than the previous one?
- Uno: IF you only assume player's initial choice is correct with probability 1/3 THEN you know that switching gives him the car with probability 2/3
- Due: IF you actually assume that *all* doors are initially equally likely to hide the car THEN not only is the preceding true, but we also are guaranteed that all conditional probabilities are at least 1/2 (and on average 2/3 because of "uno"). Not only is "always switching" a good idea, but also there is no strategy (sometimes switching, sometimes not, depending on the door numbers in question) which is better.
- Tre: IF you not only assume that all doors are initially equally likely to hide the car but ALSO assume that either choice of Monty is equally likely when he has a choice, THEN the door numbers are irrelevant by symmetry. The car is behind the other door with probability 2/3. Both unconditionally, and conditionally given the specific doors. You know in advance that you can ignore the door numbers. You can also ignore the day of the week - it doesn't matter if it's Tuesday or Wednesday, either.
That's basically all there is to say. All that remains is to add the references to the sources, say that they tend to quarrel about the right assumptions to make, and to organise the whole thing into a nice story for wikipedia readers. Gill110951 (talk) 12:19, 23 September 2010 (UTC)
- Exactly. Uno: True (because there is only one car, but the player had no knowledge of where it is. That is known only by the host and his staff, and nobody else knows.)
Due: True (probability from at least 1/2 (in 2/3) to max. 1 (in 1/3), so on average 2/3, as long as nothing else is evident.)
Tre: True (for as long as you have no knowledge whatsoever about the host's behaviour, you impossibly can declare this behaviour to be "evident". So in opening of one door, no matter which "number", you are absolutely out of position to "assume better information".) – But you can speculate about it (if the host should be extremely biased either in 2/3 the odds of both doors are min. 1/2 each, or in 1/3 the chance for the second closed door will be max "1". But, for this one single game, in the situation given, you have no knowledge whatsoever on that, and so all you can say is: Pws will double your chance to 2/3. Anything more would be an arrogant presumption. Gerhardvalentin (talk) 14:42, 23 September 2010 (UTC)
- Exactly. Uno: True (because there is only one car, but the player had no knowledge of where it is. That is known only by the host and his staff, and nobody else knows.)
Suspending discussion here during mediation
Article talk pages are intended for editorial decisions concerning the article. During a mediation, the editing process is usually suspended while issues are sorted out. I know it is hard, as we have a limit on length of discussion on the mediation talk page, however, would editors please refrain from discussion here? If participants would like to hold particular discussions during the mediation, I would suggest creating a subpage (or pages) of the mediation talk page. That way any consensus reached can be related back to the mediation. Sunray (talk) 14:18, 23 September 2010 (UTC)
Need for page protection
The article was recently placed under temporary protection due to edit waring. That protection has since expired. Further edit waring will result in the article being protected indefinitely. Hopefully the mediation will produce a climate of collaborative editing. In the mean time, would editors be willing to refrain from editing the article? Thank you. Sunray (talk) 21:17, 6 October 2010 (UTC)
- Would you categorize the deletions I made last week that were reverted and led to the page protection as 'major edits'? Glkanter (talk) 21:40, 6 October 2010 (UTC)
- Or replacing the paraphrased solution with the exact quote? Glkanter (talk) 17:33, 7 October 2010 (UTC)
I regret to inform you that I have a 'policy' of not agreeing to self-censorship on Wikipedia. I suggest you have the page protection restored. I'll wait a reasonable amount of time. Glkanter (talk) 21:18, 7 October 2010 (UTC)