>>>>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 two labeling systems are completely different.
>Your system corresponnds to always switch for both doors
>which averages out to 2/3. You can't relabel door C door
>B and say its the same!
>
The fact situation is the same. The contestant chooses A and Monty opens door C with no car behind it, the probability of the car being behind B is then 1 - 1/3 - 0 = 2/3. Note that B can also win if the contestant chooses door C and switches to B but that is not the fact situation. In the one third time that the car is behind A, neither B or C can win regardless of whether the contestant switches to them.

>Suppose the contestant chooses A and Monty never chooses
>C with a car behind A under "my" system. What are the odds
>on B and C? Door C can never lose....think about it...

The question is not whether a contestant wins or looses with a particular door but what are the probabilities of the car being behind B given the fact situation. Having Monty open door B is not the fact situation at issue.

Have a Good Day
KJ Ramsey