The two key facts that keep this from being a 50-50 are:

1. The host is constrained to always open a door revealing a goat

2. The host always makes the offer to switch.

That leaves you with exactly 3 paths to 2 different outcomes. Each path has an equal probability, but since 2 paths lead to the same result, one result happens exactly 2/3 of the time while the other happens 1/3 of the time:

1. choose goat A, monty hall opens door to goat B --> SWAP, win car
2. choose goat B, monty hall opens door to goat A --> SWAP, win car
3. choose car, monty hall opens door to goat (A or B) --> SWAP, lose car

If it helps keep your head from hurting too much, the second choice would be a 50-50 chance of winning if you switch, but the host is CHEATING by knowing where the car is, so you have a 2/3 chance of winning if you switch in the second choice.

Show-off...


In reply to your comment:
“1. choose goat A, monty hall opens door to goat B --> SWAP, win car
2. choose goat B, monty hall opens door to goat A --> SWAP, win car
3. choose car, monty hall opens door to goat (A or B) --> SWAP, lose car

that's 2 out of 3...”

wow...

here's a well-written book explaining the development of statistics and probability thinking:
http://www.amazon.com/Chances-Are-Adventures-Probability/dp/0670034878

for the monty hall problem, just remember there are 2 different ways you can choose a goat door initially (where you'll win if you swap and lose if you don't), and only 1 way you choose a car (where you'll lose if you swap and win if you don't).

1. choose goat A, monty hall opens door to goat B --> SWAP, win car
2. choose goat B, monty hall opens door to goat A --> SWAP, win car
3. choose car, monty hall opens door to goat (A or B) --> SWAP, lose car

that's 2 out of 3...

They'd love you on Deal or No Deal, Payback. ;-) Seriously though, with 100 doors, Monty knows which of the hundred doors has the car behind it. The chance you picked the correct door at the start is 1%. From Monty's perspective, the chance you picked the wrong door is 99%. He knows where the car is. He is not opening doors at random. He is eliminating 98 wrong doors. Now, the door you picked at the start had a 1% chance of being the car. Then, after eliminating 98 wrong doors, there is a 99% chance his remaining door has the car. If, from his perspective, the last door he opens will have the car behind it 99% of the time, why wouldn't you switch doors? With 100 doors, 99% of the time Monty will have a car behind his last door. With 3 doors, 66% of the time Monty will have the car behind his unopened door.

Or, something like that.

Bedtime...

Oh wait! Think of it this way... If you play the 3-door game, Monty has a 66% chance of having the car behind one of his two doors. You would trade your one door for his two doors if given the chance. Him eliminating a known bad one of his two doors doesn't change the odds that the chance of him ultimately having the car is 66%. You are still effectively trading your one door for his two, which is what Deano and rembar and others pointed out above, so if you read this far... sorry for the repetitive redundancy.

Ok... I just know I'm gonna be dreaming of goats. What's new, huh?

Hehehe, Payback, that´s true. Many films even include the death scream of doom or the opening-door-sound. Doom is everywhere ;-)