The Monty Hall problem goes back to Hall's show "Let's Make a Deal." Hall would offer a contestant three doors. Behind one, a wonderful prize, behind the others, goats. The contestant would choose a door, and sometimes Hall would open another door and offer the contestant a chance to switch.
Assuming that Hall doesn't choose to open a door for any particular reason, what should you do? The intuitive answer is that it doesn't matter. Switch or don't switch, it's 50:50.
But this is wrong. You should switch. Your odds of winning go up to 2/3 if you switch.
Jason explains it well enough, but it's so much fun, I have to join in.
First, a simple point. Some people object to the asymmetry in principle, arguing that probability doesn't move around. Opening the door doesn't really give you any information. False! Probability does switch around as you open doors. What if Hall opens the door with the good behind it? Your probability of winning if you switch or if you don't goes to zero.
The solution to the problem is to break down the conditional probabilities. Conditional probability works like this. In baseball, most batters can swing as righties or as lefties, but some can hit from either side. Such a player has a batting average of P(H), the probability of a hit. For a regular player, that's where the story ends. A switch hitter has two batting averages, P(H|L) and P(H|R), the probability of a hit given (that's what the vertical bar means) that he bats left or right. His overall average is based on the probability he bats lefty – P(L), and the probability he bats righty – P(R). The equation is P(H) = P(H|L)P(L)+P(H|R)P(R)
To solve Monty Hall's problem, we do the same.
When we first picked the door, there was 2/3 chance we'd pick the wrong door – P(wrong first choice) = 2/3. After we chose, Hall had two doors he could open. If we picked wrong, he can only open one door. He would have eliminated the other wrong door, and if we switch, we're guaranteed to switch to the right door – P(win after switch|wrong first choice)=1.
The 1/3 of the time when we picked right at first, any switch will make us lose – P(win after switch|right first choice) = 0.
The total probability that we'd win after a switch is therefore 1/3*0+2/3*1, or 2/3. It all hinges on the initial probability of picking the right door. If Monty only offers you the choice when you picked right, you should never switch, and if he only offers it when you're wrong, you should always switch. There's a large literature that developed around this problem given different assumptions about how Hall behaved and how he decided whether to open a door.
In later years, he denied understanding the formal probability behind the phenomenon, but agreed from experience that switching doors was the smart choice.