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Hey doesn't the Monty Fall problem also have the same answer as the Monty Hall problem - since the host falls and opens that door which always contains the goat

Hey doesn't the Monty Fall problem also have the same answer as the Monty Hall problem - since the host falls and opens that door which always contains the goat, its same as saying that he knows which door contains the goat and opens that only. Please correct me if I am wrong ! EDIT: Saw the soln in the pdf and not able to digest it though. Well if Erdos wasn't able to see it...

harpastum 17y ago
Well, I think most of the confusion comes from the way that Rosenthal ignores a key part of the Monty Fall problem. In the monty fall problem, you first choose a random door. Assuming a random placement of the car and a random guess, you should have a 1/3 chance of choosing the correct door on your first guess, and therefore a 1/3 chance of winning no matter what else happens. I believe what's really happening in this example is that the Rosenthal is allowing for a possibility that he doesn't explicitly mention: the chance that the host opens the wrong door, nullifying the game and ruining the show. Since there is a 1/3 chance that you picked the right door, it's a probability of 2/3 that it is in one of the other two doors. The host accidentally opens one of the two remaining doors. Since the host randomly chooses one of the remaining doors, the chance that he picks the door with the car is 1/3. If he does that, the game is invalidated, so Rosenthal ignores that result. Therefore, the chance that you picked the correct door is 1/3, and the chance that the remaining door is the correct one is 1/3 (the remaining third is the chance that the host ruined the game). Since those are equal probabilities, you end up with an even chance whether or not you switch.
- rottencupcakes 17y ago
  What's actually interesting is linking this result back to Deal or No Deal. This implies that if you got down to two briefcases on the show, with both the $1,000,000 and $1 left, it's a 50/50 chance of picking the million dollar case whether you switch or not.

harpastum 17y ago

Well, I think most of the confusion comes from the way that Rosenthal ignores a key part of the Monty Fall problem. In the monty fall problem, you first choose a random door. Assuming a random placement of the car and a random guess, you should have a 1/3 chance of choosing the correct door on your first guess, and therefore a 1/3 chance of winning no matter what else happens. I believe what's really happening in this example is that the Rosenthal is allowing for a possibility that he doesn't explicitly mention: the chance that the host opens the wrong door, nullifying the game and ruining the show. Since there is a 1/3 chance that you picked the right door, it's a probability of 2/3 that it is in one of the other two doors. The host accidentally opens one of the two remaining doors. Since the host randomly chooses one of the remaining doors, the chance that he picks the door with the car is 1/3. If he does that, the game is invalidated, so Rosenthal ignores that result. Therefore, the chance that you picked the correct door is 1/3, and the chance that the remaining door is the correct one is 1/3 (the remaining third is the chance that the host ruined the game). Since those are equal probabilities, you end up with an even chance whether or not you switch.

rottencupcakes 17y ago
What's actually interesting is linking this result back to Deal or No Deal. This implies that if you got down to two briefcases on the show, with both the $1,000,000 and $1 left, it's a 50/50 chance of picking the million dollar case whether you switch or not.