The Monty Hall Problem.

The problem, simplified, states the following:
Three choices are presented, one choice wins and the other two lose. If, after a choice is made, a remaining, non-winning choice is revealed (leaving two choices), changing the choice will increase your chance of winning. Intuitively, it doesn’t seem like it would matter

The problem colorfully explained:

Despite the overwhelming editorial explaining that this is the case, I doubted it! (Partcularly since the existing simulation stats are f*&%ed up from bad programming, how can I have confidence that the simulators or authors aren't also skewed) ...So I wrote a little simulator to prove (or disprove) it.

Sure 'nough, changing the answer wins far more frequently than staying with the first. Then, after proving it to be true, I sitting in Atlanta traffic, listening to String Cheese Incident, and the math came to me. It's really simple, actually. Then again, I could be wrong. On the first selection, you have a 33% chance of winning. When a choice is removed, your chance for losing is narrowed. Your chance for winning conversely increases. By staying with the origianl choice, you are staying with your original chances for losing. By switching after remaining losing choices have reduced, you take advantage of the change in odds. It is more clear if you consider 10 doors. If you choose 1, then 8 are removed, you know one of the two is the right door. Your chances on that first contest are basically 50% (you can esentially disregard the first choice of 10) but upon repeating the game, your chances increase dramatically by switching to the other door when only two remain.

here is a good explanation as well