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

When you first pick, you only have a 1/3 chance of beeing right. So you know that 2/3 of the time, the car will be behind one of the other doors. By switching you basically get to choose both other doors at once, because the host gives you one for free by telling you it’s not the car.

You can also just simply list all possibilities:

1. You pick the door with the car, you switch and loose.

2. You pick the door with goat number 1, you switch and win.

3. You pick the door with goat number 2, you switch and win.

So in 2 out of 3 scenarios you win.

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