TITLE: The Monty Hall problem
QUESTION [80 upvotes]: 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.

REPLY [22 votes]: Here is an equivalent formulation of the Monty Hall problem, which folks on twitter seem to like. 
If you solve this problem, you have solved the original. To see the connection, you can imagine boxers hidden behind the doors.

There are three boxers. Two of the boxers are evenly matched (i.e. 50-50, no
  draws!); the other boxer will beat either them, always.
You blindly guess that Boxer A is the best and let the other two fight.
Boxer B beats Boxer C.
Do you want to stick with Boxer A in a match-up with Boxer B, or do you want to switch?

Intuitively witnessing Boxer B beating Boxer C gives you information about Boxer B. The advantage of switching no longer feels "paradoxical."
Details: If Boxer B is best, then the probability of witnessing Boxer B beat Boxer C is equal to 1. By contrast, if the initially chosen Boxer A is best, then this probability is equal to 1/2 (B&C are evenly matched with no draws). This means that the event you witnessed is twice as likely if Boxer B were best than if Boxer A were best (i.e. 1 vs. 1/2).  This is the information you have received. Bayes rule now tells us that the initial 1-to-1 odds in favor of Boxer B in a match-up against Boxer A must double, i.e. there are now 2-to-1 odds in favor of Boxer B being the best boxer. This means there are now 2 chances in favor of Boxer B for every 1 chance in favor of Boxer A, i.e. 2 chances out of 3 total chances. This is equivalent to saying that the probability is now 2/3 that Boxer B is the best boxer.  You should switch.
Notes:
This formulation has the advantage of providing clear intuition for the solution to the problem as presented, which is equivalent to the famous version originally presented by Marilyn vos Savant.
By contrast, the current top response since 2012 on math.stackexchange correctly solves a different problem, in particular the prior probability of winning if you commit to switching regardless of which door Monty opens.  What that response is missing is an argument connecting the posterior (conditional) probability to the prior. For a justification, see my comment there, or you can read this paper, page 151--153.
Update: I googled and it appears that the first to point out a framing like this were Burns & Wieth, Journal of Experimental Psychology--General 2004.