The car is behind either the originally chosen door or the door that remains. The chance that the car is behind the originally chosen door is 1/3. The solution to the problem is quite straightforward under this assumption, though the straightforward solution doesn’t convince everyone. It’s reasonable to assume that there is a 1/3 chance that the contestant’s initial choice is the door that has the car behind it. In any problem involving chances, the assumptions about randomness are important. Which door should she choose to open, if she wants the car? Should she stick with her initial choice, or switch to the other door? That is the Monty Hall problem. The contestant now gets to choose which of the two doors to open. One of the doors has the car behind it, and the other one has a goat. There are two doors left, one of which was the contestant’s original choice. Monty opens one of these doors to reveal a goat, displayed in all its glory in Wikipedia: The contestant makes an initial choice, but that door isn’t opened.Īt least one of the other two doors must have a goat behind it. The contestant doesn’t know where the car is, and has to attempt to find it under the following rules. Behind one of the doors is a fancy car, and behind each of the other two there is a goat. The setting is a game show in which the contestant is faced with three closed doors. This is the basis for what is now known as the Monty Hall problem. An exciting part of the show was that while the contestants had the chance to win great prizes, they might instead end up with “zonks” that were less desirable. Monty Hall hosted this show in the 1960’s, and it has since led to a number of spin-offs. The setting is derived from a television game show called “Let’s Make a Deal”. Let’s see if we can work it out by simulation. This problem has flummoxed many people over the years, mathematicians included.
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