Monly Hall 3 doors problem

[Guest]

Hello,

I guess you all are familiar with the Monty Hall problem https://en.wikipedia.org/wiki/Monty_Hall_problem.

The problem goes like this:
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

I disagree with the typical conclusion that it would be better to switch doors at the second choice. I don't think it would be beneficial or harmful because the probability to win on the last two doors will always be 50 % when the third door is eliminated. Yes, you initially had a 1/3 chance to pick the right door, so your door is chosen with a 1/3 probability to win. But when one door is eliminated, your chances increase to 50 %. Would that change if you switched to the other remaining door? No.

I know there is a mathematical argument that explains why I am wrong, and you should actually switch to the renaming door.

But if it is true that it would be better to switch, then this must mean that in 1 000 of these games, you would consistently end up with > 500 wins in the 50% probability problem which occurs at the second choice. And that is practically impossible.

Am I wrong?

If you think I am wrong, are there any practical applications to this problem that would illustrate it better so I can understand?

[Guest]

The only reason why the 2/3 works is because the game has a rule that usually is not clear from the statement: the host must always reveal a goat door regardless of which content the contestant picked. He is not acting by random; he knows where each content is located and, moreover, he has the restriction that he cannot reveal the contestant's selection and neither the door that hides the car. He must always reveal a door that is not any of those two. That means that it is guaranteed that everytime the player fails to hit the car door at first, the other door the host leaves closed is which has the car. Now, the contestant hits a goat door 2 out of 3 times on average, not 1 out of 2, and since the car can never be revealed, in those same 2 out of 3 times the host is who is forced to avoid revealing it from the other two doors, in order that it can remain hidden, and so it will be in the switching door.

In this way, there will be always two doors remaining, but one was randomly left by the contestant while the other was purposely left by the host, who knew the positions. One person had advantage over the other. The contestant's selection will be the correct 1 out of 3 times, and the other one will be the correct 2 out of 3.

This can be seen better imagining there were more doors, like 1000, only one has a car. You must select one and then the host must reveal 998 doors having goats, and he cannot reveal your door. Suppose that in a particular game the correct door is number 150:

1) If you pick door 1, the host must reveal all but doors 1 and 150. You win by switching.
2) If you pick door 2, the host must reveal all but doors 2 and 150. You win by switching.
3) If you pick door 3, the host must reveal all but doors 3 and 150. You win by switching.
...
999) If you pick door 999, the host must reveal all but doors 999 and 150. You win by switching.
1000) If you pick door 1000, the host must reveal all but doors 1000 and 150. You win by switching.

The only way you could win by staying is if at first you selected door 150, so you must have been very lucky to hit the correct having a total of 1000 possibilities. If you made 1000 attemps, you only expect to have that luck in about only one time. Instead, you only need to have hit any of the 999 incorrect ones, and if that is the case you will automatically force the host to show you which is the correct one (the other he does not reveal).

[Guest]

Thank you. Good explanation. This makes sense.

[elenasimons]

[quote="Guest"]Thank you. Good explanation. This makes sense.
Thanks to