You Pick Door 1, Car in door 1, you stick -> Win

You Pick Door 1, Car in door 1, you switch -> Lose

You Pick Door 1, Car in door 2, you stick -> Lose

You Pick Door 1, Car in door 2, you switch -> Win

You Pick Door 1, Car in door 3, you stick -> Lose

You Pick Door 1, Car in door 3, you switch -> Win

You Pick Door 2, Car in door 2, you stick -> Win

You Pick Door 2, Car in door 2, you switch -> Lose

You Pick Door 2, Car in door 1, you stick -> Lose

You Pick Door 2, Car in door 1, you switch -> Win

You Pick Door 2, Car in door 3, you stick -> Lose

You Pick Door 2, Car in door 3, you switch -> Win

You Pick Door 3, Car in door 3, you stick -> Win

You Pick Door 3, Car in door 3, you switch -> Lose

You Pick Door 3, Car in door 1, you stick -> Lose

You Pick Door 3, Car in door 1, you switch -> Win

You Pick Door 3, Car in door 2, you stick -> Lose

You Pick Door 3, Car in door 2, you switch -> Win

Using the method that was proposed a few posts ago, but getting every scenerio properly weighted, we see 18 possible scenerios. In 12 of them switch wins the car and 6 of them switch loses the car. In 12 of them stick loses the car and 6 stick wins the car.

Switching changes the odds of winning from 1/3 to 2/3. For some of you, I know accepting this hurts your brain but believe it's the right answer. Google Monte Hall Problem all you want. Any thorough treatment of the problem will reach this conclusion.

If you still don't believe me, run it out with the Monte Carlo method (the mathematical term for computing propbability by actually running samples). You could write a program (or pull the Perl one off Wikipedia) if you understand programming. If not, find a partner and run through a bunch of samples using a deck of cards.

Trust me on this one. Switch is 2/3. Trust me.

-L2T