Murray... "I have to say that I am totally amazed by gnu's inability to grasp the essence of this problem."

Likewise, I am sure. You are talking about probabilities based on a simple ruse... offering two choices, the first of which DOES NOT COUNT and therefore CANNOT influence determination of subsequent probabilities. You can think it does, you can "prove" it does, but it simply cannot. Allow me, again... just one last time...

THERE ARE ONLY TWO DOORS! Now, IF Monty doesn't know where the car is, there ARE three doors. But, Monty KNOWS where the car is, so truth and probability tables and fancy theorems based on there being three doors (FALSE!) are inapplicable. You cannot apply conditions and analyses from the first choice to the second choice because they are inherently different... there are only two doors.

Anyone who thinks there are three doors should be locked in a dark room with a small black and white TV with poor reception and forced to watch EVERY Monty Hall car-goat segment ever aired over and over and over and...

Ya know, if I ever get a chance to win that car, I am gonna switch... and if I get the goat, I am gonna tie it up in front of yer house. Don't worry... there is only a 50/50 chance of that happening.