# Game Show Probability

Hello,

I need help getting my head around this practice question:

In your favorite noontime show, the player reaches the final round and faces the challenge of choosing one out of three boxes to win a \$2M prize. Should she choose the wrong box, she would receive nothing. The player selects the First Box. The host opens the Second Box, which is revealed to have nothing in it. The player is theportunity to switch her box from the First one to the Third. What should the player do?

A) She should agree to the hostâ€™s offer
B) She should not agree to the hosts offer
C) Doesnâ€™t matter what she does

I answered C. The right answer is A and the solution states â€śThe player is twice more likely to win the \$2M should she switch to box 3.â€ť Iâ€™m floored. I appreciate the help!

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This is a classic Monty Hall problem as noted above

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By all means, look up Monty Hall problem; itâ€™s a good education.

Perhaps a starker problem will drive home the point more clearly. Suppose that there were one million boxes from which to choose. You choose one, and the host opens 999,998 others to reveal 999,998 nothings. Do you truly think that the probability of the \$2 million being in the box you chose is 50%?

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Thanks everyone. I was thinking of this in the context of Deal or No Deal, where the contestant is choosing the curtain the host would reveal instead of the game show knowingly selecting the box without the prize, so then itâ€™s basically the chance of you originally selecting correctly or not, which is 1/3 to 2/3. Wonâ€™t forget now!