Re: [math-fun] best published explanation of the Monty Hall Paradox?
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
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. --Joshua
There are two conditional probabilities here: P(B|A) and P(B|~A). In the first case we're asking what's the probability that Monty's door has a goat given that the first door has a car. As I said, it's obviously 1. For the second we're asking what's the probability that Monty's door has a goat given that the first door has a goat. If Monty picks at random, it's 1/2. In the instance that we've seen he happens to open a door with a goat. On the other hand Monty's procedure would be to always open a door with a goat (since he knows what's behind the doors), which clearly gives no information. In that case P(B|~A) = 1. On Thu, Oct 29, 2009 at 12:41 AM, Dan Asimov <dasimov@earthlink.net> wrote:
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
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participants (3)
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Dan Asimov -
Joshua Zucker -
victor miller