fantastic contraption

http://FantasticContraption.com/?designId=3688163

"Let's Make a Deal" Riddle

I'm pretty sure most everyone is wrong about the "Let's Make a Deal" riddle. Have you heard this one? Check it out if you haven't, and just think on it for a while. I maintain that there is no advantage to switching doors after the host opens one of the doors. It's abuse of probability mathematics. In reality, your choice before the host opens a door has no consequence. Your choice after the host opens a door is the only one that matters, and that choice is 50/50. Please enlighten me if I am being a complete fool.

UPDATE: I like this guy's explanation, if you insist on cohering the first and second choices. Since the host will always pick a goat, it's as though there were only 2 doors available in the first place: 1 with a goat, and 1 with a car. The 3rd choice is only an illusion. Let's try this with 4 doors, labeled @, #, *, and Q. 25%, 25%, 25%, and 25%. Then, Monty opens @ and *, revealing a cardboard cut-out of Michael Richards and a poorly-drawn portrait of Snoopy. Now, take into account that the desired prize is behind door #. Your chances of picking Q were no greater than your chances of picking # in the beginning. It was 25% and 25%. Therefore, at this stage of the game, you are no more likely to have chosen the losing door than the prize door (50/50), and your next choice is 50/50 as well. Lots of good discussion here, along with plenty of bogus conclusions.

UPDATE 2 (the last): Ah, here are some people with empirical evidence proving me wrong, and an interesting statement from someone as stubborn as I am. My head is still spinning....