Rick Block (talk | contribs) →Comments: about compromising |
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* My counter proposal, which is a compromise, is to include a single solution section presenting both unconditional and conditional solutions in an NPOV manner, more or less like the draft still currently above ([http://en.wikipedia.org/w/index.php?title=Talk:Monty_Hall_problem&oldid=348338883#Sources_present_three_different_kinds_of_solutions permanent link]). IMO, Martin's proposal creates a structural NPOV violation (see [[WP:NPOV#Article structure]]). -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 16:16, 7 March 2010 (UTC) |
* My counter proposal, which is a compromise, is to include a single solution section presenting both unconditional and conditional solutions in an NPOV manner, more or less like the draft still currently above ([http://en.wikipedia.org/w/index.php?title=Talk:Monty_Hall_problem&oldid=348338883#Sources_present_three_different_kinds_of_solutions permanent link]). IMO, Martin's proposal creates a structural NPOV violation (see [[WP:NPOV#Article structure]]). -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 16:16, 7 March 2010 (UTC) |
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::While your proposal might be more npov or better than Martin's, I really fail to see how it is a compromise in anyway (the same somewhat holds for Martin's suggestion). You both simply reiterate your point of view, you've been arguing the whole time. Imho there's nothing compromising here at all.--[[User:Kmhkmh|Kmhkmh]] ([[User talk:Kmhkmh|talk]]) 16:24, 7 March 2010 (UTC) |
::While your proposal might be more npov or better than Martin's, I really fail to see how it is a compromise in anyway (the same somewhat holds for Martin's suggestion). You both simply reiterate your point of view, you've been arguing the whole time. Imho there's nothing compromising here at all.--[[User:Kmhkmh|Kmhkmh]] ([[User talk:Kmhkmh|talk]]) 16:24, 7 March 2010 (UTC) |
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:::I completely agree Martin's suggestion is no compromise at all (in any respect). The proposal I'm making is a compromise between presenting on the one hand simple solutions as correct and sufficient vs. on the other hand saying the problem is inherently conditional and that these solutions fail to address it. The tone I'm aiming for is "here's a solution ..., here's another kind of solution ..." - not "here's a solution ..., that solution is wrong here's the real solution ...". In a personal sense, the compromise is keeping my individual POV out of the article - which I think is the compromise we must all make. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 17:15, 7 March 2010 (UTC) |
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Sources present three different kinds of solutions
Does anyone disagree that most solutions presented in sources are one of the following three types:
1) Completely unconditional, i.e chance of initially picking the car is 1/3 and a goat 2/3, and if you switch these flip.
2) Assuming the player has picked (for example) door 1, i.e. vos Savant's table:
Door 1 | Door 2 | Door 3 | result if switching |
---|---|---|---|
Car | Goat | Goat | Goat |
Goat | Car | Goat | Car |
Goat | Goat | Car | Car |
3) Assuming the player has picked (for example) door 1, conditional given the host has opened (for example) door 3, e.g. any of the "conditionalists". These end up as (1/3) / (1/3 + 1/6) = 2/3.
There is clearly conflict among sources about these solutions and clearly conflict among editors about these solutions, so how about a single solution section somewhat like this:
- Solution
- Different sources present solutions to the problem
that directly address slightly different mathematical questionsusing a variety of approaches.
The average probability of winning by switchingSimplest approach
This is the simplest kind of solution.The player initially has a 1/3 chance of picking the car. The host always opens a door revealing a goat, so if the playerignores what the host does anddoesn't switch the player has a 1/3 chance of winning the car. Similarly, the player has a 2/3 chance of initially picking a goat and if the player switches after the host has revealed the other goat the player has a 2/3 chance of winning the car. (some appropriate reference, perhaps Grinstead and Snell)
- What this solution is saying is that if 900 contestants all switch, regardless of which door they initially pick and which door the host opens about 600 would win the car.
The probability of winning by switching givenEnumeration of all cases where the player picks Door 1
- If the player has picked, say, Door 1, there are three equally likely cases.
Door 1 | Door 2 | Door 3 | result if switching |
---|---|---|---|
Car | Goat | Goat | Goat |
Goat | Car | Goat | Car |
Goat | Goat | Car | Car |
- A player who switches ends up with a goat in only one of these cases but ends up with the car in two, so the probability of winning the car by switching is 2/3. (some appropriate reference, perhaps vos Savant)
- What this solution is saying is that if 900 contestants are on the show and roughly 1/3 pick Door 1 and they all switch, of these 300 players about 200 would win the car.
- The probability of winning by switching given the player picks Door 1 and the host opens Door 3
- This is a more complicated type of solution involving conditional probability. The difference between this approach and the previous one can be expressed as whether the player must decide to switch before the host opens a door or is allowed to decide after seeing which door the host opens (Gillman 1992).
- The probabilities in all cases where the player has initially picked Door 1 can be determined by referring to the figure below or to an equivalent decision tree as shown to the right (Chun 1991; Grinstead and Snell 2006:137-138 presents an expanded tree showing all initial player picks). Given the player has picked Door 1, the player has a 1/3 chance of having selected the car. Referring to either the figure or the tree, if the host then opens Door 3, switching wins with probability 1/3 if the car is behind Door 2 but loses only with probability 1/6 if the car is behind Door 1. The sum of these probabilities is 1/2, meaning the host opens Door 3 only 1/2 of the time. The conditional probability of winning by switching for players who pick Door 1 and see the host open Door 3 is computed by dividing the total probability (1/3) by the probability of the case of interest (host opens Door 3), therefore this probability is (1/3)/(1/2)=2/3.
- Although this is the same answer as the simpler solutions for the unambiguous problem statement as presented above, in some variations of the problem the conditional probability may differ from the average probability and the probability given only that the player initially picks Door 1, see Variants below. Some proponents of solutions using conditional probability consider the simpler solutions to be incomplete, since the simpler solutions do not explicitly use the constraint in the problem statement that the host must choose which door to open randomly if both hide goats (multiple references, e.g. Morgan et al., Gillman, ...).
- What this type of solution is saying is that if 900 contestants are on the show and roughly 1/3 pick Door 1, of these 300 players about 150 will see the host open Door 3. If they all switch, about 100 would win the car.
- A formal proof that the conditional probability of winning by switching is 2/3 is presented below, see Bayesian analysis.
I'm not overly attached to any of the specific wording used, but I think presenting these as three different types of solutions and including with the last one the essence of the controversy is an NPOV approach. -- Rick Block (talk) 15:41, 2 February 2010 (UTC)
Responses
It is just the same old thing again. Nobody is interested in the average probability of winning by switching. The only question to be answered is the probability after the player has chosen a door (say door 1) and the host has opened another door to reveal a goat (say door 3). If the host chooses randomly which legal door to open then this probability is always exactly 2/3 and, by reason of symmetry, the simple table at the top of this section is a mathematically valid solution. The, so called, condition, that the host has opened a specific door is irrelevant because it can be shown that the probability of interest is independent of the door opened in the symmetrical case.
- That's not quite true for 2 reasons in particular:
- 1.) the average probability provides a reasonable heuristic (in particular if no exact knowledge regarding the influence of the conditions is given). Generally due to the lack of better information people assume the total probability (average) to hold for subsets as well (=being approximately identically with the true conditional probability) unless you have specific data for a subset telling you so or you have some very good reason to believe that this subset differs from the total set regarding a property influencing the probability.
- 2.) Many (probably even most or all) unconditional approaches in reputable literature do make an argument based on the average (total) probability (this includes in particular vos Savant, Devlin, Henze, Behrens). and we need to summarize what the sources do and not how we might improve or extend on them as the latter would be WP:OR.
- --Kmhkmh (talk) 14:09, 4 February 2010 (UTC)
- That's not quite true for 2 reasons in particular:
- If he host chooses a legal door randomly, the probability of winning by switching given the host has opened door 3 is 2/3, given the host has opened door 2 it is 2/3, and the average is also 2/3. There is no condition. Martin Hogbin (talk) 22:41, 5 February 2010 (UTC)
- I don't quite see what this has to do with the 2 points above.--Kmhkmh (talk) 01:18, 6 February 2010 (UTC)
- I assumed that the subsets that you wre referring to above are the sets where the host opens door 3 and the set where the host opens door 2. If this is not what you mean, per hap you could explain further. Martin Hogbin (talk) 23:21, 6 February 2010 (UTC)
- I don't quite see what this has to do with the 2 points above.--Kmhkmh (talk) 01:18, 6 February 2010 (UTC)
- If he host chooses a legal door randomly, the probability of winning by switching given the host has opened door 3 is 2/3, given the host has opened door 2 it is 2/3, and the average is also 2/3. There is no condition. Martin Hogbin (talk) 22:41, 5 February 2010 (UTC)
- Somehow it looks as if you, like others, desperately are seeking a way of reasoning, to avoid conditioning. But there is none! Try to formulate your above reasoning in formulas, and you will see. BTW. what do you mean with: "the probability of interest is independent of the door opened in the symmetrical case". Nijdam (talk) 23:25, 2 February 2010 (UTC)
- Is this not the same discussion that we are having elsewhere? If the host chooses a legal door uniformly at random then you agree that the probability of interest (probability of winning by switching) is exactly 2/3 whichever door the host opens. In other words the probability of interest is independent of the door number opened by the host. I can show this more formally if you like.
- Since the probability of interest is independent of the door number opened by the host, the door number opened by the host need not be a condition of the problem, by your own argument. Martin Hogbin (talk) 23:35, 2 February 2010 (UTC)
Regarding what sources say, the sources that give the simple solutions do not in any way say that their solutions apply to a different problem or that they are only average solutions. The sources that present simple solutions present them as complete solutions to the question as asked. Martin Hogbin (talk) 16:09, 2 February 2010 (UTC)
- Are you denying that the simplest (fully unconditional) solutions literally address the average probability of winning by switching, or that (say) vos Savant's solution literally addresses the probability of winning by switching given the player picks Door 1? The solution vos Savant presents (for example) includes cases where the host opens Door 2, so it is clearly not directly addressing the conditional probability where the player has chosen a door (say door 1) and the host has opened another door to reveal a goat (say door 3). As far as I know, that the sources offering these solutions present them as "complete" solutions is only implied by these sources. I'm attempting to do the same thing here - they're in the "Solution" section, hence these are by implication complete solutions.
- I think it's fairly clear these are the three typical sorts of solutions that are presented. If you find the words I've used to describe them POV, can you suggest a way to describe them that you don't find POV? -- Rick Block (talk) 21:24, 2 February 2010 (UTC)
- I do agree that the simplest solutions address the average probability of winning by switching, it is just that nobody cares about this. Vos Savant, in common with many editors here, obviously believes that, in the symmetrical case, the door numbers are of no relevance to the problem. Thus a solution presented in which the host opens door 2 is equally valid if the host opens door 3, which of course it is, by symmetry. This especially true as it is not clear that Whitaker even intends to specify door numbers.
- I am not sure what you mean by your second point. Several reliable sources present simple solutions without any form of reservation or restriction. Thus they clearly are asserting that they are answering the question as asked. Some other sources say, to varying degrees, that they are wrong. Martin Hogbin (talk) 23:26, 2 February 2010 (UTC)
- The second point? About vos Savant's solution including cases where the host opens Door 2? The question asks about the case where the player has picked, say, Door 1 and the host has opened, say, Door 3. The table shows the (unconditional) probabilities where the player picks Door 1 and the host opens Door 2 and (not or) the host opens Door 3. What this table literally shows are the unconditional probabilities where the player has initially selected Door 1 - i.e. the player is not looking at either door the host has opened but is thinking about all cases. To analyze the situation after the host has opened a door without caring about the door numbers, call them A, B, and C. The player has selected Door A and the host has opened Door C while Door B remains closed. With this set up, the probability of winning where the host has opened Door B is "this case didn't happen", not 1/3. There's only one case, the player picked Door A and the host opened Door C. This is a conditional case. This is the point Nijdam keeps making that you keep ignoring. The probability of the car being behind A,B, or C is initially 1/3 for each. After the host opens Door C its probability is now 0 (no longer 1/3). What are the other door's probabilities? The question is asking what are their conditional probabilities. -- Rick Block (talk) 01:13, 3 February 2010 (UTC)
- This is about where we started some years ago. In the symmetrical case, the fact that a specific door has been opened reduces the options available to the player but does not affect the probability of interest. It is therefore not a condition. I suggest that we continue this discussion in the section that I have set up on the arguments page. Martin Hogbin (talk) 09:46, 3 February 2010 (UTC)
- Indeed, and you still don't get it. You're still talking about the value 2/3, where we have explained over and over it is not the value that counts, but the nature of the probability. We demonstrated that opening of a door is a condition, and still you persist in saying it is not. What is your intention? Nijdam (talk) 12:53, 3 February 2010 (UTC)
- 'demonstrated that opening of a door is a condition' is exactly what you have not done. You have asserted this several times and also suggested reasons why the event should be considered a condition. I have shown all these reasons to be invalid. I suggest we continue on the appropriate section on the arguments page. Martin Hogbin (talk) 13:58, 3 February 2010 (UTC)
- Indeed, and you still don't get it. You're still talking about the value 2/3, where we have explained over and over it is not the value that counts, but the nature of the probability. We demonstrated that opening of a door is a condition, and still you persist in saying it is not. What is your intention? Nijdam (talk) 12:53, 3 February 2010 (UTC)
- This perhaps should go elsewhere, but Martin if I've picked Door 1 and the host has opened Door 3 what lines of the table do I look at to figure out my probability of winning by switching? The table looks wrong to me since it says the probability of the car being behind Door 3 is 1/3. It clearly isn't anymore. The probabilities inferred from this table are prior probabilities - those that are in effect before the host opens a door. To convey what you're saying it conveys, this table should really only have two lines, a Door 1 line and a "Door 2 or Door 3" line. This makes the probabilities inarguably unconditional, which I think is what you're saying - but then it's clearly showing the probability of winning given the player has picked Door 1 (which you seem to object characterizing it as). -- Rick Block (talk) 18:23, 3 February 2010 (UTC)
- Thus we should, in the article, present the simple solutions in the same way that they are presented by the sources that give them, as complete solutions without reservation or restriction to the exact question that was asked. Martin Hogbin (talk) 23:37, 2 February 2010 (UTC)
- How about the changes as above? Better? -- Rick Block (talk) 01:13, 3 February 2010 (UTC)
- Better, but you still have, 'so if the player ignores what the host does', and 'What this solution is saying is that if 900 contestants all switch, regardless of which door they initially pick and which door the host opens about 600 would win the car'. Later on, in the conditional solution section, you say, 'This is a more complicated type of solution involving conditional probability. The difference between this approach and the previous one can be expressed as whether the player must decide to switch before the host opens a door or is allowed to decide after seeing which door the host opens (Gillman 1992)'.
- My first two quotes should go completely, they are your POV concerning sources that give simple solutions, there are several sources giving a simple solution that do not say what you say.
- Regarding, 'This is a more complicated type of solution involving conditional probability....'. We are now talking about the link between the sections and sources. This should therefore start with something like, 'Some sources treat the problem as requiring the use of conditional probability...' preferably including 'this is important if the host may not choose randomly which door to open, when he has a choice...' Martin Hogbin (talk) 09:46, 3 February 2010 (UTC)
- I've struck "player ignores what the host does". I'm reluctant to strike the 600 out of 900 win bit. You agreed above this solution is literally addressing the average probability. Is it the numbers you object to or the "regardless of which door ..." wording?
- I am not sure what you mean by average probability here. What you say is quite correct, but why mention it?. In the symmetrical case the probability of winning by switching is 2/3 if the host opens door 3, 2/3 if the host opens door 2, and 2/3 on average. Martin Hogbin (talk) 22:44, 3 February 2010 (UTC)
- The "difference between this approach" needs to go somewhere. The point is not to say conditional probability is required, but to explain what it means. The tone I'm trying to hit is "an alternate approach", not "the correct approach". Your suggested addition about the importance is your POV, which as far as I recollect no one's ever offered up a source for (just WP:OR) - and addressing the problem using conditional probability is certainly a valid approach whether the host chooses randomly or not. -- Rick Block (talk) 18:23, 3 February 2010 (UTC)
- Your last point is the one to focus on. The simple solution only applies if the host is taken to choose a legal door randomly, it fails if the host chooses non-randomly. Martin Hogbin (talk) 22:44, 3 February 2010 (UTC)
- Yes I agree, but you do not need the conditional approach just to show that the simple argument is valid in the symmetrical case. See my discussion with Nijdam on the arguments page. Martin Hogbin (talk) 15:06, 4 February 2010 (UTC)
- No, no, no. The simple argument is never valid, i.e. for the MHP where the offer to switch is given AFTER the goat door is opened. The probability of interest is the conditional probability. It may be calculated in a lot of ways, with Bayes', with symmetry, etc., but it IS a conditional probability! Nijdam (talk) 23:49, 5 February 2010 (UTC)
- This remains just an assertion on your part. We are currently discussing this subject on the arguments page. Martin Hogbin (talk) 00:44, 6 February 2010 (UTC)
- I entirely agree with Martin here against Nijdam: there is nothing in the Marilyn vos Savant question saying "you must use a conditional probabilty". It is also not the case that because you are to decide *after* this and that event, you *must* decide by computing a conditional probability. I realise that if the question is posed as an examination question in Probability 101, this might be "understood" by the student eager to get full marks. But I am not interested in Probability 101 questions. I am interested in a famous question posed in a popular weekly magazine in which any technical code words like probability, equally likely, identical... did not occur. Gill110951 (talk) 05:39, 7 February 2010 (UTC)
- I will get along with Gill in so far the academic discussion goes about what is meant by the MHP. But I don't follow his reasoning about the decision to be made. If an event has happened, it is almost compulsery to take this into account. In most cases it is even of vital importance, as I showed in examples. And anyone I gave the choice to decide on basis of the unconditional (average) probability, or on the specific conditional, did not hesitate to choose the conditional probability. Only because the answers i the symmetric MHP will be the same, one might have the idea it is not needed to consider the occurence. But ... to know the answers do not differ, one has to consider the conditioning on the ocurence of the event. Nijdam (talk) 21:48, 8 February 2010 (UTC)
Nice discussion. My present point-of-view is written up (for mathematicians) in http://arxiv.org/abs/1002.0651 (eprint/prepublication archive). I have also submitted it to a peer-reviewed journal. But I already see that I want to improve the symmetry solution, and to add equivalent verbal proofs next to each mathematical proof, so that ordinary people can read it too. Suggestions for improvement are welcome. Gill110951 (talk) 08:09, 4 February 2010 (UTC)
Rick wrote: "this table should really only have two lines, a Door 1 line and a "Door 2 or Door 3" line. This makes the probabilities inarguably unconditional". Rick, given that "door2 or door3" is opened, it's a conditional probability just the same? Heptalogos (talk) 21:54, 5 February 2010 (UTC)
- Well, sort of, but with only two lines (the 2nd line is "car behind Door 2 or Door 3") the condition "door 2 or door 3 opened" (more) obviously does not reduce the sample space. A table with number of samples might help clarify what I'm saying:
Door 1 | Door 2 | Door 3 | result if switching | total cases | cases if host opens Door 2 or Door 3 |
---|---|---|---|---|---|
Car | Goat | Goat | Goat | 100 | ? |
Goat | Car | Goat | Car | 100 | ? |
Goat | Goat | Car | Car | 100 | ? |
- What "in two out of three equally likely cases switching wins" means is in 200 out of 300 cases switching wins - it's not saying anything at all about the column with the question marks. The cases are only obviously equally likely before the host opens a door. Looking at this table can you fill in the ? entries based on knowing the host opens "Door 2 or Door 3"?
- I'll give you a hint: focus on the row where the car is behind Door 1. In how many of these 100 cases does the host open Door 2 or Door 3?
- The point is if all you know is the host opens "Door 2 or Door 3", the numbers are ALL still 100 (!). The host opens Door 2 or Door 3 in ALL 300 cases. Asking what happens after the host opens "Door 2 or Door 3" is the same thing as asking what the situation is before the host opens a door. It's the same set of cases, i.e. the same sample set. All the before and after probabilities are the same.
- If we do the same thing with the 2 line table, here's what happens
Where's the car | result if switching | total cases | cases if host opens Door 2 or Door 3 |
---|---|---|---|
Behind Door 1 | Goat | 100 | ? |
Behind Door 2 or Door 3 | Car | 200 | ? |
- The ? entries in this table are (more) clearly 100/200 (right?). Wherever the car is, the host always opens Door 2 or Door 3. The confusion about whether the ? in the row where the car is behind door 1 should be 100 or 50 pretty much goes away. This confusion is exactly what Morgan et al. are referring to when they say "The distinction between the conditional and unconditional situations here seems to confound many".
- In my opinion (some WP:OR), a better table would be like this:
Door 1 | Door 2 | Door 3 | total cases | host opens Door 2 | host opens Door 3 | ||
---|---|---|---|---|---|---|---|
cases | result if switching | cases | result if switching | ||||
Car | Goat | Goat | 100 | 50 | Goat | 50 | Goat |
Goat | Car | Goat | 100 | 0 | N/A | 100 | Car |
Goat | Goat | Car | 100 | 100 | Car | 0 | N/A |
- And the explanation becomes, "in two equally likely cases (host opens Door 2 or host opens Door 3), switching wins twice as often as staying". The key to the solution is not that the car is equally likely to be behind each door before the host opens a door (this is obvious to everyone), but that it's equally likely that the host open either door and in each case you're twice as likely to win by switching. -- Rick Block (talk) 04:26, 6 February 2010 (UTC)
- The sample space is anyway reduced by opening a door:
Door picked | Door unopened | Door opened | result if switching | total cases | cases if door is opened |
---|---|---|---|---|---|
Car | Goat | Goat | Goat | 100 | 50 |
Goat | Car | Goat | Car | 100 | 100 |
Goat | Goat | Car | Car | 100 | 0 |
- Heptalogos (talk) 12:36, 6 February 2010 (UTC)
- You're so close! However there's a problem in your table. In the "Door opened" (3rd) column, the last row says it has a car. This cannot happen. So, in this row, the number in the "total cases" column must be 0. The basic problem is if you want to show a 100/100/100 column (before the host opens a door) and a 50/100/0 column (after the host opens a door) you have to individually identify the doors. You could call them doors A,B,C if you'd like, but you might as well just stick with 1,2,3. The 50/100/0 column can only exist if you know which door the host has opened. It is a conditional case. I suspect what you're really going for is something sort of like this
Door picked | Unpicked door A | Unpicked door B | result if switching | total cases | cases if host opens Door B |
---|---|---|---|---|---|
Car | Goat | Goat | Goat | 100 | 50 |
Goat | Car | Goat | Car | 100 | 100 |
Goat | Goat | Car | Car | 100 | 0 |
- The point is the so-called condition "host opens door 2 or door 3" does not have an "exclusive or" sort of sense (meaning a specific one of these doors) but only a "union" sort of sense (meaning all cases where the host opens either door 2 or door 3, i.e. all cases). This is why the ? entries in the first table above are all 100. If you want a conditional case (a 50/100/0 column), the condition has to be "host opens a specific door". Given the problem statement, you might as well stick with Door 3 - it's representative of any other specific case.
- If you're really aiming for the Door A/Door B table, what is your description for why the probability of winning by switching is 2/3 and why the top number in the "cases if host opens Door B" (last) column is 50 and not 100? I assume your explanation doesn't start with "in 2 out of 3 equally likely cases ..." And, BTW, if this had been vos Savant's table I'd imagine the Morgan et al. paper would have never been written. -- Rick Block (talk) 17:46, 6 February 2010 (UTC)
- It's hopeful to see that you don't like 'impossibilities' in the sample space. For now, let's keep focused on the given method. In your example door3 has a car, which is of course just as impossible. You are confusing the situation before with the situation after. Heptalogos (talk) 18:20, 6 February 2010 (UTC)
- In which example? My WP:OR table above? I thought it was obvious the first several columns are the before state, and the last were the after state. My point all along has been the 100/100/100 (or 1/3:1/3:1/3) distribution is obviously the case only in the before state. We could explicitly label the columns if that helps. This makes the table equivalent to the large figure at the end of the Probabilistic solution section of the article.
Situation BEFORE the host opens a door | Situation AFTER the host opens a door | ||||||
---|---|---|---|---|---|---|---|
Door 1 | Door 2 | Door 3 | total cases | host opens Door 2 | host opens Door 3 | ||
cases | result if switching | cases | result if switching | ||||
Car | Goat | Goat | 100 | 50 | Goat | 50 | Goat |
Goat | Car | Goat | 100 | 0 | N/A | 100 | Car |
Goat | Goat | Car | 100 | 100 | Car | 0 | N/A |
- Anything talking about "2 out of 3 equally likely cases" must be talking about the before state. Are you disagreeing with this? -- Rick Block (talk) 18:40, 6 February 2010 (UTC)
- Yes. But what I mean is that my "door opened" which has a car is the same as your door3 with a car. This is only impossible after we know it has been opened. Before that, I don't identify doors 2 and 3 individually. One can look at the number 3 and always open it, or one can open a door and always number it as 3; there's no relevance in the difference. Unless you assume that a "number bias" may exist. Heptalogos (talk) 19:07, 6 February 2010 (UTC)
- But then, to be consistent, the before condition must be 100/200, not 100/100/100. The door you're opening is one of the 200 that is not door 1, not one of either of the 100 that is door 2 or door 3. You're talking about the 2-line table. -- Rick Block (talk) 19:52, 6 February 2010 (UTC)
- It is known that the 200 cases are equally divided over the two doors without the need to identify them individually. It's conditional because opening one of the doors reduces the sample space. Heptalogos (talk) 20:17, 6 February 2010 (UTC)
- Are you calling the door that is opened (whichever one it is) door 3 or not? If yes, you're not reducing the sample space. We start with 300 cases with an obvious 1/3:1/3:1/3 distribution, one of the doors is opened, whichever door is opened is now called door 3 - leaving us still with 300 cases (like the table below).
Situation BEFORE the host opens a door | Situation AFTER the host opens a door | ||||
---|---|---|---|---|---|
Door picked | Unpicked door A | Unpicked door B | total cases | result if switching | cases if host opens a door |
Car | Goat | Goat | 100 | Goat | 100 |
Goat | Car | Goat | 100 | Car | 200 |
Goat | Goat | Car | 100 |
- The opened door3 is either A or B, not both. The doors are unique, although we don't identify them statically BEFORE. So AFTER we do have 150 cases instead of 300. If you state door3 may be both A or B, it makes no sense to create 2 columns for them with certain, different distributions. Heptalogos (talk) 22:26, 6 February 2010 (UTC)
- BEFORE they are identified vertically; AFTER they are also identified horizontally (statically). Vertically means that each door has a certain distribution (GCG, GGC). Horizontally means that any opened door (last column) is one of those vertical columns. Heptalogos (talk) 23:02, 6 February 2010 (UTC)
- If you are reducing the sample space, you're talking about the Door A/Door B table above (where the first 5 columns are the before state and only the last column is the after state).
- If we cut through all the pointless arguing here, what the simple solutions mean to be saying is that the average probability of winning by switching is 2/3, and assuming any particular case (for example the case where the player has picked Door 1 and the host has opened Door 3) has the same probability as any other particular case (which is true only if the host picks randomly between two goats) then in all cases the probability must be the same as the average (this is the "symmetry" argument). The issue is most sources presenting simple solutions don't include the "and assuming" part of this - and certainly never mention that how the host picks between two goats matters. They say what the average probability is (which is NOT affected by how the host chooses between two goats), and then stop. Morgan et al. (and others) are saying because they stop after saying what the average probability is they haven't quite answered the question.
- You (and Martin and Jeff) are saying any idiot should understand what they mean, but since they're not explicit about it many idiots walk away with the misunderstanding that "what the host does can never affect the player's initial probability" - vos Savant said it ("The odds that your choice contains a pea are 1/3, agreed? Then I simply lift up an empty shell from the remaining other two. As I can (and will) do this regardless of what you've chosen, we've learned nothing to allow us to revise the odds on the shell under your finger.") it must be true! It is true, but only if we're talking about unconditional probabilities. Conditional probabilities (like what are asked about in this problem) are always affected by what the host does. The conditional probability of the player's door MAY end up numerically the same as its initial probability, but it's always a different probability (in the tables above it's out of a sample set of 150 cases, not 300 cases). The only conditional probability that's obvious from the problem statement is the one for the opened door. Both of the others are always a direct function of how the host chooses between two goats. -- Rick Block (talk) 21:24, 6 February 2010 (UTC)
- Why is the average probability not affected by the host's goat-door prevalence? It may be. Heptalogos (talk) 22:35, 6 February 2010 (UTC)
- No, it may not be. If we're talking about 300 cases where the player has picked Door 1, and no player switches, about 1/3 of them win regardless of what the host does (assuming the host opens a door). Similarly, about 2/3 of them win if they all switch. This is the average chance of winning by staying and by switching and it is independent of any host goat-door preference. If the host picks randomly the 300 cases split into two equivalent samples of 150 cases. The stickers and switchers win 1/3:2/3 in each sample, i.e. 100 out of 150 win by switching in each case. With an extreme host preference, say the host always opens Door 3 if possible, the 300 cases split unevenly - 100 players will see the host open Door 2 and 100% (all 100) of these players win if they switch, while 200 players will see the host open Door 3 and only 50% (also 100) of these players will win if they switch. The average chance of winning by switching is (100+100)/300 = 2/3. Using formulas, if the host preference is p, (1/3)+(1/3)p players see the host open his preferred door and (1/3)+(1/3)(1-p) see the host open his non-preferred door. The first group has a 1/(1+p) probability of winning by switching, the second group has a 1/(1+1-p) probability. The composite probability (which is the average) is
- ( (1/3)+(1/3)p )( 1/(1+p) ) + ( (1/3)+(1/3)(1-p) )(1/(2-p) )
- which simplifies down to
- 1/3 + 1/3 = 2/3
- -- Rick Block (talk) 00:56, 7 February 2010 (UTC)
- You assume several players. In the problem statement you are the only player, and you may have a bias with a significant relation to the host's bias. Heptalogos (talk) 12:57, 7 February 2010 (UTC)
- Conditional probabilities are always affected by the conditions, yes. The condition is the opened door, not the number. Heptalogos (talk) 22:35, 6 February 2010 (UTC)
- I have no idea what distinction you're trying to make here. -- Rick Block (talk) 00:56, 7 February 2010 (UTC)
- @ Rick, please try to understand the distinction. It is vitally important. Gill110951 (talk) 05:42, 7 February 2010 (UTC)
- I think I understand the "indistinguishable doors" POV, and that in this POV the host opening a door does not change the sample space (so the prior probability of the car being behind the player's door is unchanged by the host opening a door since the posterior sample space is identical to the prior sample space). I believe this is the POV Jeff, Martin, and Glkanter (and possibly Heptalogos, but I'm not sure) think is "correct".
- I think I understand the "distinguishable doors" POV, and that in this POV the host opening a door does change the sample set (i.e. there are different prior and posterior probabilities, where the posterior probability depends on how the host selects between two goats). I believe this is the POV Nijdam thinks is "correct". I think I also understand that in the "distinguishable doors" POV, the host's preference may or may not be accessible to the player, and if it's not accessible the most sensible assumption is to take it as random.
- If somebody could please clarify the distinction Heptalogos is making between the condition being the "opened door" as opposed to "the number" I'd appreciate it. -- Rick Block (talk) 19:51, 7 February 2010 (UTC)
- Of course what a wikipedia editor *thinks* is strictly speaking irrelevant, but just for the record, I do not think that it is a sensible assumption at all, to take something as uniformly at random when you don't know it. That's the whole point of the game theoretic approach: what is sensible to do, when you don't know the other's strategy!!! Regarding the distinction: Prob(car is not behind your door|you chose 1,MH opened 3) might well be different from Prob(car is not behind your door|you chose 1, MH opened a door). Isn't that obvious???? Gill110951 (talk) 22:48, 7 February 2010 (UTC)
- Yes, of course P(not your door|you chose 1, host opens 3) is not the same as P(not your door|you chose 1, host opens a door) - that's the exact point I've been trying to explain this entire thread. Have you read this whole thread? In addition P(not your door|you chose 1, host opens a door) is exactly equivalent to P(not your door|you chose 1) - the conditions "you chose 1, host opens a door" and "you chose 1" define the same sample space, so all probabilities are the same with regard to these two conditions. My point is that vos Savant's solution addresses probabilities in this sample space, i.e. she's addressing P(not your door|you chose 1) (whether you add "host opens a door" or not changes nothing), not P(not your door|you chose 1, host opens 3). Martin and Heptalogos seem to be objecting to characterizing vos Savant's solution as addressing P(not your door|you chose 1). This is what it literally does, doesn't it? -- Rick Block (talk) 23:44, 7 February 2010 (UTC)
Table showing the host opening an unnamed door
Thread moved to /Arguments#Table showing the host opening an unnamed door. -- Rick Block (talk) 17:54, 7 February 2010 (UTC)
I love it
Can I just say Monte Halperin turns 89 this summer and I think he would never have imagined the role he has assumed in this fascinating mathematical back-and-forth. I've watched this page for some time, and very much enjoy the discussions. My humble suggestion to involved editors is to remember what brought you here in the first place, which I'm guessing are the deliciously captivating counter-intuitive qualities of the Monty Hall problem. I hope you can edit collaboratively to communicate and elucidate this to readers! Respectfully, RomaC (talk) 17:54, 6 February 2010 (UTC)
- Was Monty the first in the world to reveal small prices and offer to switch? Or did he copy the act? I heard the puzzle on television and indeed had my 50/50 answer ready. But that was because of the problem statement which didn't make clear that the host would always do that. So I checked it out on Wikipedia and understood the paradox, but there I met some people who were convinced that the number of the door had any significant information or relevance. I am still trying to show them that it hasn't, and for sure wasn't meant to have any relevance. Heptalogos (talk) 18:31, 6 February 2010 (UTC)
- That's what I think too, @ Heptalogos! But lots of authoritative sources have said that the numbers do have significance and that the problem *has* to be solved by conditional probabilty and that you *have* to make a load of extra assumptions too (their arguments are often pretty shaky or absent)... One is not obliged to agree with these sources. One can find equally authoritative sources who think otherwise. Both points of view have to be covered by wikipedia since they are all out there, part of the living Monty_Hall_problem problem. We the editors have to look down on all these little ants running around in circles dispassionately, and give a wise and clear overview so that future new ants who fall into this sand-hill don't have to run in all the same circles all over again. Gill110951 (talk) 06:10, 7 February 2010 (UTC)
I love it too
The discussions going on here are nothing else than the eternal rerun of the Monty_Hall_Problem problem. The problem to decide what is the problem. That is what the wikipedia page should be about. There are authoritative sources for all kinds of particular versions. There are authoritative sources that people see the problem in many different ways. A small problem here is that many authoritative sources, especially those by mathematicians, are authoritarian, dogmatic. And often the more ignorant the mathematician, the more dogmatic.
Can't wikipedia present a number of points of view, can't it present a history?
I too have a point of view and as a mathematician I often appear to write in dogmatic authoritarian style. I strongly want us to structure the article by starting with "Craig's problem" as quoted by Marilyn. If that can be agreed, then we can move on to a next step, structuring the material that we agree needs to be put on the page. Gill110951 (talk) 05:58, 7 February 2010 (UTC)
- I agree with you except for one thing, The order of presentation. I, at one time, suggested historical ordering of the article, but the problem is that the history of the subject is mainly one of argument and dispute. This is not conducive to writing a good article.
- A more logical order, in my opinion, is one of complexity, start with the simplest understanding and solution of the problem first and then proceed to add complications. This is how most good text books work, they start with the basics, possibly glossing over some of the subtleties for the sake of clarity, then, when the reader is assumed to understand the basics, they go on to say that the initial exposition was, maybe, a little over simplified and deal with the complicating factors.
- The logical order for me would be:
- Simple (non-conditional) problem and solution - fully explained with aids to understanding and no confusing mention of conditions.
- Conditional problem
- Game theory
- Variants
- History Martin Hogbin (talk) 10:31, 7 February 2010 (UTC)
- Agreed, to order by increasing complexity is very wise. Seems we two agree about almost everything. Those who disagree, seem to want to erase some parts of the Monty Hall story because of their personal taste or because of the dogmatic/authoritarian outloook of their preferred source.Gill110951 (talk) 22:39, 7 February 2010 (UTC)
The justification given does not justify the original problem but only justifies the revised problem
When i saw this i originally thought the math was ridiculous. If a newcomer came onto the game show the chance of the car being behind the doors would be 50:50 so why not with the original contestant! To explain the problem wiki extrapolated the problem and likened it to 1 in a million chance. This altered my thinking for a moment. But only for a moment. I quickly realised that this was a different argument. True what are the chances of your initial prediction being the 1 in a million. In this scenario there would be definately an advantage to switching. But if this argument is extrapolated again and we say what if monty did this simuteaneously with 1 million other contestants clearly it would not always be advantageous to switch even in these evidentally advantageous conditions. One may argue that this is not fair and not the same problem - in this case its the literal truth - but then that is my point. Exaggerating the problem changes the problem and it is no longer the same problem.
Matt Cauthon —Preceding unsigned comment added by 194.106.220.83 (talk) 12:58, 11 February 2010 (UTC)
- You're talking about two different issues.
- 1) In the standard problem a newcomer comes onto the game show after the door has been opened and does not know the situation, this newcomer has a 50:50 chance of correctly picking the door hiding the car but the original contestant has a 2/3 chance of winning by switching.
- Read this statement very carefully. It says the newcomer has a 50:50 chance of correctly picking the door. This does not mean the "fully aware" probability of the car being behind the two doors is 50:50, but if you make a random choice between them you end up with the car 50% of the time. What's 50:50 here is not the distribution of the car behind the doors, but the result of a random choice. Think about 150 shows where the original contestant picked door 1 and the host opened door 3. We'd expect the car to be behind door 1 about 50 times and behind door 2 about 100 times reflecting the contestant's 1/3:2/3 chances. If the newcomer randomly picks door 1 or door 2, she ends up with the car how many times? The answer is if she picks between the doors randomly she'll pick each door about 75 times and win the car about 1/3 of the times she picks door 1 and 2/3 of the times she picks door 2. So, the number of times she'll win the car is (about) 75*1/3 + 75*2/3 = 75(1/3+2/3) = 75, which is 1/2 of 150.
- 2) What if different contestants pick different doors - how can it be advantageous for them all to switch?
- I think this is the other issue you're talking about. For example in the 3 door case what if there were two contestants, one picked door 1 and the other picked door 2 and the host opens door 3. This is a completely different problem and requires some extra rules. To make it possible for the host to always open a door that doesn't reveal the car and isn't one of the doors one of the players picked, there must be at least two more doors than the number of players - so this 3-door example doesn't really work (what happens if the car is behind door 3?). You could say if there are m contestants there must be n >= m+2 doors and, after each contestant picks a unique door, the host (randomly) opens all but 1 of the unpicked losing doors - leaving the m doors the contestants picked and 1 more. In this case the contestants should all switch to the unopened unpicked door, but switching between themselves doesn't make any difference. The probability for each contestant's door is 1/n and the probability for the other door is (n-m)/n.
- Your point might be that the million door example requires extra rules also. This is true, but I think the general assumption for the 1-player, n-door case is that the host (randomly) opens n-2 losing doors none of which is the door the player initially picked, leaving the player's door and one more. Under these rules, the player's door has a 1/n chance and the other door has a (n-1)/n chance. -- Rick Block (talk) 17:05, 11 February 2010 (UTC)
- Yes, anything correct and clear, just want to once more underline this one detail:
"The newcomer chooses randomly between the two still closed doors" says: The newcomer can not distinguish these two still closed doors. He does NOT know which one had been chosen by the guest (out of three doors with a chance of 1/3), and so he does NOT know which one has been offered as an alternative for switching with a chance of 2/3. He has no information about that, and so has to choose "randomly". But if he could distinguish, e.g. if the doors were marked, he could decide for the originally chosen door with a chance of 1/3, or for the alternatively offered second closed door with a chance of 2/3. Just wanted to underline that detail once more. Regards, -- Gerhardvalentin (talk) 18:17, 11 February 2010 (UTC)
- Yes, anything correct and clear, just want to once more underline this one detail:
2b) What if different contestants pick different doors - 3 contestant, 3 door variation
As in the original style MH game, there are 3 doors, only one of which has a new car.
But here, Monty has a large crowd of potential contestants; each one has picked one of the 3 doors.
For each of the 3 doors, Monty randomly selects one contestant who has picked that door.
Monty separately isolates the 3 contestants so that none of them know the others are playing.
Monty randomly opens one of the doors with a goat, and the contestant for that door looses.
Monty separately allows each of the 2 remaining contestants to switch to their previously unchosen door. (Davrids (talk) 00:55, 16 February 2010 (UTC))
(If both contestants pick the winning door, they both get a new car.)
Should each remaining contestant switch to their previously unchosen door?
Each contestant's situation seems to be the same as if he were the only contestant in the original style (single contestant) MH game.
If so, the arguments that prove that the player in the original style MH game should switch, would also prove that both remaining contestants in this game should switch to get their 2/3 chance of winning.
But they can't both have a 2/3 winning chance if they switch, can they? Wouldn't the combined chance, for the two remaining doors together to have a car, be 4/3? What is wrong?
Davrids (talk) 00:34, 16 February 2010 (UTC)
- Monty randomly opens one of the doors with a goat, and the contestant for that door looses.
- Let's say you pick Door 1 and the player who picks Door 3 is eliminated (the same analysis applies for any particular door you pick and door the host opens). The prior probability of each door (before the host opens a door) is 1/3. Like in the normal MHP the posterior probability that Door 1 hides the car (after the host opens Door 3) is
- P(car is behind Door 1 and host opens Door 3) / ( P(car is behind Door 1 and host opens Door 3) + P(car is behind Door 2 and host opens Door 3) )
- Because of the rule I've put in bold above, P(car is behind Door 1 and host opens Door 3) is (1/3)(1/2) = 1/6 and this is the same as P(car is behind Door 2 and host opens Door 3). So the posterior probability the car is behind Door 1 is (1/6) / (1/6 + 1/6) = 1/2. There are only two unopened doors, so the probability the other one hides the car must also be 1/2.
- In the normal MHP the host MUST open Door 3 if the car is behind Door 2, so P(car is behind Door 2 and host opens Door 3) is 1/3, not 1/6. -- Rick Block (talk) 03:11, 16 February 2010 (UTC)
Remove references to "The Monty Hall Trap" by Phil Martin
I would like to remove references to "The Monty Hall Trap" by Phil Martin because I believe his article contains numerous false claims about both probability and the game of bridge. I don't feel that it's right for me to unilaterally remove the references without first discussing this with the person(s) who posted those references. How should I proceed?Secondfoxbat (talk) 07:57, 26 February 2010 (UTC)
- Why not explain here what you think the false claims are? Martin Hogbin (talk) 09:02, 26 February 2010 (UTC)
I am currently writing a paper on the problems that I see, so I am reluctant to say too much here. Briefly, Mr. Martin's idea of the Monty hall trap is not one which anyone falls for. His trap, even if true, does not apply to bridge. As far as I can tell, his concept of "biased" data is new with him, but it impugns the idea of declarer drawing an inference from the opening lead. His argument that playing the same deal first as a South declarer, then as a North declarer is based on a fallacy. And there's more.... Secondfoxbat (talk) 01:14, 27 February 2010 (UTC)
- It's in the "History" section, so the question should be whether this article has any historical importance. It is one of only a few references before the Parade articles appeared. It also shows the multi-disciplinary appeal of the problem. Rather than delete it, you might abbreviate the paragraph a bit. I believe this reference is far less significant than the Nalebuff one immediately before (which gets only a single sentence as opposed to a paragraph for the Martin one). -- Rick Block (talk) 02:49, 27 February 2010 (UTC)
Thanks for giving me the proper perspective. I agree that the Nalebuff reference is much more significant. I'll shorten the Martin reference so it is shorter.Secondfoxbat (talk) 04:27, 28 February 2010 (UTC)
Someone changed the reference to "The Monty Hall Trap" to give more credit than is due. The best that can be said for Mr. Martin's paper is that it attempted to show a connection between the Monty Hall problem and the game of bridge. Anything more is not factual.Secondfoxbat (talk) 17:50, 5 March 2010 (UTC)
- I'm the one who changed it. My intent was to make it more flatly descriptive. "Attempted to show" conveys a critical, rather than neutral, POV (see WP:NPOV). What I changed it to was "presented Selvin's problem as an example of what Martin calls the probability trap of treating non-random information as if it were random, and relates this to concepts in the game of bridge such as ...". Do you agree that this is more or less literally what Martin's article says (as opposed to agree with the content of what Martin is saying)? We need to say what the article says, without adding any "spin" indicating we either agree or disagree with the content. -- Rick Block (talk) 01:28, 6 March 2010 (UTC)
Thanks for pointing out the Wikipedia perspective. I suppose we could agree on a shortened version of your change: "presented Selvin's problem as an example of what Martin calls the probability trap of treating non-random information as if it were random, and relates this to concepts in the game of bridge." If anything more is said about how he relates the probability trap to bridge, it should not be about restricted choice, but about not drawing inferences from the opening lead and not using the Vacant Places probability calculation. Secondfoxbat (talk) 06:47, 7 March 2010 (UTC)
Bayes section is REALLY bad: propose to replace by application of Bayes' law in odds form
I think the article is not bad at all at present. However the section on the Bayesian approach to it is totally out of proportion to the rest. It seems to imply that a philosophical Bayesian viewpoint, ie the viewpoint that all probabilities are subjective and represent degrees of belief (as reflected by your willingness to take bets at certain odds) helps understand the problem. Whereas what is actually done here is simply to apply the standard definitions of conditional probability, which hold for anybody's probability, whether subjective or "objective" ("frequentist"). Having defined the basic events and written out the probabilties which we are told in the full problem statement, we simply calculate Prob(car is behind door 2 | player chose 1, host opened 3).
In other words, there is a confusion between use of Bayes theorem aka Bayes formula (which is proved in one line from the definition of conditional probability), and a Bayesian philosophy. The philosophy is irrelevant here, since the calculation is the same whatever philosophy of probability you adhere to. You don't even have to have a unique philosophy, you can just apply probability calculus to whatever kind of probability seems most appropriate in a particular context.
Secondly, *if* we are going to do this by a calculation by Bayes theorem/formula (it is not the only way to get to the answer), *then* it is more attractive to use Bayes' theorem in its form: posterior odds equal prior odds times likelihoods.
In case you didn't know this before, I'll briefly run through a numerical example, then apply the technology to MHP.
We want to decide about some proposition A and we are given some data D. A priori, we think of A as having a certain probability p, the probability that it was not true was therefore 1-p, the odds on its being true were p:1-p. For instance, if A had a priori probability 1/10, then initially the odds were 9 to 1 against it. (Which is the same as odds of 90 to 10 against, or 180 to 20 against..). If the odds are 90 to 10 against, then the probabilities for and against are 10/(10+90) and 90/(10+90), ie (of course) 1/10 and 9/10.
OK, now we get some data D. The data D can be more or less likely to be produced if A is true, or if A is not true. Suppose there are probabilities that D is observed in both situations: Prob(D|A) and Prob(D| not A). These two numbers are called "the likelihood for/against A, given the data D". Again, only their ratio is important. For instance, it could be that Prob(D|A)=0.05 while Prob(D|not A)=0.0025. In both cases (A true, A not true), D is pretty unlikely, but D is 20 times more likely to occur if A is true than if A is not true.
So we have so far: prior odds of A versus not A are 1 : 10
Likelihood ratio for A versus not A, given data D are 20 : 1
Bayes formula states that the posterior odds for A versus not A are now 1 x 20 : 10 x 1 , or 20 to 10, or 2:1. In other words, having seen D, A is now twice as likely as not A, while before it was 10 times less likely. Simply because the data D is 20 times more likely to happen if A is true than when it is not true.
Three door problem. Suppose we choose door 1. The data is going to be "host opens 3"
Prior odds on car was actually behind door 1 : door 2 : door 3 = 1:1:1 (initially all doors equally likely)
Likelihood ratio for doors 1, 2, 3 given door 3 opened = 1/2 : 1 : 0
The 1/2 for door 1, because if the car is behind door 1, the probability is half that door 3 (and not door 2) is opened
The 1 for door 2, because if the car is behind door 2, the probability is one that door 3 will be opened (you chose 1)
The 0 for door 3, because if the car is behind door 3, the host is not going to open door 3 and show you a goat!
The posterior odds are therefore 1/2 : 1 : 0 which is the same as 1 : 2 : 0 which means probabilities 1/3, 2/3 and 0 on doors 1, 2 and 3. Gill110951 (talk) 18:43, 6 March 2010 (UTC)
Bayes section is REALLY bad: replace by link to Bayes Theorem, Example 2
We can cut the whole section and replace it by a link to:
http://en.wikipedia.org/wiki/Bayes%27_theorem#Example_3:_The_Monty_Hall_problem
Gill110951 (talk) 18:55, 6 March 2010 (UTC)
- The point of the section is to present a fully rigorous proof using a specific formalism, which is why it is more or less presented as an appendix. -- Rick Block (talk) 00:08, 7 March 2010 (UTC)
- The problem with the current version is, that it might insinuate to less versed readers that a Bayesian philosophy is required to provide a full formal proof based on Bayes' formula, which of course is nonsense. The (same) formal proof with the Bayes formula holds for both a Bayesian and a frequentist viewpoint, i.e. though the philosophical underpinnings or interpretation are different for both viewpoints, they more or less produce the same formal proof and of course Bayes' formula is valid in both.--Kmhkmh (talk) 13:59, 7 March 2010 (UTC)
- I also strongly object the section called "Bayesian approach". As I mentioned before, it hardly can be considered a real Bayesian approach, and it only shows a unnecessarily complex way of just using Bayes' formula to calculate the solution. Many textbooks just give the straightforward calculation, and I made a proposal some time ago. Look in Talk:Monty Hall problem/Construction#Mathematical formulation.Nijdam (talk) 15:15, 7 March 2010 (UTC)
Proposal
Mediation seem to be proceeding painfully slowly and many of us here want to get on with improving the article. I have suggested this before but I am going to propose it again here as a compromise between differing viewpoints. I therefore ask all those on both sides of the fence to give their support for this basic proposal to try to reach an end to the argument on this page.
Andrevan, if you would like to move this discussion to your mediation page and act as a mediator for it please feel free to do so.
I propose changing the order of sections to be as follows:
- Basic (non-conditional) problem and solution
- Much as it does now this section should deal with the MHP (mathematical puzzle) as it is understood by the vast majority of the general public (who are aware of it) and the majority of sources (of all types) on the subject. It is the notable MHP and almost certainly what the reader has come here to read about. No mention of conditional/unconditional probability should be made here.
- Aids to understanding
This section should, as it is currently, be aimed a helping the reader to see why the answer is 2/3 and not 1/2. No mention of conditional/unconditional probability should be made here.
- Conditional solution
Here the view of those who think the problem is essentially one of conditional probability (such as Morgan) should be put forward, including the reasons that they believe simple solutions do not address the true problem.
This is essentially an academic section (although in the interests of neutrality I do not call it that) is which what some sources say what about other sources, and the value of studying the problem in this way for students of probability, can be discussed in a proper manner. By fully discussing these issue here we ensure that the academic credentials of WP are maintained.
- Game theory
This a new section which I think should be added. Once you get past the simple (puzzle) problem and consider the characters to be acting as more than agents of chance it is reasonable to ask what happens if they try to maximise their gain or minimise their loss.
- Sources of confusion
Here we discuss what sources say about why people cannot answer the simple problem correctly. The tests on pigeons might fit in here
- Variants
Intentional complications, generalisations etc as currently.
- History
Much as now.
I would request all editors, on both sides of the fence, to subscribe to this proposal. Either way it would help if people try to restrict their replies to 'support' or 'oppose' at this stage. There will always be plenty to discuss later. I strongly believe that these changes will help us all to work together. Martin Hogbin (talk) 12:15, 7 March 2010 (UTC)
Support
Oppose
Comments
Please try to keep these at a minimum.
- My counter proposal, which is a compromise, is to include a single solution section presenting both unconditional and conditional solutions in an NPOV manner, more or less like the draft still currently above (permanent link). IMO, Martin's proposal creates a structural NPOV violation (see WP:NPOV#Article structure). -- Rick Block (talk) 16:16, 7 March 2010 (UTC)
- While your proposal might be more npov or better than Martin's, I really fail to see how it is a compromise in anyway (the same somewhat holds for Martin's suggestion). You both simply reiterate your point of view, you've been arguing the whole time. Imho there's nothing compromising here at all.--Kmhkmh (talk) 16:24, 7 March 2010 (UTC)
- I completely agree Martin's suggestion is no compromise at all (in any respect). The proposal I'm making is a compromise between presenting on the one hand simple solutions as correct and sufficient vs. on the other hand saying the problem is inherently conditional and that these solutions fail to address it. The tone I'm aiming for is "here's a solution ..., here's another kind of solution ..." - not "here's a solution ..., that solution is wrong here's the real solution ...". In a personal sense, the compromise is keeping my individual POV out of the article - which I think is the compromise we must all make. -- Rick Block (talk) 17:15, 7 March 2010 (UTC)
- While your proposal might be more npov or better than Martin's, I really fail to see how it is a compromise in anyway (the same somewhat holds for Martin's suggestion). You both simply reiterate your point of view, you've been arguing the whole time. Imho there's nothing compromising here at all.--Kmhkmh (talk) 16:24, 7 March 2010 (UTC)