.

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.

But one can lose more than the other depending on which
door Monty picks, suppose he always opens B, switching
to B never loses..

>>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.
>

Consider the following 6 cases, where Monty never opens Door
C with the car behind A. Initially

X00
X00

0X0
0X0

00X
00X

Now after Monty opens his door given by an M

XM0
XM0

0XM
0XM

0MX
0MX

Notice that in 4 of the cases door B is opened with
a win rate of 50% for switch.

XM0
XM0

0MX
0MX

In only two of the cases was door C opened with a win
rate of 100%

0XM
OXM

However the average win rate is still 2/3
because (4/6)(50%) + 2/6(100%) = 1/3 + 1/3 = 2/3

So be very careful when talking about the win rate
for a particular door without knowing Monty's actions
in the car behind door A case.