>Read Devlin's column again. A B and C are used by Devlin as
>follows. A is the door that the contestant chooses, C is the
>"wrong" door that Monty opens and B is the other door that
>"was not opened". You are arguing apples instead of oranges
>because using Devlin's lingo, door B is never the door
>opened by Monty. You must use Devlin's lingo if you are to
>argue an error in his logic. You did not. You are right only
>if you do not follow the lingo in Devlin's column, but that
>does'nt mean that Devlin's column was wrong. The probability
>that the car is behind door B (using Devlin's lingo) is
>truely 2 in 3 because it can't be behind door C which is a
>wrong door "The car is not behind door C".

Remember in the (2/3) cases when the car is behind door B or C
the chance of winning is 100%. Your reasoning is that 2/3
of the time (when the car is behind B or C) the chances of winning is 2/3.

The error in your reasoning is to associate a probablity of losing
with "those instances when the car is not behind A".

Indeed 2/3 of the time the contestant will win 100% of the
time. Half the time being the cases when Monty opens door C.
Considering the 2/3 of the time the car is behind B or C,
those cases in which door C is opened do nothing to alter
the winning percentage. It's still 100%.

To figure out the overaall winning percentage for Door C you have to
figure out how many times door C lOSES. The losing cases come
from when the car is behind door A. Here Monty has a choice,
he can open door B or C. If he always opens door B he shifts
all the losing cases to door B and door C always wins.

half of that (2/3) or 1/2(2/3)=1/3 Monty opens door C
and the car is behind B so we know that the contestant
wins 1/3 of the time.

All we know then is how many times the contestant wins.
What fraction of the time Does