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[ChairmanPoo]

For the benefit of Kagus:
I think the confusion lies in the fact that it's not your door choice what determines the odds, but the other guy's:
The problem lies in that you're not picking a door the first time. Rather, you are giving a choice of two doors to the door-man
Since there are two doors without a price, there's a 2/3 chance that you'll pick one of those. If you do, then in effect you have left the door man with the price door, and one door without price.

Now, the door-man ALWAYS must open an unpriced door, after your initial pick. So if he has the price in one of his two doors (and there's a 2/3 chance of this, as explained above), the one he doesn't open will have the price.

Thus, in effect you can choose between your door (which has 1/3 chance of being the correct one) vs the other door (which has 2/3 chance of being the correct one). The key is that the man has to discard a fake door ALWAYS, and odds are in favor that he is left with the price and a fake door, rather than two fake doors.

[Cheeetar]

Kagus:


(Wikipedia's image)

[Frajic]

Easier way of explaining it: there's a 2/3 chance you'll hit a goat. If you hit a goat, the host will open the door that shows the other goat. Therefore, the last door must be the car.
There is a 2/3 chance you'll win by switching.