Re: [math-fun] best published explanation of the Monty Hall Paradox?
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
On Wed, Oct 28, 2009 at 10:34 PM, Dan Asimov <dasimov@earthlink.net> wrote:
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.
Well, yes: 1/3 of the time the original pick hides a car and Monty shows a goat, 1/3 of the time Monty shows a car, and 1/3 of the time Monty shows a goat and the car can be obtained by switching. So in this case there is no advantage to switching. On the other hand, if Monty knows where the car is and avoids showing it, then we have a 1/3 vs 2/3 situation, and there is a huge advantage to switching. So I agree that the 1/3 probability doesn't change, but the question is not "what is the probability that the originally selected door contains the car?" but rather "is it advantageous to switch?" --Joshua
As I tried to say before the problem as given is ill-posed, in that it depends on what Monty's strategy is. In particular you could imagine that what we observe is the result of some random process, but that we only get to see the outcomes in which Monty opens a door with a goat. If you assume that we get to see ALL outcomes, then in the notation I gave before, we have P(B|~A) = 1, and, the probability that the first door has the car is still 1/3. However, imagine the following: Fix a positive integer n. Before each random trial, we choose one of (A,B,C) uniformly and independently of this we choose j from {1,...,n} uniformly. There are 3n equally likely outcomes. Now, if we choose A we put the car behind door 1, B -- door 2, and C -- door 3. If the car is behind door 1, then Monty always opens door 2. If the car is behind door 2 or door 3, then Monty will open the door with a goat if and only if j=1, otherwise he opens the door with the car. So we get to see n+2 outcomes: (A,j) for all j, (B,1) and (C,1), and in n of those the car is behind door 1. So now the probability that the car is behind door 1 is n/(n+2). If n = 2, then we're in the situation that I spoke of before -- Monty flips a coin, and we have probability 1/2. However, if n is really large the probability approaches 1. On Thu, Oct 29, 2009 at 1:56 AM, Joshua Zucker <joshua.zucker@gmail.com> wrote:
On Wed, Oct 28, 2009 at 10:34 PM, Dan Asimov <dasimov@earthlink.net> wrote:
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.
Well, yes: 1/3 of the time the original pick hides a car and Monty shows a goat, 1/3 of the time Monty shows a car, and 1/3 of the time Monty shows a goat and the car can be obtained by switching. So in this case there is no advantage to switching.
On the other hand, if Monty knows where the car is and avoids showing it, then we have a 1/3 vs 2/3 situation, and there is a huge advantage to switching.
So I agree that the 1/3 probability doesn't change, but the question is not "what is the probability that the originally selected door contains the car?" but rather "is it advantageous to switch?"
--Joshua
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participants (3)
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Dan Asimov -
Joshua Zucker -
victor miller