I maintain that as long as ----- the player has no knowledge of Monty Hall's algorithm to pick a door to open and show a goat ----- (e.g., even if Monty doesn't know where the car is, and just lucks out in picking a goat door to open), then from the player's point of view, which is what matters here, the probability that the original pick hides a car remains 1/3. --Dan << On Wed, Oct 28, 2009 at 9:41 PM, Dan Asimov <dasimov@earthlink.net> wrote:
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?
OR, maybe your definition of the game isn't clear enough? You say "opens a door to show a goat", but we don't know if that's because he knows where the goat is AND always shows a door with the goat, or because he opened a door which happened to have a goat.
_____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele