Monty Hall problem

Michael Godin Arlotto (154aamga@pic.ucla.edu)
Thu, 16 Nov 95 08:47:24 -0800

Again assuming the host is nuetral, the host opens up one of
the doors, and you are left with the door you chose and the remaining door. So
you either have the car or you don't. In other words you have a probability
equal to one or zero, not a 50/50 chance. The contestant will be following one
strategy, so when he reaches the "second round" he is not choosing between two
options; instead he is following a predetermined strategy of winning. The
contestant is not asked to pick one of the remaining two doors. He is asked if
he would like to keep his original door or switch. This is a very subtle
difference but a significant one that affects the probabilities for this
problem. Tell me what you think.

Thanks,
Mike

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