Problem

[Dilshod]

This is sort of advanced math problem that I can't understand.

There are 3 doors. Behind 1 of them there's a prize. So you select one of them, but don't open. Then 1 of the remaining 2 doors is opened and the prize is not there. Question is: would you stay with your first choice or choose the second door to increase your chances to get the prize?

For me, it's logical to stay with the first choice, cause you still have 1/2 probability of getting the right door, and changing the door will not increase your chances.

But the guy who asked me this question says that theoretically it would be wiser to change your 1st choice and choose the 2nd door. Explanation: the probability that you chose the wrong door before one of the doors was opened is 2/3, which means that you more than likely than not chose a wrong door. So after one of the doors is opened, and you more likely than not chose a wrong door before, it's logical to change your selection. ???

This principle has to do with some concept in Algebra or statistics, and at some point theoretically it seems to be right, but I can't live with the idea that if there are 2 doors, I somehow get less than 1/2 probability of holding the right door.

What do you think?

[Peshehod]

let's put it this way: the probability that you chose the right door before one of the doors was opened is 1/3, which means that you less likely than not chose a right door.After one of the doors is opened, you less likely than not chose a right door before...If you combine this explanation with the one that your friend gave you, you are gonna sill end up with 50/50 probability that the door you chose is the right one. I wouldn't go so far and use algebraic concepts to make sense out of it...

[Komil]

you were right ,
that you picked up the door doesn't play anything
in this problem.
Finding the probability of that there is a prize back of one of two remaining doors is the problem.
probability is 1/2.
at least theoretically it is right.