#### laurencewithau

##### New member

- Joined
- Mar 25, 2019

- Messages
- 15

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.

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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