>>Once again you are not using the lettering system used by
>>Devlin where the letters are not applied until after the
>>choices are made.
>
>That is not true here is what Delvin said:
>
>"Suppose the doors are labeled A, B, and C. Let's assume the
>contestant initially picks 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."
>
>Delvin did not say that Monty picks one of the two remaining
>doors
>and then labels that door C.
>
>
>>Monty does indeed have a choice between
>>two doors but the labels "B" and "C" are not applied until
>>after the contestant choses door "A" and Monty next opens a
>>door. Regardless of which door Monty opens that door is
>>labeled "C" and the remaining door is labeled "B".
>
>I agree that such a labeling system changes the problem
>but that is not the spirit of the orginial question in which
>three doors each have distinct identities given by A B and C
>or if you prefer 1,2,3. People can judge for themselves what
>they think Delvin really meant.
>
>
The paradox is that with either labeling system the same fact situation arises and the problem is the same, what is the probability of "B" at that instant. Using "my" labeling system one can easily see that by switching, contestants double their chances of winning. The only way to resolve the paradox is to convince yourself that at that point the probability of "A" is indeed only 1/3 since the contestant had no idea where the car is out of three choices, and the probability of "C" is indeed 0 since in fact there is no car behind it, therefore regardless of your analysis, the probability of "B" remains 1 - 1/3 - 0 = 2/3.
Have a Good Day
KJ Ramsey