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1990s / The Monty Hall Problem
In a 1990 issue of Parade Magazine, Marilyn vos Savant examined a question that was first posed to mathematicians in a statisticians' publication back in 1975. It was based on a situation that frequently came up on the TV game show Let's Make a Deal--thus it was dubbed the Monty Hall Problem in honor of the show's longtime host. Here's the situation: You are a contestant on Let's Make a Deal. You have to choose one of three doors numbered 1, 2, and 3. Behind one of the doors is a fabulous prize (a new car). Behind the other two doors are worthless prizes (a goat). Let's suppose you choose Door #1. Monty--who knows which door conceals the car--then opens Door #3 to reveal a goat. He asks you, "Do you want to change your choice of door to Door #2 or do you want to stick with Door #1?" One would think that either door is a 50:50 proposition--but that would be wrong. According to mathematicians, you have to consider the initial odds. By choosing Door #1 you had a 1/3 chance of picking the car and a 2/3 chance of picking a goat. Doors #2 and #3 offer a combined 2/3 chance of concealing the car. The fact that Door #3 is revealed does not change the original odds. It seems counterintuitive, but the math experts say you are better to make the switch to Door #2 as it will conceal the car 2/3 of the time.