The Monty Hall Paradox

Over the last couple of days I have been having a debate with KiwiGirl over what is widely known as The Monty Hall Paradox. It started with a harmless question on her blog - and ended with a flood of emails back and forth. So here it is:

'You are on a game show on television. On this game show the idea is to win a car as a prize. The game show host shows you three doors. He says that there is a car behind one of the doors and there are goats behind the other two doors. He asks you to pick a door. You pick a door but the door is not opened. Then the game show host opens one of the doors you didn't pick to show a goat (because he knows what is behind the doors). Then he says that you have one final chance to change your mind before the doors are opened and you get a car or a goat. So he asks you if you want to change your mind and pick the other unopened door instead. What should you do?

So peoples - what do you think? This has sparked all sorts of debates in mathematical circles - and I don't understand why? It seems simple to me...

Easiet way to sum up my reasoning is from an email to KiwiGirl;

Nope - to me what happened with the first door becomes completely irrelevant after it is opened. You are left with 2 doors - one has a goat, one has a car. 50/50.

Along the same lines as if you toss a coin 4 times, before you start - you would expect 2 heads, 2 tails. But if you have thrown it twice and got heads both times, the odds don't shorten on tails for the third throw based on the outcome of the first 2 throws. It is still 50 - 50 for throw three (and will always be when you toss a coin).

Does that follow?

Hhhhmmmm.