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
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.
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.
Tails and they choose the one to the right. Which means that in 1 of 3 games the host will accidentally reveal the prize.
And what happens then? I mean, I think people have said from the start, that this behavior is outside the specification of the problem - which is that the host opens a door and reveals nothing.
The situation is about only the situation where the host choose the door with nothing. The host "opens one door without a prize". The key detail is this, "opens one door without a prize", not "knows where the prize is". If the host doesn't know where the prize but still, by chance, "opens one door without a prize", then, in this situation the contestant's information remains the same and the odds remain the same.
I don't know what to tell you at this point. From the wikipedia article:
> Most people come to the conclusion that switching does not matter because there are two unopened doors and one car and that it is a 50/50 choice. This would be true if the host opens a door randomly, but that is not the case; the door opened depends on the player's initial choice, so the assumption of independence does not hold.
Random means there's a possibility the host opens the door containing the prize. The Monty Hall problem specifies that that is not a possible scenario, and that the host does not choose randomly.
> If the host doesn't know where the prize but still, by chance, "opens one door without a prize", then, in this situation the contestant's information remains the same and the odds remain the same.
The host cannon open a door by chance if there's no chance of opening the door with a prize behind it.
The host doesn't know exactly where the prize is, but he knows that the door he's opening is a dud; that's the information that he has and the audience doesn't.
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.
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