No. Door#1 earlier had 1/3, now that it is open, it is out of equation. Only 2 doors are in race now and we have to re-assign prob. of 1/2, 1/2 for #2,#3. Both the cases you mentioned are equally likely (hence 1/2, 1/2) to happen.

Few may think that changing to #3 is better option but there is no place for finding prior prob. in this case (as #1 is removed out of equation instead of just having its prob changed).

You can read my older comment on why i still wish to stick with #2, even though it has only 1/2 of winning ..

You cannot reassign the probabilities like that.

Let me tell you in the way that makes the most sense to me.

Let's say we have two people playing this game, Switcher and Stayer.

Switcher has firmly decided to always switch doors, while Stayer has decided to never switch doors.

Let's start with Stayer.

At the outset of the game Stayer has a 1/3 probability of picking the door with the car, and 2/3 probability of picking a door with a goat.

If he picks right, which happens about 1/3 of all the times he's taken part in this silly competition, then when Monty shows the other door, Stayer will win, since he initially picked right and stayed with it.

If Stayer picks wrong initially, which happens 2/3 of the time, he will lose after Monty opens a door, because he was initially wrong and stuck with it.

We can therefore say that Stayer will win 1/3 of the time and lose 2/3 of the time.

Now let's look at Switcher.

Just like Stayer, at the outset of the game, Switcher has a 1/3 probability of picking the door with the car, and 2/3 probability of picking a door with a goat.

If he picks right, which happens about 1/3 of the time, then after Monty opens a goat door, switcher will switch doors to the other goat door and lose, since he switched from the correct door to the particular goat door that Monty didn't open.

If Switcher picks wrong initially, which happens 2/3 of the time then he will be pointing at one of the goat doors. Monty, knowing where everything is, will open the other goat door, and Switcher will switch doors to the car door and win. (Remember this happens 2/3 of the time)

We can therefore say that Switcher will win 2/3 of the time and lose 1/3 of the time.

Let me recap, based on the strategies of the two players and a logical analysis of all possibilities,

Stayer will win with probability 1/3 and lose with probability 2/3

Switcher will win with probability 2/3 and lose with probability 1/3.

Who would you rather be?

Think of it this way, Switcher is betting his initial guess is wrong, while Stayer is betting his initial guess is right. Wrong and right have probabilities 2/3 and 1/3 each, so which would you bet happened?

You bet you were wrong! Sometimes that's the best way to be right.

I think looking at the problem like this, causes it to agree with intuition rather than disagreeing with it.

**Edited by mmiguel1, 19 October 2010 - 04:12 AM.**