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Whether or not the host knows where the prize is absolutely makes a difference. There's a reason why it's stated twice in TFA, very unambiguously, that the host

Whether or not the host knows where the prize is absolutely makes a difference. There's a reason why it's stated twice in TFA, very unambiguously, that the host is intentionally opening a door they know does not contain the prize. In the case of a host that doesn't know where the prize is, your odds don't improve by switching. Think of it slightly differently: Imagine you pick a door, and then the host, who has no idea what's behind the doors, also picks one. Then the remaining door opens and reveals that there's nothing behind it. There's really no difference between your choice and the hosts.

joe_the_user 5y ago
Think of it slightly differently: Imagine you pick a door, and then the host, who has no idea what's behind the doors, also picks one. Then the remaining door opens and reveals that there's nothing behind it. Yes, if the host were to pick a door, not tell you which it was or reveal anything else and then offer to switch your choice for their choice, there would be no difference in the odds of each choice. That just happens not to be the statement of the situation. The statement of situation is: you pick a door. The host picks a door, other than what you picked, then opens the door and reveals there's nothing there. This excludes the situation of the host picking the prize or the host picking the same door as you. How that exclusion happened doesn't matter. At that point, you get to choose your original pick or switch to the remaining door. At that point, what the host knew before irrelevant. You had a 1/3 chance of picking the door with the prize before regardless of the host, and now you can make a choice that gives every other possibility and so a 2/3 chance of getting the prize. Edit: I think that the problem is stated as "the host knows" because that means the host is guaranteed to open a door without the prize. It's not that the host's knowledge matters to the strategy, it's that host's knowledge makes it certain the situation will happen as described. See mtlogstdo's comment. https://news.ycombinator.com/item?id=29846498 https://news.ycombinator.com/item?id=29846498
- tawan 5y ago
  I disagree with your conclusion that the odds remain 2/3 even if the host doesn’t know where the prize is. If that were the case, it would mean that if the same scenario is repeated many many times, then in 2/3 of the cases you would still pick the door with the prize with the same strategy. But what about the instances where the host picks the door with prize before you even get a chance to pick the other door? These cases happen with a probability of 2/3 * 1/2 = 1/3. Therefore, you only get to choose in 2/3 of the cases and in 1/2 of those cases you have already picked the correct door the first time.
  - 8note 5y ago
    Do you still have the opportunity to switch if he picks the winning door? Even if you could, you already know that you're losing, unless he doesn't show you what's behind it. If the game is still going, you know that he picked an empty door regardless of whether he knew he was doing it
- mikenew 5y ago
  Here, I wrote a simulation in nim: https://gitlab.com/mikenew/monty_hall/-/blob/main/monty_hall.nim https://gitlab.com/mikenew/monty_hall/-/blob/main/monty_hall... If you have nim installed you can run it yourself with `nim c -r monty_hall.nim`. I ran 1 million iterations, and here are the results: Running the simulation 1000000 times where the host chooses doors randomly 500193 # of times the contestant's originally chosen door contained the prize 499807 # of times the remaining door contained the prize Running the simulation 1000000 times where the host intentionally chooses empty doors 334432 # of times the contestan'ts originally chosen door contained the prize 665568 # of times the remaining door contained the prize
  - joe_the_user 5y ago
    I'm not going to create an account on that site just for your thing but since you essentially haven't defined what "the host chooses randomly" means in this circumstance", I don't see how a print-out or whatever it gives would enlighten me. Edit: plus your claim is senseless on it's face. Why would what the host knew about the door influence the probability of the contestant's choice being correct? The contestant picks first and the host doesn't influence the contestant.
    - mikenew 5y ago
      Didn't realize that gitlab.com doesn't allow you to view things if you're not logged in. Here it is on github: https://github.com/michaelnew/monty_hall/blob/main/monty_hall.nim https://github.com/michaelnew/monty_hall/blob/main/monty_hal... By "the host chooses randomly", I mean the host picks from the remaining two doors without any knowledge of whether or not the contestant chose the prize door, or which of the two remaining doors contains a prize. The host flips a coin and if it's heads, they choose the remaining door to the left. Tails and they choose the one to the right. Which means that in 1 of 3 games the host will accidentally reveal the prize. Whether or not the host is acting with knowledge to filter out incorrect choices or they are just randomly revealing doors is what makes the difference between a 50/50 probability between the two remaining door, or a 1/3 vs 2/3 probability.

joe_the_user 5y ago

Think of it slightly differently: Imagine you pick a door, and then the host, who has no idea what's behind the doors, also picks one. Then the remaining door opens and reveals that there's nothing behind it. Yes, if the host were to pick a door, not tell you which it was or reveal anything else and then offer to switch your choice for their choice, there would be no difference in the odds of each choice. That just happens not to be the statement of the situation. The statement of situation is: you pick a door. The host picks a door, other than what you picked, then opens the door and reveals there's nothing there. This excludes the situation of the host picking the prize or the host picking the same door as you. How that exclusion happened doesn't matter. At that point, you get to choose your original pick or switch to the remaining door. At that point, what the host knew before irrelevant. You had a 1/3 chance of picking the door with the prize before regardless of the host, and now you can make a choice that gives every other possibility and so a 2/3 chance of getting the prize. Edit: I think that the problem is stated as "the host knows" because that means the host is guaranteed to open a door without the prize. It's not that the host's knowledge matters to the strategy, it's that host's knowledge makes it certain the situation will happen as described. See mtlogstdo's comment. https://news.ycombinator.com/item?id=29846498 https://news.ycombinator.com/item?id=29846498

tawan 5y ago
I disagree with your conclusion that the odds remain 2/3 even if the host doesn’t know where the prize is. If that were the case, it would mean that if the same scenario is repeated many many times, then in 2/3 of the cases you would still pick the door with the prize with the same strategy. But what about the instances where the host picks the door with prize before you even get a chance to pick the other door? These cases happen with a probability of 2/3 * 1/2 = 1/3. Therefore, you only get to choose in 2/3 of the cases and in 1/2 of those cases you have already picked the correct door the first time.
- 8note 5y ago
  Do you still have the opportunity to switch if he picks the winning door? Even if you could, you already know that you're losing, unless he doesn't show you what's behind it. If the game is still going, you know that he picked an empty door regardless of whether he knew he was doing it
mikenew 5y ago
Here, I wrote a simulation in nim: https://gitlab.com/mikenew/monty_hall/-/blob/main/monty_hall.nim https://gitlab.com/mikenew/monty_hall/-/blob/main/monty_hall... If you have nim installed you can run it yourself with `nim c -r monty_hall.nim`. I ran 1 million iterations, and here are the results: Running the simulation 1000000 times where the host chooses doors randomly 500193 # of times the contestant's originally chosen door contained the prize 499807 # of times the remaining door contained the prize Running the simulation 1000000 times where the host intentionally chooses empty doors 334432 # of times the contestan'ts originally chosen door contained the prize 665568 # of times the remaining door contained the prize
- joe_the_user 5y ago
  I'm not going to create an account on that site just for your thing but since you essentially haven't defined what "the host chooses randomly" means in this circumstance", I don't see how a print-out or whatever it gives would enlighten me. Edit: plus your claim is senseless on it's face. Why would what the host knew about the door influence the probability of the contestant's choice being correct? The contestant picks first and the host doesn't influence the contestant.
  - mikenew 5y ago
    Didn't realize that gitlab.com doesn't allow you to view things if you're not logged in. Here it is on github: https://github.com/michaelnew/monty_hall/blob/main/monty_hall.nim https://github.com/michaelnew/monty_hall/blob/main/monty_hal... By "the host chooses randomly", I mean the host picks from the remaining two doors without any knowledge of whether or not the contestant chose the prize door, or which of the two remaining doors contains a prize. The host flips a coin and if it's heads, they choose the remaining door to the left. Tails and they choose the one to the right. Which means that in 1 of 3 games the host will accidentally reveal the prize. Whether or not the host is acting with knowledge to filter out incorrect choices or they are just randomly revealing doors is what makes the difference between a 50/50 probability between the two remaining door, or a 1/3 vs 2/3 probability.