**The Monty Hall problem is a probability puzzle, based loosely on the American television game show Let’s Make a Deal and named after its original host, Monty Hall.**

In the problem, you are on a game show, being asked to choose between three doors. Behind each door, there is either a car or a goat. You select a door. The host, Monty Hall, picks one of the other doors, which he knows has a goat behind it, and opens it, showing you the goat. (You know, by the rules of the game, that Monty will always reveal a goat.) Monty then asks whether you would like to switch your choice of door to the other remaining door. Assuming you prefer having a car more than having a goat, **do you choose to switch or not to switch?** The solution is that switching will let you win twice as often as sticking with the original choice, a result that seems counterintuitive to many. But why is this?

A player who stays with the initial choice wins in only one out of three of these equally likely possibilities, while a player who switches wins in two out of three.

An intuitive explanation is that, if the contestant initially picks a goat (2 of 3 doors), the contestant will win the car by switching because the other goat can no longer be picked, whereas if the contestant initially picks the car (1 of 3 doors), the contestant will not win the car by switching. The fact that the host subsequently reveals a goat in one of the unchosen doors changes nothing about the initial probability.

Most people come to the conclusion that switching does not matter because there are two unopened doors and one car and that it is a 50/50 choice. This would be true if the host opens a door randomly, but that is not the case; the door opened depends on the player’s initial choice, so the assumption of independence does not hold. Before the host opens a door there is a 1/3 probability the car is behind each door. If the car is behind door 1 the host can open either door 2 or door 3, so the probability the car is behind door 1 and the host opens door 3 is 1/3* 1/2 * 1/6. If the car is behind door 2 (and the player has picked door 1) the host must open door 3, so the probability the car is behind door 2 andthe host opens door 3 is . These are the only cases where the host opens door 3, so if the player has picked door 1 and the host opens door 3 the car is twice as likely to be behind door 2. The key is that if the car is behind door 2 the host must open door 3, but if the car is behind door 1 the host can open either door.

So, the next time you find yourself on a gameshow where the host offers you the chance to switch doors, take the opportunity and you might be thankful that you did!

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