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Wrong.
Right. I don't watch Mythbusters, so I'd be interested in how they went about it. In any case, it's a simple proof.
Proof:
Since there's one car and three doors, the odd are 2/3 that you are going to choose an empty door. Regardless of whether you were initially right or wrong, the host always has an empty door they can show after you choose. Therefore, choosing to switch is equivalent to originally selecting the two doors you didn't choose together instead of the one. So you odds at winning the car double from 1/3 to 2/3.
This problem generalizes to the case of having x number of doors and being shown empty doors until only the one you chose and one more remain. Switching then increases your odds from 1/x to (x-1)/x.
That's one of my all-time favorites. totally unrelated to the discussion, but I think it'd be damned cool if you started a thread where you and others pose the best math/word problems that they know for others to answer. That'd be fun.