Monty Hall variation

laurencewithau

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Hi, the usual Monty Hall problem runs as follows. On a TV show Monty stands in front of three doors behind one of which there's a prize, perhaps a probability mathematics textbook. and invites the subject/contestant to choose a door.
Clearly the chance of winning is a third. After the choice has been made Monty throws open one of the two remaining doors to reveal no prize behind it. He then invites the subject to stick with her choice or to switch to the unopened door.
Everyday intuition insists that there is no difference between the alternatives, the chance of guessing the correct door being half whether or not one sticks or switches. Even professors of mathematics in this field have sided with common sense, at least until apprised of the correct reasoning, the solution being two-thirds, not half.
This will start to become apparent, perhaps, if it's noted that if the subject initially chooses door A, call it, then there is a two-thirds chance that one of the other two doors, B and C, is the winning door. But Monty opens either B or C, so the remaining door has a two-thirds chance of being the winner.
A philosopher with an interest in probability theory, however, would argue that the matter is not so clear cut; because the theory is itself based on intuition, albeit systematised, so that there would seem to be a conflict of intuitions rather than an obvious case of everyday fallacious probability reasoning.
Our professor, after all, sided with this latter. That being the case, here is a variant of the problem, the conventional solution to which is again counter-intuitive.
This time let the subject, before making her initial choice, be surreptitiously informed that door C has been inadvertently locked, unknown to Monty, this being the full content of the information passed to her; so, she knows only that it is locked, not whether or not it is the winning door. How should she choose? Let her choose door A and wait to see what happens. The possibilities are as follows:
1/ Monty tries door C, clearly with the intention of opening it, but is then forced, hoping no-one has noticed his confusion, to open door B instead. I make it, even without doing the maths, to be certain that the winning door is A.

2/ Monty tries door C, which he cannot open, and immediately asks the subject whether she wants to switch to door B. Again we have certainty, the winning door being B.

3/ Monty goes directly to door B, opens it and asks the question. Should the subject switch to door C? Assume that there's a key somewhere.
But surely, one might think, the answer is the same as with the standard version: there is a two-thirds chance that door C is the winning door. Monty, after all, knows exactly the same as in the standard version; and the subject is not able, in the present case, to use her special knowledge, unlike in cases 1/ and 2/.
But this time, I suggest, we should do the maths.This, however, is where I encounter a difficulty, which will become apparent if I first of all set out the solution to the standard problem.
Let my use of block letters and lower case be such that "B", for instance, refers to B being the winning door, whereas "b" designates the door, B, that Monty has opened.
Since it is given that the subject initially chooses door A, and since by symmetry it does not then matter whether we take Monty to open door B or door C, let it be B.
Then P(b|A) =1/2
P(b|B) =0
Pb|C = 1
So p(C|b) =P(b|C) x P(C)/[P(b|C) x P(C) + P(b|A) x P(A)] =2/3.

That leaves the variant problem, which I am not sure how to work out. Should the subject, in case 3/, switch to door C? What are the probabilities? Does P = 1/2? I'd be grateful for any help.
 
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Dr.Peterson

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Hi, the usual Monty Hall problem runs as follows. On a TV show Monty stands in front of three doors behind one of which there's a prize, perhaps a probability mathematics textbook. and invites the subject/contestant to choose a door.
Clearly the chance of winning is a third. After the choice has been made Monty throws open one of the two remaining doors to reveal no prize behind it. He then invites the subject to stick with her choice or to switch to the unopened door.
Everyday intuition insists that there is no difference between the alternatives, the chance of guessing the correct door being half whether or not one sticks or switches. Even professors of mathematics in this field have sided with common sense, at least until apprised of the correct reasoning, the solution being two-thirds, not half.
What question is that the answer to? The answer to the problem as usually stated is just, "He should switch". I suppose you are saying the probability of winning if he switches is 2/3. Right?

This will start to become apparent, perhaps, if it's noted that if the subject initially chooses door A, call it, then there is a two-thirds chance that one of the other two doors, B and C, is the winning door. But Monty opens either B or C, so the remaining door has a two-thirds chance of being the winner.
A philosopher with an interest in probability theory, however, would argue that the matter is not so clear cut; because the theory is itself based on intuition, albeit systematised, so that there would seem to be a conflict of intuitions rather than an obvious case of everyday fallacious probability reasoning.
Our professor, after all, sided with this latter. That being the case, here is a variant of the problem, the conventional solution to which is again counter-intuitive.
I'm not at all sure what this means. There are various philosophical opinions about probability, but for this problem I think there's no trouble saying that if this game were played many times, in 2/3 of them he would win by switching. What does your professor say, exactly?

This time let the subject, before making her initial choice, be surreptitiously informed that door C has been inadvertently locked, unknown to Monty, this being the full content of the information passed to her; so, she knows only that it is locked, not whether or not it is the winning door. How should she choose? Let her choose door A and wait to see what happens. The possibilities are as follows:
1/ Monty tries door C, clearly with the intention of opening it, but is then forced, hoping no-one has noticed his confusion, to open door B instead. I make it, even without doing the maths, to be certain that the winning door is A.

2/ Monty tries door C, which he cannot open, and immediately asks the subject whether she wants to switch to door B. Again we have certainty, the winning door being B.

3/ Monty goes directly to door B, opens it and asks the question. Should the subject switch to door C? Assume that there's a key somewhere.
But surely, one might think, the answer is the same as with the standard version: there is a two-thirds chance that door C is the winning door. Monty, after all, knows exactly the same as in the standard version; and the subject is not able, in the present case, to use her special knowledge, unlike in cases 1/ and 2/.
But this time, I suggest, we should do the maths.This, however, is where I encounter a difficulty, which will become apparent if I first of all set out the solution to the standard problem.
This is a bigger change to the problem than you make it sound. There is significant new information given by Monty's actions, that is not possible in the original. There are also new rules for Monty to follow, which are not specified in the original. How will he react to a surprise?

My "intuitive" reaction is that each case is different. In case (1), we can tell that Monty knows that he can choose either door B or C, so we know A is the winner. In case (2), we know that Monty knows that door C is not the winner, and we can guess (based on our subjective knowledge of how Monty would think) that he knows that door B is the winner, though it's possible that he just changed strategies and asked without opening anything, merely to keep you guessing. I wouldn't say we're certain, but it seems likely. Note that a proper calculation of probability depends on knowing Monty's rules.

In case (3), the existence of a locked door has no effect on what has happened. So, yes, I would say that in this case, the probability is 2/3 that C is the winner.

Now I'll look at the rest of what you say.

Let my use of block letters and lower case be such that "B", for instance, refers to B being the winning door, whereas "b" designates the door, B, that Monty has opened.
Since it is given that the subject initially chooses door A, and since by symmetry it does not then matter whether we take Monty to open door B or door C, let it be B.
Then P(b|A) =1/2
P(b|B) =0
Pb|C = 1
So p(C|b) =P(b|C) x P(C)/[P(b|C) x P(C) + P(b|A) x P(A)] =2/3.

That leaves the variant problem, which I am not sure how to work out. Should the subject, in case 3/, switch to door C? What are the probabilities? Does P = 1/2? I'd be grateful for any help.
Okay, I thought you said you were going to show that something is different in your scenario. Now I see you are just saying you haven't been able to work out a solution and show that it is still 2/3.

Here's your work, written out a little more fully to make sure I follow:

\(\displaystyle P(\text{C has the prize} | \text{Monty opens B}) =\\ \frac{P(\text{Monty opens B} | \text{C has the prize}) * P(\text{C has the prize})}{P(\text{Monty opens B} | \text{C has the prize}) * P(\text{C has the prize})+ P(\text{Monty opens B} | \text{A has the prize}) * P(\text{A has the prize})}\\ = \frac{1 * 1/3}{1 * 1/3 + 1/2 * 1/3} = 2/3\)​

To apply the same sort of reasoning to the new problem, I would break it further into cases according to which door is locked, so you'd have things like "Monty chooses B | B has the prize and B is locked". But you'd have to specify definitively what he will do in each case.

Give it a try! But realize that you have added genuine subjectivity into the problem.
 

Jomo

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Hi, the usual Monty Hall problem runs as follows. On a TV show Monty stands in front of three doors behind one of which there's a prize, perhaps a probability mathematics textbook. and invites the subject/contestant to choose a door.
Clearly the chance of winning is a third. After the choice has been made Monty throws open one of the two remaining doors to reveal no prize behind it. He then invites the subject to stick with her choice or to switch to the unopened door.
Everyday intuition insists that there is no difference between the alternatives, the chance of guessing the correct door being half whether or not one sticks or switches. Even professors of mathematics in this field have sided with common sense, at least until apprised of the correct reasoning, the solution being two-thirds, not half.
This will start to become apparent, perhaps, if it's noted that if the subject initially chooses door A, call it, then there is a two-thirds chance that one of the other two doors, B and C, is the winning door. But Monty opens either B or C, so the remaining door has a two-thirds chance of being the winner.
A philosopher with an interest in probability theory, however, would argue that the matter is not so clear cut; because the theory is itself based on intuition, albeit systematised, so that there would seem to be a conflict of intuitions rather than an obvious case of everyday fallacious probability reasoning.
Our professor, after all, sided with this latter. That being the case, here is a variant of the problem, the conventional solution to which is again counter-intuitive.
This time let the subject, before making her initial choice, be surreptitiously informed that door C has been inadvertently locked, unknown to Monty, this being the full content of the information passed to her; so, she knows only that it is locked, not whether or not it is the winning door. How should she choose? Let her choose door A and wait to see what happens. The possibilities are as follows:
1/ Monty tries door C, clearly with the intention of opening it, but is then forced, hoping no-one has noticed his confusion, to open door B instead. I make it, even without doing the maths, to be certain that the winning door is A.

2/ Monty tries door C, which he cannot open, and immediately asks the subject whether she wants to switch to door B. Again we have certainty, the winning door being B.

3/ Monty goes directly to door B, opens it and asks the question. Should the subject switch to door C? Assume that there's a key somewhere.
But surely, one might think, the answer is the same as with the standard version: there is a two-thirds chance that door C is the winning door. Monty, after all, knows exactly the same as in the standard version; and the subject is not able, in the present case, to use her special knowledge, unlike in cases 1/ and 2/.
But this time, I suggest, we should do the maths.This, however, is where I encounter a difficulty, which will become apparent if I first of all set out the solution to the standard problem.
Let my use of block letters and lower case be such that "B", for instance, refers to B being the winning door, whereas "b" designates the door, B, that Monty has opened.
Since it is given that the subject initially chooses door A, and since by symmetry it does not then matter whether we take Monty to open door B or door C, let it be B.
Then P(b|A) =1/2
P(b|B) =0
Pb|C = 1
So p(C|b) =P(b|C) x P(C)/[P(b|C) x P(C) + P(b|A) x P(A)] =2/3.

That leaves the variant problem, which I am not sure how to work out. Should the subject, in case 3/, switch to door C? What are the probabilities? Does P = 1/2? I'd be grateful for any help.
First, I think that you need to state clearly that Monty ALWAYS opens up a losing curtain in the real game scenario.


In the newer version you say:
1/ Monty tries door C, clearly with the intention of opening it, but is then forced, hoping no-one has noticed his confusion, to open door B instead. I make it, even without doing the maths, to be certain that the winning door is A. Not true!. The question is will Monty open door B if it has the winning prize or not? If Monty opens door B and it is a losing door then the prize is behind door A, but why can't the prize be behind door B?

2/ Monty tries door C, which he cannot open, and immediately asks the subject whether she wants to switch to door B. Again we have certainty, the winning door being B. Not true at all. Why can't A be the winning curtain and B & C be the two losing curtains?
 

laurencewithau

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What question is that the answer to? The answer to the problem as usually stated is just, "He should switch". I suppose you are saying the probability of winning if he switches is 2/3. Right?


I'm not at all sure what this means. There are various philosophical opinions about probability, but for this problem I think there's no trouble saying that if this game were played many times, in 2/3 of them he would win by switching. What does your professor say, exactly?


This is a bigger change to the problem than you make it sound. There is significant new information given by Monty's actions, that is not possible in the original. There are also new rules for Monty to follow, which are not specified in the original. How will he react to a surprise?

My "intuitive" reaction is that each case is different. In case (1), we can tell that Monty knows that he can choose either door B or C, so we know A is the winner. In case (2), we know that Monty knows that door C is not the winner, and we can guess (based on our subjective knowledge of how Monty would think) that he knows that door B is the winner, though it's possible that he just changed strategies and asked without opening anything, merely to keep you guessing. I wouldn't say we're certain, but it seems likely. Note that a proper calculation of probability depends on knowing Monty's rules.

In case (3), the existence of a locked door has no effect on what has happened. So, yes, I would say that in this case, the probability is 2/3 that C is the winner.

Now I'll look at the rest of what you say.


Okay, I thought you said you were going to show that something is different in your scenario. Now I see you are just saying you haven't been able to work out a solution and show that it is still 2/3.

Here's your work, written out a little more fully to make sure I follow:

\(\displaystyle P(\text{C has the prize} | \text{Monty opens B}) =\\ \frac{P(\text{Monty opens B} | \text{C has the prize}) * P(\text{C has the prize})}{P(\text{Monty opens B} | \text{C has the prize}) * P(\text{C has the prize})+ P(\text{Monty opens B} | \text{A has the prize}) * P(\text{A has the prize})}\\ = \frac{1 * 1/3}{1 * 1/3 + 1/2 * 1/3} = 2/3\)​

To apply the same sort of reasoning to the new problem, I would break it further into cases according to which door is locked, so you'd have things like "Monty chooses B | B has the prize and B is locked". But you'd have to specify definitively what he will do in each case.

Give it a try! But realize that you have added genuine subjectivity into the problem.
Hi Dr Peterson. Thanks very much for your detailed reply. Your "Dr." is American usage, by the way, in contrast to the British "Dr", with no full stop.
Firstly: yes, I'm saying that 2/3 is the probability of winning if she switches.
Second, Marilyn vos Savant, in her "Ask Marilyn" column in Parade Magazine (hence my American reference) worked out the Monty Hall problem and arrived at a probability of 2/3 that the subject will win if she switches doors. Several thousand readers disagreed with her or even complained about her being allowed to print such an obviously wrong answer. These dissidents included very many professional mathematicians and postgraduates with mathematics PhDs.
Foremost among them was Paul Erdos, one of the most distinguished mathematicians of the 20th century, who realised his mistake, or so I've read, only after trying to prove that she was wrong and finding that she was right.
So what? you might say, but the fact is that in philosophy there are very many aspects of probability theory that are problematic on some accounts, depending on whether one is a frequentist, a Bayesian, a proponent of the a priori interpretation of the probability relation between evidence and conclusion, and so on.
Of particular relevance here is the continuing dispute about Bayesian conditionalisation and its correct analysis, with some theorists denying that it has any valid probability application. They sometimes adduce the apparent counter-intuitiveness of the accredited solution to the Monty Hall problem, though I do not know whether any attempt has been made to construct a probability theory or analysis such that the solution is P=1/2, not 2/3.
Perhaps you would say that this cannot be done, because quite clearly the conventional answer can be tested by conducting a series of trials in which the switch is made, the aim being to demonstrate that switching is a winning move in two-thirds of the trials.
But such tests are statistical, which is to say that they are creatures of probability theory applied to the data that the testing throws up, so that the data has to be interpreted. A value of 2/3 may, for instance, be construed as a relative frequency limit, the notion of which has problems of its own.
We are back, then, with disputes about analysis and interpretation.
The way around it, you might suggest, is to establish by these tests not that P=2/3, which is clearly a theoretical construct, but only that switching succeeds much more often than sticking. No doubt it does, but the Monty Hall problem is probabilistic, and the solution has to be an exact value, not the exhibiting of a vague preponderance of one type of outcome over another.
Thirdly, you say that case 2/ is unclear, for I have not specified the rule by which Monty deals with a locked door. But I did not need to, for if, in case 1/, Monty finds door C locked, whereupon he opens door B, then in case2/he would have done the same if door B had been a losing door. He did not, so it must be the winning door.
That leaves case3/, which you think reduces to the standard case. So do I, which is to say that we agree; but thousands of people agree that the Monty Hall solution is 1/2, not 2/3, and the consensus is that they are wrong.
My point is that in this case it's best to do the maths, except that we would have to decide which terms to substitute into the conditional probability formula or the Bayes expression for it, and we would have to specify what they mean, with some indication as to who knows what when. So we are back, really, with the philosophy of probability theory and its problems.
Or, on the other hand, perhaps it can be shown very easily that case3/ is indeed the standard case, the locked door being irrelevant. But what if a lot of people disagreed? Bring on the relative frequency test via numerous trials; But such a test itself belongs within probability theory. Thanks again.
 

laurencewithau

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First, I think that you need to state clearly that Monty ALWAYS opens up a losing curtain in the real game scenario.


In the newer version you say:
1/ Monty tries door C, clearly with the intention of opening it, but is then forced, hoping no-one has noticed his confusion, to open door B instead. I make it, even without doing the maths, to be certain that the winning door is A. Not true!. The question is will Monty open door B if it has the winning prize or not? If Monty opens door B and it is a losing door then the prize is behind door A, but why can't the prize be behind door B?

2/ Monty tries door C, which he cannot open, and immediately asks the subject whether she wants to switch to door B. Again we have certainty, the winning door being B. Not true at all. Why can't A be the winning curtain and B & C be the two losing curtains?
Hi Jomo. Thanks very much for your reply. Yes, in the standard problem Monty always opens a losing door.
Case/. You ask why, if Monty opens door B, it cannot be the winning door. The answer is that the rule is that he opens only a losing door. So door C, which he tried to open, is a losing door, as is door B, which he did open. Therefore door A is the winning door.
Case2/. Monty tries door C, which is therefore a losing door. He does not try door B, which is therefore the winning door. Thanks again.
 

Dr.Peterson

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My response was essentially Jomo's: In order to properly analyze the new problem, Monty's behavior has to be fully defined. I assume, as you do, that it is given that Monty will never open a winning door; but you opened the additional possibility that he opens no door, so you have to define whether he can do that when it is not necessary (and especially if he is taken by surprise). Without that, we have to guess at his psychology: What percent of the time would he decide to go ahead and open the other door, knowing that it is a loser? Might he try to cover his accidental revelation by randomly deciding whether to open the other?

With a clear statement that he will always open a door if possible, we get the answers I aid are "likely". But I haven't gone through the formal work to confirm that; that's your job ...

What I said, that I don't think Jomo mentioned, is that the additional information we get by seeing an abortive attempt to open a door changes the problem so that in that case, we would not follow the general strategy of always switching; so our overall odds of winning improve. You didn't state whether your answer of 2/3 is the probability of winning by always switching, or our overall probability of winning following an optimal strategy.

As for the philosophical issues, I am aware of all that, though I haven't dug into it deeply; I asked whether that was the basis of your professor's objection, and you haven't quite answered that. Clearly, though, it's not worth doing the math if you think it's all subjective, so my assumption is that you don't take the philosophical considerations too seriously, and weren't trying to answer them.

As for Dr vs Dr., it happens that the only reason I put the period in my name here is that I initially tried making an account as DrPeterson, and something went wrong so I tried again!
 

laurencewithau

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My response was essentially Jomo's: In order to properly analyze the new problem, Monty's behavior has to be fully defined. I assume, as you do, that it is given that Monty will never open a winning door; but you opened the additional possibility that he opens no door, so you have to define whether he can do that when it is not necessary (and especially if he is taken by surprise). Without that, we have to guess at his psychology: What percent of the time would he decide to go ahead and open the other door, knowing that it is a loser? Might he try to cover his accidental revelation by randomly deciding whether to open the other?

With a clear statement that he will always open a door if possible, we get the answers I aid are "likely". But I haven't gone through the formal work to confirm that; that's your job ...

What I said, that I don't think Jomo mentioned, is that the additional information we get by seeing an abortive attempt to open a door changes the problem so that in that case, we would not follow the general strategy of always switching; so our overall odds of winning improve. You didn't state whether your answer of 2/3 is the probability of winning by always switching, or our overall probability of winning following an optimal strategy.

As for the philosophical issues, I am aware of all that, though I haven't dug into it deeply; I asked whether that was the basis of your professor's objection, and you haven't quite answered that. Clearly, though, it's not worth doing the math if you think it's all subjective, so my assumption is that you don't take the philosophical considerations too seriously, and weren't trying to answer them.

As for Dr vs Dr., it happens that the only reason I put the period in my name here is that I initially tried making an account as DrPeterson, and something went wrong so I tried again!
Hi Dr Peterson,
Thanks for making this thread so interesting. I don't agree that Monty's psychology is relevant to case 1. What is clearly to be understood is that Monty's actions mean that both door C and door B are losing doors. My concern, anyway, is with case 3, and what I was trying to say was that although intuitively we believe that door C being locked is irrelevant, how can we be sure, given that so many clever people were sure that the solution to the Monty Hall problem was that P = 1/2. Well, it's just obvious, but so was P=1/2. Thanks again.
 

Dr.Peterson

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I thought your main goal was to carry out the work for your problem the way you showed for the original. I suggested how I might do it, but you don't seem to be pursuing it.

Break your case 3 (Monty opens B) into not just 2 parts (A has the prize, C has the prize) but 6 (A has prize, A is locked; A has prize, C is locked; B has prize, A is locked; B has prize, C is locked; C has prize, A is locked; C has prize, C is locked). Then apply Bayes' Theorem to that.

My main point, to repeat, is not that case 1 depends on psychology; I agreed with you that that case is "open and shut" (given the usual assumption that Monty never opens a winning door). Rather, I said that the other cases are not definite until you clearly define Monty's behavior in order to make the problem clear. I've suggested how to do that: stipulate that he always opens a losing door when there is one available. That fills in the gaps, I think, and agrees with what you are assuming without stating it.

Now you can work on the problem.
 

Jomo

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Hi Jomo. Thanks very much for your reply. Yes, in the standard problem Monty always opens a losing door.
Case/. You ask why, if Monty opens door B, it cannot be the winning door. The answer is that the rule is that he opens only a losing door. So door C, which he tried to open, is a losing door, as is door B, which he did open. Therefore door A is the winning door.
Case2/. Monty tries door C, which is therefore a losing door. He does not try door B, which is therefore the winning door. Thanks again.
You ask why, if Monty opens door B, it cannot be the winning door. The answer is that the rule is that he opens only a losing door. So door C, which he tried to open, is a losing door, as is door B, which he did open. Therefore door A is the winning door. I am not usually this cold but your response is complete nonsense. You talk about the rules but you never did say what happens if the winning door is door B. The rules do say that Monty MUST open a losing door that is not the one which the contestant picked. Well, since door C is unable to be open, then Monty can't follow the rules of the game. So what happens?

Case2/. Monty tries door C, which is therefore a losing door. He does not try door B, which is therefore the winning door. The rule says that Monty triesto open up a losing door that is not a door chosen by the contestant. I guess that you miss (even though you stated in your OP) the point that 1/3 of the time the contestant picked the winning door. This implies that there are two losing doors that the contestant did not choose. This means that Monty get to choose one of the two door losing doors! You said he tried (unsuccessfully) to open door C. Just because he tried to open door C does not mean A is the winning door!
 

laurencewithau

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I thought your main goal was to carry out the work for your problem the way you showed for the original. I suggested how I might do it, but you don't seem to be pursuing it.

Break your case 3 (Monty opens B) into not just 2 parts (A has the prize, C has the prize) but 6 (A has prize, A is locked; A has prize, C is locked; B has prize, A is locked; B has prize, C is locked; C has prize, A is locked; C has prize, C is locked). Then apply Bayes' Theorem to that.

My main point, to repeat, is not that case 1 depends on psychology; I agreed with you that that case is "open and shut" (given the usual assumption that Monty never opens a winning door). Rather, I said that the other cases are not definite until you clearly define Monty's behavior in order to make the problem clear. I've suggested how to do that: stipulate that he always opens a losing door when there is one available. That fills in the gaps, I think, and agrees with what you are assuming without stating it.

Now you can work on the problem.
Hi Dr Peterson.' yes, I do intend to do the maths in order to see if our intuitions bear scrutiny, and I did assume, of course, that Monty always opens a losing door, and never a winning one, which would be pointless. I'll do what you say and break the problem down. Thanks again for all your help.
 

laurencewithau

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You ask why, if Monty opens door B, it cannot be the winning door. The answer is that the rule is that he opens only a losing door. So door C, which he tried to open, is a losing door, as is door B, which he did open. Therefore door A is the winning door. I am not usually this cold but your response is complete nonsense. You talk about the rules but you never did say what happens if the winning door is door B. The rules do say that Monty MUST open a losing door that is not the one which the contestant picked. Well, since door C is unable to be open, then Monty can't follow the rules of the game. So what happens?

Case2/. Monty tries door C, which is therefore a losing door. He does not try door B, which is therefore the winning door. The rule says that Monty triesto open up a losing door that is not a door chosen by the contestant. I guess that you miss (even though you stated in your OP) the point that 1/3 of the time the contestant picked the winning door. This implies that there are two losing doors that the contestant did not choose. This means that Monty get to choose one of the two door losing doors! You said he tried (unsuccessfully) to open door C. Just because he tried to open door C does not mean A is the winning door!
Hi Jomo. Thanks for your reply. The very fact that you think my reasoning is nonsense, whereas it makes perfect sense to me, signifies quite a lot about one of the ways in which probability theory and application is problematic.
But in the present instance the point at issue concerns the rules of Monty's game, and it is clear to me that in case 1 they prescribe that he open a losing door and never a winning one. Since he tries to open door C, it must be a losing door, and since, having failed, he then opens door B, it must also be a losing door; therefore door A is the winning door.
He might, in reality, cancel that particular game, send for a handyman and then play the game again;
but the Monty Hall problem is a probability problem, not a question about a real game show featuring real people and, with regard to the present variant, a door that actually cannot be opened.
The game show is just the vehicle by which one travels from one point of reasoning to the next, and it is in that spirit that my argument in relation to case 1 is to be understood.
If you are trying to work out one of the numerous problems about truth tellers and liars, you treat the problem as a puzzle in logic and reasoning, without which you would not be able to understand it.
If the completely mendacious individual says on New Year's Day that his New Year Resolution is that he will always tell the truth, rather than always lie, you do not wonder whether he really means it, for you know that it must be a lie. Nor, as a matter of interest, do you take issue with the problem itself, its having occurred to you that if he always lies then he cannot claim to make a New Year's Resolution, since this would, by implication, be to tell the truth: not about the resolution he says he will make but about the date.
Clearly, this would not be a valid criticism of the puzzle or its creator, for it is understood, in the spirit of the puzzle, that the lie must be a direct lie. Otherwise, there could be no puzzles about truth tellers and liars.
If, on the other hand, the puzzle creator has the liar saying " I am lying", a statement which clearly is meant to refer to itself, not to another statement, then the puzzle cannot be solved, for it now incorporates a genuine paradox, the statement being neither true nor false without contradiction.
In the spirit of the thing, then, my reasoning with regard to case 1 seems to me to be both valid and perspicuous.
As regards case 2, I concluded that door B was the winning door, not, as you said, door A. I concede that one would have to be fully in the spirit of the thing for it to be clear that door B must be the winning door. Thanks again for all your help.
 

Jomo

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Messages
3,698
Hi Jomo. Thanks for your reply. The very fact that you think my reasoning is nonsense, whereas it makes perfect sense to me, signifies quite a lot about one of the ways in which probability theory and application is problematic.
But in the present instance the point at issue concerns the rules of Monty's game, and it is clear to me that in case 1 they prescribe that he open a losing door and never a winning one. Since he tries to open door C, it must be a losing door, and since, having failed, he then opens door B, it must also be a losing door; therefore door A is the winning door.
He might, in reality, cancel that particular game, send for a handyman and then play the game again;
but the Monty Hall problem is a probability problem, not a question about a real game show featuring real people and, with regard to the present variant, a door that actually cannot be opened.
The game show is just the vehicle by which one travels from one point of reasoning to the next, and it is in that spirit that my argument in relation to case 1 is to be understood.
If you are trying to work out one of the numerous problems about truth tellers and liars, you treat the problem as a puzzle in logic and reasoning, without which you would not be able to understand it.
If the completely mendacious individual says on New Year's Day that his New Year Resolution is that he will always tell the truth, rather than always lie, you do not wonder whether he really means it, for you know that it must be a lie. Nor, as a matter of interest, do you take issue with the problem itself, its having occurred to you that if he always lies then he cannot claim to make a New Year's Resolution, since this would, by implication, be to tell the truth: not about the resolution he says he will make but about the date.
Clearly, this would not be a valid criticism of the puzzle or its creator, for it is understood, in the spirit of the puzzle, that the lie must be a direct lie. Otherwise, there could be no puzzles about truth tellers and liars.
If, on the other hand, the puzzle creator has the liar saying " I am lying", a statement which clearly is meant to refer to itself, not to another statement, then the puzzle cannot be solved, for it now incorporates a genuine paradox, the statement being neither true nor false without contradiction.
In the spirit of the thing, then, my reasoning with regard to case 1 seems to me to be both valid and perspicuous.
As regards case 2, I concluded that door B was the winning door, not, as you said, door A. I concede that one would have to be fully in the spirit of the thing for it to be clear that door B must be the winning door. Thanks again for all your help.
Lawrence,
All I am really trying to say is that you can't ask questions about the alternate version of the game until you give a clear set of rules. You never said what would happen if door B is the winning door if the contestant picked door A. Sure if Monty will not open door B (only if it is the winning door) then the player will always win if Monty 1st tries to open door C. Why? Because if he opens door B (after trying to open door C and the player see this), then Door A is the winning door and the player should not switch. If he shows door B (so it is a losing door) after failing at opening door C (and the player sees this) then Door A is the winning Door.
This part of the game that you left out is crucial to determine the probability of certain things happening. Do you see this?
 

laurencewithau

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Joined
Mar 25, 2019
Messages
15
Lawrence,
All I am really trying to say is that you can't ask questions about the alternate version of the game until you give a clear set of rules. You never said what would happen if door B is the winning door if the contestant picked door A. Sure if Monty will not open door B (only if it is the winning door) then the player will always win if Monty 1st tries to open door C. Why? Because if he opens door B (after trying to open door C and the player see this), then Door A is the winning door and the player should not switch. If he shows door B (so it is a losing door) after failing at opening door C (and the player sees this) then Door A is the winning Door.
This part of the game that you left out is crucial to determine the probability of certain things happening. Do you see this?
Hi Jomo, Thanks again for your reply. You are being very patient with me.
I was a bit confused by the account you gave in your latest post of the first two cases, so I'll try to make it as clear as possible.
If the game goes according to plan, Monty will get the contestant to pick a door, say door A, and then he will open one of the other doors to show that there is no prize behind it.
If B and C are both losing doors, because A is the winning door, then Monty will arbitrarily choose one to open. If there is only one losing door, either B or C , the other being a winning door,, then Monty will open the losing door.
He then invites the contestant to stick or switch. What makes the probability problem for which the game show is a vehicle so interesting is that many people, even probability theorists, statisticians and other experts in the mathematical field, believe intuitively that the probability of winning is 1/2, irrespective of whether one sticks or switches.
This is a mistake, but of further interest is that exposing that mistake and showing that the right answer is 2/3 is not an easy task, given that one's intuition strongly suggests otherwise. I'm told that the maths experts in particular have difficulty in explaining why the answer is 2/3, because they unconsciously assume that they are talking to other experts. Since I am not an expert, I perhaps have an advantage, so permit me to put it to some use by providing an explanation that will help other non-experts to understand the Monty Hall problem.
Let the doors be A, B and C, the contestant picking door A. Her chance of its being the winning door, since it is 1/3, corresponds to a relative frequency of winning of about one third in a series of trials. Some theorists would perhaps dispute this, but it is intuitively plausible and hard to deny.
If a hundred trials were conducted, with the subject picking a door, which was then opened, she would be correct about a third of the time. For about two-thirds of the trials, then, either door B or door C would be the winning door. But this means that for about two-thirds of the number of trials, if the losing door, either A or B, is opened, then the remaining door, either B or A, is the winning door, in which case one shroud switch when invited to, because the door that one switches to will be the winning door for about two-thirds of the time.
If we now go back to Jomo's doubts about my reasoning in cases 1 and 2, then it may be partly my fault for speaking of a variant or altered version of the problem when clearly it is nothing of the kind.
Rather, it concerns what happens when the game goes wrong, and I should have made clear that Monty then tries to make adjustments in order for the show to continue.
I should have said, as Dr Peterson suggested, that Monty will always open a door if possible, where it goes without saying that this must be a losing door. What I deny is that anything else needs to be made clear, for instance that if Monty tries door C and turns away, unable to open it, then this means that it is a losing door.
This is already clear, one reason for which is that on any other interpretation there would not be a problem to solve.
This is what I meant when I spoke of the spirit of the thing, for the understanding must always be that in a puzzle or problem all the given information is relevant, a requirement which determines how that information is to be understood.
Thus it is that case 2 also makes perfect sense, or would have done if I had made clear what I have just said that I should have made clear.
In case 2 Monty tries door C, fails to open it and immediately invites the contestant to stick or to switch to door B. This tells the contestant, or rather the reader, that door B is the winning door, for otherwise Monty would have opened it, as in case 1.
I cannot put it any plainer than that, and in any case I was also trying to draw attention to case 3, in which Monty goes straight to door B, opens it and invites the contestant to stick or to switch to door C.
Monty does not know, of course, that door C is locked, but the contestant does. The really interesting question is this: although it is intuitively obvious that door C being locked, and the contestant knowing it, but not Monty, makes no difference to anything, the probability of winning still being 2/3 if the switch is made, provided a locksmith or key can be found, how can we be sure without doing the maths, and how, in any case is that to be done? Jomo,Thanks again,
 
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