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If the host is choosing a door randomly and it doesn't happen to contain the prize, your odds don't improve if you switch your answer. I'm not sure what you me
by joe_the_user 5y ago
If the host is choosing a door randomly and it doesn't happen to contain the prize, your odds don't improve if you switch your answer.
I'm not sure what you mean here. If you pick 1 door in 3, the chance you've picked the door with the prize is 1/3 and so the chance the prize is behind another of the 2 remaining door is 2/3. If one of the doors you didn't pick is opened and reveals no prize and the situation is as-described, the chance of the prize being behind one of the non-picked doors remains the same, 2/3. And since you know now which of these door doesn't have the prize, the chance of the final unpicked door having the prize is now 2/3.
The host has to pick a door with no prize for the described situation to happen but if the situation happens, your knowledge doesn't depend on the host's knowledge.
Maybe there's some world where the host picks at random and so 1/3 of the time, the door with the prize open and you then know with 100% certainty where the prize (though whether anyone gets a prize is ambiguous, to say the least). But that revealed-prize situation isn't what the problem describes.
Edit: The only way that host knowledge matters is if the host can choose whether or not to offer this particular deal based on whether the contestant picked the correct door originally. If we assume some optimal game theoretic counter-play on the part of host, then I would guess the host's action give no information and so there's no reason to switch. But that's a bit far from your comment.
- mikenew 5y agoWhether 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 agoThink 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 agoI 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 agoDo 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 agoHere, 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 agoI'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.