The Monty Hall Problem is a probability puzzle based on a game that contestants played on the popular old TV game show Let's Make a Deal, hosted by Monty Hall. On the show, contestants were given a choice of three doors. Behind one door is a car, behind each of the other two, goats. The contestant got to choose a door, and won whatever was revealed to be behind that door. The twist was that after the contestant had selected a door, Monty would show them what was behind one of the two doors not selected, and give them the option of changing their mind.
The game was deceptively simple, in that it seemed as though the contestant always had one chance in three of winning the car, whether they switched their choice or not. There was a strategy that would double the odds of winning though. It turns out that the winning strategy is as simple as the game.
All a contestant had to do to double their odds of winning was always switch their guess after Monty opened one of the doors. This simple strategy worked because the contestant had only one chance in three of guessing the right door initially. Once they had made their guess, Monty would open one of the other doors that did not conceal the car. Since the player's initial guess was wrong 2/3 of the time, Monty was opening the only other losing door 2/3 of the time (the car was behind one door, so he wouldn't open that).
This strategy isn't intuitive to a lot of people. In fact, when the strategy was first published in Parade magazine, thousands of people wrote in to claim the solution was wrong. It becomes more clear if you look at a bigger variation of the same problem.
Imagine you're on a game show where there are 100 doors. Behind one of them is a new car, while the other 99 conceal goats. Your odds of selecting the right door are 1 in 100. Now imagine that after you've made your selection, the host of the show opens 98 of the doors, revealing 98 of the goats. (Keep in mind that he doesn't randomly open doors. He knows where all the goats are.) Would you switch your original guess to the one remaining door? You only had a 1% chance of winning with your original guess, so does it seem advantageous to switch after the choices are narrowed down to only two?
With a bigger set of doors to choose from, it becomes much easier to see that you have a distinct advantage when you switch from your initial guess. The same holds true for the original problem. By always switching after one door was opened in the original game, the odds of winning were improved from 1/3 to 2/3. That's certainly not a sure thing, but it's not bad odds on a brand new car.
Additional Note: If you want to try out an online interactive version of the original game, you can play it here. Make sure you give yourself enough trials to confirm that the optimal strategy wins 2/3 of the time.