[elitegimp]

This is a simple counting proof -- here are all the possible outcomes of the game, all equally likely (important for figuring out the probability)

Call the door you choose A:
If the goat is in A, then it doesn't matter which door is chosen. You want to stay (if door B is opened, you win my staying... if door C is opened, you win by staying)
If the goat is in B, there is a 1/2 chance you lose automatically (door B opened), or a 1/2 chance you win if you switch (door C is opened)
If the goat is in C, there is a 1/2 chance you lose automatically (door C opened), or a 1/2 chance you win if you switch (door B is opened)

So you win by staying 2/6 times, you win by switching 2/6 times, and you lose because the door with the prize is opened 2/6 times.

Not surprising, because no information is gained by having a door opened so you still have a 1/3 chance of picking the door.

Note: if the door is opened and you don't lose automatically, there is now a 2/4 chance you win by staying and a 2/4 chance you win by switching... i.e. it's 50-50.