>or consider this approach: when you nominate a door you
>split the objects into His and Mine. He has two and you have
>one, so he probably has the car ('probably' here meaning
>'more likely than not'). Now he shows you a goat - you
>always knew that he had at least one, so you have learned
>nothing and therefore still think he probably has the car.
>The only difference now is that you know WHERE the car is if
>he has it. So you argue: he probably has the car, and he
>certainly would put it THERE..... so it probably is THERE.
>Clearly you must change your choice.

That's assuming that you KNOW you chose incorrectly from the beginning.

The initial pick, and the choice to stay or switch, are independent events. The first "pick" is a throwaway, unless sometimes he does not offer you the ability to switch. Since that's not the rules of the game, we don't have to worry about that. Think through it like this.

Regular "Let's Make a Deal" (LMAD): three doors, one car, two goats. You make a pick, then Monty throws out a bad door showing you it's a loser. You then determine if you'll switch or stay.

Now let's test out two more versions of the game--
LMAD version 2.0: three doors, one car, two goats. Monty throws out a bad door, and then you select what door you want. You can switch or stay, shouldn't matter either way.

LMAD version 2.1: only two doors, one car, one goat. Pick a door.

Those of you that argue that switching wins 2/3 of the time will still say that regular LMAD has a 66% win rate when switching. You MUST agree that LMAD version 2.0 has a 50% win rate. Same goes for version 2.1-- 50% win rate.

Guess what: all three are the same game. You're just being confused with the extra information in the original game. The game becomes 50/50 when you only have two choices. Don't mix that up with the 1/3 probability of initial success.

Let's analyze this one: LMAD version 2.2. One car, two goats. You are ASSIGNED a door (let's say door A) and told that there is a goat behind it. You can now keep the goat, or switch to one of the other doors. Well, anyone with an IQ over 85 will switch, but now the dilemma is door B or door C. Those are 50% odds, my friend, and this is the same game as the original.

Here's the probability if it helps:
Start of game
Given: doors A, B, and C; prize is behind one door
P(A) = 1/3
P(B) = 1/3
P(C) = 1/3
P(A) + P(B) + P(C) = 1

Assume one door is always the loser, which it is, we just don't know which one. For sake of example, we'll say it's C. Nobody ever chooses door C in our game, and Monty always opens door C. It's a waste of time to even put three doors out there (much as is a person making an initial pick, if you'll see it). Once we get to phase two of our game, here are the new probabilities, and the only ones that matter.
P(A) = 1/2
P(B) = 1/2
P(C) = 0
P(A) + P(B) + P(C) = 1

The fallacy that the 66% crowd believes is that there is not an equal chance of the car being behind doors A and B. They try to draw it up like this...
initial choice is A, P(A) = 1/3
*** THAT'S RIGHT, if there's not a second phase to the game ***
but they leave P(A) at 1/3, and since P(C) is 0, then P(B) must be 2/3, and that's why it pays to switch. In reality, we are guaranteed that P(A) = 1/2 and P(B) = 1/2, which means this is a 50/50 game.