If Monty opens the door by chance there are 3 equally likely cases: There's a 1/3 chance you picked the car and Monty shows you a goat, 1/3 chance you picked a goat and Monty shows you the other goat, 1/3 chance you picked a goat and Monty shows you a car. So if Monty shows you a goat you have equal probability of being in one of the first two cases.
If Monty doesn't open the door by chance then he never shows you a car. So 2/3 of the time you picked a goat and Monty shows you the other goat.
Yup, the fact that you only have an advantage in switching when Monty "leaks information" in explicitly choosing _not_ to open a certain door as [1] pointed out, is likely the crux of what makes this unintuitive, since it is a very unusual prior.
It changes. Monty's behaviour influenced by his knowledge were the car is. He is leaking information. It is a probabilistic leak: if you first picked by a chance the only door with a car, then Monty is free to open any of remaining doors.
But if you had chosen a door with a goat, then Monty has no choice at all, he must open the only door with a goat that you didn't pick. It is a leak.
From other hand if Monty picked a door by random, he would not leak his secret, but he might open a door with the car accidentally.
> but he might open a door with the car accidentally.
Yes, but the fact that the problems never mention this possibility makes it pretty clear to me that, from the contestant's perspective, it is guaranteed that this will not happen. The original problem even mentions that Monty Hall knows what's behind the doors, which gives a clear idea of how this guarantee is implemented (versus, say, the contestant's memory being wiped and the game reset every time a car is revealed).
The language of the problem is still ambiguous, of course, because all human language is ambiguous. It could be that the car is a Hot Wheels car and the goat is actually a more valuable prize. We could quibble endlessly about the ambiguity of the problem statement, but I personally find the mathematical problem of the traditional intended interpretation more interesting.
You had a 1/1000 chance of picking the right door at once.
Monty had a 1/1000 chance of opening only empty doors.
He had a 998/1000 chance of opening a door to a car, but that didn't happen.
So it's either of the other two cases, with equal probabilities.
If he used his knowledge to never open a car, the 998/1000 case wouldn't exist, and there would be a 999/1000 chance that the door he left unopened had the car.
full_monty|2 years ago
If Monty opens the door by chance there are 3 equally likely cases: There's a 1/3 chance you picked the car and Monty shows you a goat, 1/3 chance you picked a goat and Monty shows you the other goat, 1/3 chance you picked a goat and Monty shows you a car. So if Monty shows you a goat you have equal probability of being in one of the first two cases.
If Monty doesn't open the door by chance then he never shows you a car. So 2/3 of the time you picked a goat and Monty shows you the other goat.
krackers|2 years ago
[1] https://news.ycombinator.com/item?id=39514463
ordu|2 years ago
But if you had chosen a door with a goat, then Monty has no choice at all, he must open the only door with a goat that you didn't pick. It is a leak.
From other hand if Monty picked a door by random, he would not leak his secret, but he might open a door with the car accidentally.
tshaddox|2 years ago
Yes, but the fact that the problems never mention this possibility makes it pretty clear to me that, from the contestant's perspective, it is guaranteed that this will not happen. The original problem even mentions that Monty Hall knows what's behind the doors, which gives a clear idea of how this guarantee is implemented (versus, say, the contestant's memory being wiped and the game reset every time a car is revealed).
The language of the problem is still ambiguous, of course, because all human language is ambiguous. It could be that the car is a Hot Wheels car and the goat is actually a more valuable prize. We could quibble endlessly about the ambiguity of the problem statement, but I personally find the mathematical problem of the traditional intended interpretation more interesting.
Scarblac|2 years ago
You had a 1/1000 chance of picking the right door at once.
Monty had a 1/1000 chance of opening only empty doors.
He had a 998/1000 chance of opening a door to a car, but that didn't happen.
So it's either of the other two cases, with equal probabilities.
If he used his knowledge to never open a car, the 998/1000 case wouldn't exist, and there would be a 999/1000 chance that the door he left unopened had the car.
Scubabear68|2 years ago