Trade one door for two

Question

Three doors hide one prize. You pick door 1. Before opening anything, the host makes an offer: you may keep door 1, or trade it for both of the other two doors, winning if the prize is behind either one.

What is your probability of winning if you take the trade?

Accepted Answer

No reveal is involved, so this is a clean counting problem. The prize is equally likely behind each door. - Keeping door 1 wins with probability 	frac{1}{3}. - Trading for doors 2 and 3 wins if the prize is behind either, which is probability 	frac{2}{3}. P(	ext{win by trading}) = \frac{2}{3}. This is exactly the engine behind the usual Monty Hall answer. In the standard game the host opens one empty door among the two others, which does not change their combined 	frac{2}{3} chance, it just tells you which of the two to point at. Switching is really "take both other doors," with the empty…

Trade one door for two

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