User talk:Martin Hogbin/MHP - Fixing Devlin

Here is Devlin's original explanation:

Imagine you are the contestant. Suppose the doors are labeled A, B, and C. Let's assume you (the contestant) initially pick door A. The probability that the prize is behind door A is 1/3. That means that the probability it is behind one of the other two doors (B or C) is 2/3. Monty now opens one of the doors B and C to reveal that there is no prize there. Let's suppose he opens door C. (Notice that he can always do this because he knows where the prize is located.) You (the contestant) now have two relevant pieces of information: [1] The probability that the prize is behind door B or C (i.e., not behind door A) is 2/3. [2] The prize is not behind door C. Combining these two pieces of information, you conclude that the probability that the prize is behind door B is 2/3.

The problem

Do we all agree that Devlin's omission is assuming that the (value of the) probability that the prize is behind door B or C (i.e., not behind door A) remains 2/3 after the host has opened door 3 to reveal a goat? Martin Hogbin (talk) 11:37, 14 October 2012 (UTC)

Fixing the problem

Richard's fix

The contestant initially chooses door number 1. Initially, his odds that the car is behind this door are 2 to 1 against: it is two times as likely for him that his choice is wrong as that it is right.The host opens one of the other two doors, revealing a goat. Let’s suppose that for the moment, the contestant doesn’t take any notice of which door was opened. Since the host is certain to open a door revealing a goat whether or not the car is behind door 1, the information that an unspecified door is opened revealing a goat cannot change the contestant’s odds that the car is indeed behind door 1; they are still 2 to 1 against. Now here comes the further detail which we will take account of in this solution: the contestant also gets informed which specific door was opened by the host – let’s say it was door 3. Does this piece of information influence his odds that the car is behind door 1? No: from the contestant’s point of view, the chance that the car is behind door 1 obviously can’t depend on whether the host opens door 2 or door 3 – the door numbers are arbitrary, exchangeable. Therefore, also knowing that the host opened specifically door 3 to reveal a goat, the contestant’s odds on the car being behind his initially chosen door 1 still remain 2 to 1 against. He had better switch to door 2.

Comments and suggestions

I believe that there are two problems with Devlin's explanation. The minor one, which Richard addresses above is to show that the host's choice of door does not alter the probability that the car is behind the originally chosen door. The more important problem is to show that the revealing of a goat does not alter the probability that the car is behind the originally chosen door.

Devlin's explanation, as it stands, does not distinguish against the ignorant host. This is a major flaw. The real problem is that, if the host were to open an unchosen door at random and this were to happen to show a goat, Devlin's explanation, as it is above, would still apply and would, of course, give the wrong answer. We need to show that the host's actions, in their entirety, do not change the (value of the) probability that the car is behind the originally chosen door.

I think the best way to show this is to start with the ignorant host case. Here, it is reasonably intuitive to accept that the host, having taken a random sample from behind the two unchosen doors (which could have revealed the car), is giving us a clue as to what is likely to be behind our originally chosen door. If he happens to show a goat, it is more likely that you have initially chosen the car.

This can then be contrasted with the case where we know for certain, before the event, that the host will reveal a goat, so no information about the content of our originally chosen door is revealed. (Except by door number, which can be dealt with later).

If we could find a source for this argument, would everyone be happy? Martin Hogbin (talk) 12:04, 14 October 2012 (UTC)