>>>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!

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...