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Monty Hall problem





A puzzle in probability that was inspired by the American game show "Let's Make a Deal," hosted by Monty Hall. In its original form it goes like this: at the end of the show, you, the player, are shown three doors. Behind one of them is a new car, behind the other two are goats. Monty knows where the car is, but you don't. You choose a door. Before that door is opened however, Monty opens one of the two other doors with a goat behind it. He then gives you the option of switching to the other closed door. Should you switch or stick? At first glance, it seems as if it shouldn't make any difference. But the answer is surprising.

Suppose you stick. Your original choice made when all three doors were equally likely gives you a probability of winning the car of 1/3. Now suppose you switch. In other words, you choose a door, wait for Monty to expose a goat, then switch to the other remaining door. This means that you win if the door you chose to begin with had a goat behind it. The odds that your initial choice had a goat is two thirds, so you are twice as likely to win the car if you switch! This can be hard to grasp. To make it easier, suppose there are 100 doors to choose from, but still only one car. You pick a door, Monty opens 98 that have goats behind them, then he gives you the option of switch to the other remaining closed door. Should you? Of course – it's almost certain that the car is behind the other door, and very unlikely that it's behind your original choice. In a generalization of the original problem there are n doors. In the first step, you choose a door. Monty then opens some other door that's a loser. If you want, you may then switch your choice to another door. Monty will then open an as yet unopened losing door, different from your current preference. Then you may switch again, and so on. This carries on until there are only two unopened doors left: your current choice and another one. How many times should you switch, and when, if at all? The answer is: stick all the way through with your first choice but then switch at the very end.

In another variation of the problem, consider that in the actual game show there were two contestants. Both of them were allowed to pick a door but not the same one. Monty then eliminated a player with a goat behind their door (if both players had a goat, one was eliminated randomly, without letting the players know about it), opened the door and then offered the remaining player a chance to switch. Should the remaining player switch? The answer is no. The reason: a switcher in this game will lose if and only if either of two initial choices of the two contestants was correct. How likely is that? Two-thirds. A sticker will win in those 2/3 of the cases. So stickers will win twice as often as switchers.


Related category

   • PROBABILITY AND STATISTICS