Isn’t the Monty Hall Problem two separate games?

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Game One: You are given a choice of three doors. You pick number one. The host opens one of the other two doors, having been given instructions that, if you pick the car, the host is to open one of the other doors, and if you pick a goat, the host opens the other door with a goat. Stalemate. It is a predetermined outcome.

Game Two: The prior game’s outcome stands. The new choice you have is do you keep door number one, or do you switch?

How do you have a 2/3 chance of winning if you switch?

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34 Answers

Anonymous 0 Comments

It all comes down to the fact that the host knows which door has the car behind it and would never accidentally open that door.

When picking the first door, you have a 1/3 chance of picking correctly. Then the host eliminates a door that he knows has a goat behind it. The odds that you initially picked correctly don’t change from 1/3 as the host reveals a door with a goat.

When given the option to switch, you still only have a 1/3 chance of being right initially, meaning that the other door has a 2/3 chance of having the car.

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