Victor wrote: << . . . However, another possibility is that Monty chooses a door at random in which case b = 1/2, yielding the answer of 1/2. So the correct answer depends on what procedure Monty uses (which we don't know). . . .
Huh? If the original pick hides a goat, the door Monty opens is forced to be the only other goat, of course. If the original pick hides the car, then it makes absolutely no difference what procedure Monty uses to pick one of the goats. Because the probability that the original pick hides a goat remains 1/3, and so the remaining probability of 2/3 covers both other doors. Since Monty has eliminated one of them, that 2/3 applies to the door the player is allowed to switch to. OR, maybe you're referring to some *different* game from how I defined the game in a previous post? --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele