The Monty Hall problem I was watching the movie $21$ yesterday, and in the first 15 minutes or so the main character is in a classroom, being asked a "trick" question (in the sense that the teacher believes that he'll get the wrong answer) which revolves around theoretical probability.
The question goes a little something like this (I'm paraphrasing, but the numbers are all exact):
You're on a game show, and you're given three doors. Behind one of the doors is a brand new car, behind the other two are donkeys. With each door you have a $1/3$ chance of winning. Which door would you pick?
The character picks A, as the odds are all equally in his favor.
The teacher then opens door C, revealing a donkey to be behind there, and asks him if he would like to change his choice. At this point he also explains that most people change their choices out of fear; paranoia; emotion and such.
The character does change his answer to B, but because (according to the movie), the odds are now in favor of door B with a $1/3$ chance of winning if door A is picked and $2/3$ if door B is picked.
What I don't understand is how removing the final door increases the odds of winning if door B is picked only. Surely the split should be 50/50 now, as removal of the final door tells you nothing about the first two?
I assume that I'm wrong; as I'd really like to think that they wouldn't make a movie that's so mathematically incorrect, but I just can't seem to understand why this is the case.
So, if anyone could tell me whether I'm right; or if not explain why, I would be extremely grateful.

Suppose you and your brother play simultaneously .You and he always choose the SAME door to start. You NEVER switch,so after your initial choice you go out for a coffee and come back after all 3 doors are open. Your chance of winning is unaffected by what happens in between.It's 1/3. Your brother ALWAYS switches. EVERY TIME YOU LOSE,HE WINS. YOU LOSE 2/3 OF THE TIME.
