# The Monty Hall problem

I was watching the movie $$2121$$ 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 $$1/31/3$$ 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 $$1/31/3$$ chance of winning if door A is picked and $$2/32/3$$ 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.

At the start of the game, you are asked to pick a single door. There is a $1/3$ chance that you have picked correctly, and a $2/3$ chance that you are wrong. This does not change when one of the two doors you did not pick is opened. The second time is that you are choosing between whether your first guess was right (which has probability $1/3$) or wrong (probability $2/3$). Clearly it is more likely that your first guess was wrong, so you switch doors.
To clarify this, say you choose door $A$, and are then asked to choose between doors $A$ and $B$ (no doors have been opened yet). There is no advantage to switching in this situation. Say you are asked to choose between $A$ and $C$; again, there is no advantage in switching. However, what if you are asked to choose between a) the prize behind door $A$ and b) the better of the two prizes behind door $B$ and $C$. Clearly, in this case it is in your advantage to switch. But this is exactly the same problem as the one you’ve been confronted with! Why? Precisely because the host always opens (hence gets rid of) the door that you did not pick which has the worse prize behind it. This is what I mean when I say that the asymmetry in the situation comes from the knowledge of the host.