My wife, Amanda, came home from her course in "Discrete Mathematics" the other night eager to talk about a problem the professor had discussed in class. Turns out I had already posted on the very same problem, here. But in honor of her interest let's elaborate.
The problem concerns a television game show, popular in my youth, called "Let's Make a Deal." In one popular game, the host, Monty Hall, would show a contestant three doors. You win the prize behind the door you choose. There's always a goat behind two of the doors, but behind the third there'd be a sports car, something like that. After the contestant chose a door, Monty would direct that one of the unchosen doors be opened. Invariably a goat was revealed--Monty knew where the car was, and this was a way of building up the suspense. Here's the interesting part: Monty would also allow the contestant to switch doors. In other words, once it was known that a certain unchosen door held a goat, you could accept an offer to trade your first choice for the remaining unopened door.
Should you switch? Does it matter?
Despite what almost everyone thinks, it matters a lot. You double your chances of winning the sports car by accepting Monty's offer to switch. Let's see if I can persuade skeptics. Suppose the car is behind door 1 and you choose a door at random. If you choose door 1, Monty opens, say, door 2, revealing a goat. You then accept the offer to switch to door 3 and win--a goat. If however you choose door 2, Monty opens door 3, revealing a goat. You accept the offer to switch to door 1 and win--a car. In the third scenario, you choose door 3, Monty shows the goat behind door 2, you switch to door 1 and win--the car, again.
At the start of the game, your chance of winning is 1/3. By switching doors in the middle of the game, the probability of winning rises to 2/3--you lose if you began with the winning door (1/3 probability), but win if you began with a goat (2/3 probability).
If you don't believe me, you can model the game with playing cards and tally the result. Take, say, a joker (sports car) and the black twos (goats). Shuffle, then give "the house" two cards and yourself one (the first choice). Have "the house" discard a two--there will be at least one--and then trade your card for the house card. Repeat many times. See if you can't satisfy yourself that switching cards, or doors, is the best strategy.
Why do most people have a really hard time accepting the solution to this problem? I think in most cases it's because they feel intuitively that there is no reason to suppose the car is more likely to be behind one door than another--a supposition that is true at the start of the game only. Since Monty has determined to reveal only goats before moving to the last stage of the game, there is a nonrandom reason that the two remaining doors are this one (the one the contestant chose) and that one (the one Monty chose).
If this kind of stuff interests you, check out the Wikipedia article on "the Monty Hall problem." Turns out Amanda's math professor wrote it (at least in part)! Here's his website.
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