Seems like you are assuming he can't open the car's door.

Driver side, or passenger side?

Sorry, maybe my shorthand is unclear. The car is behind door A.
You choose door A.
You do not switch doors.
You win the car.

Hahaha good one :)
What I meant was: can this Monty guy choose the door that hides the car or does he know where the goats are?

Ok, new enigma from my youth :

A milkman on his daily run meets one of his customer and while paying him for his milk, the customer starts talking about his three sons.
When the milkman ask how old they are, the father says the product of their age is 36 and the sum equals the number on the house on the other side of the street.

The milkman looks back, sees the number of the said house, and then tells the father he needs more help to figure out their age.
The father show him a picture and tells him 'this is my oldest son'.
Then the milkman finds what are the ages of the sons.

How old are they ?

Here's a new riddle for ya:

Monty and his car are atop a trap door and underneath it, is molten hot lava. There are two gentleman standing idly by two levers and you're allowed to ask one question which will help you determine which lever will bring Monty his fiery death. The problem is, the other gentleman is a compulsive liar while the other one is basically a Jesus who always tells the truth. What is the one question you should ask so that we can finally get rid of that fucker Monty?

still missing something...

This "logic" is counting two choices as one. That's the appearance of different odds.

You say six in 9 of the switch cases you win, but you don't because you can't choose but 3 of the 6 by switching. You still have a one in three chance of winning.

You can't "cheat" math.

Originally Choose A

Possibilities:

Don't Switch:
Car Behind A: Win
Car Behind B: Lose
Car Behind C: Lose

Switch to B:
Car Behind A: Lose
Car Behind B: Win
Car Behind C: Lose

Switch to C:
Car Behind A: Lose
Car Behind B: Lose
Car Behind C: Win

Those are the only possibilities, and they always remain 1 in 3.

You can work it out by originally choosing B, and C, and they will always work out to the same odds.

2, 2, and 9

There are 8 possibilities:

36 = 1 * 1 * 36 (the sum is 38)
36 = 1 * 2 * 18 (the sum is 21)
36 = 1 * 3 * 12 (the sum is 16)
36 = 1 * 4 * 9 (the sum is 14)
36 = 1 * 6 * 6 (the sum is 13)
36 = 2 * 2 * 9 (the sum is 13)
36 = 2 * 3 * 6 (the sum is 11)
36 = 3 * 3 * 4 (the sum is 10)

Since the milkman saw the number of the house, there shouldn't have been any problem for him except in the case when the number is 13. So, the number was 13. The rest is obvious :)

I told this riddle to my teenager son today and he also said that the first thing that came to his mind was that the fastest way would be to smash the tub to pieces

Maybe I should not have allowed him to play that much Zelda Twilight Princess when he was younger :(

Ask either one of them "If I were to ask the other guy what lever to pull, what lever would he tell me?"
Whatever lever he tells you pull the opposite one.

If you ask the liar he knows the other guy always tells the truth so he will tell you the wrong lever. If you ask the truth teller he knows the other guy always lies so he will tell you the wrong lever. So asking either one that question, without even knowing who they are, and you can pull the lever to kill Monty and move on with more riddles :)
All you need to do is state in the riddle that the tub is bolted down, unbreakable, and can not be moved.

I wrote a javascript Monte Carlo simulation for the 3-door case.
It is here, if you care to review the code or run it yourself:
http://pastebin.com/8P5qHuPF

Here is some sample output.
Each of the blocks below represents one million games.

Initial means the player stuck with his initial choice.
Swap means the player switched doors when offered.


initial wins: 166537
initial losses: 333135
swap wins: 333927
swap losses: 166401

initial wins: 166142
initial losses: 333348
swap wins: 333205
swap losses: 167305

initial wins: 166702
initial losses: 333577
swap wins: 332866
swap losses: 166855

initial wins: 166430
initial losses: 332631
swap wins: 333466
swap losses: 167473

initial wins: 166791
initial losses: 333521
swap wins: 333118
swap losses: 166570

GIGO effect.

It doesn't matter how many ways you try to write it, the odds of picking the right door is always 1 in 3.

The reason you're getting the results you're getting is because you're considering BOTH options of switching at the same time, when in reality you can only pick one.

In other words, sticking with the same door has a 1 in 3 chance, while there is a 2 in 3 chance it's one of the other two. That's why you're getting the results you are. The odds are better that it's not your choice, that's the nature of having a 1 in 3 choice. But switching to one leaves you in the exact same position, with your new choice having a 1 in 3 chance of being right and a 2 out of 3 chance of it being one of the other two.

What your results show is that there is a 2 out of 3 chance that your initial choice is wrong. Which is correct, but remains correct after you switch to a single other door.

Gee, thanks. I wouldn't have bothered had I not thought highly of you. Now I feel silly.

What the results show is if you play millions of 3-door Monty Hall games, making your initial choice at random, and then deciding whether or not to swap at random:

You will win roughly 1/3 games when you stick with your initial choice.
You will win roughly 2/3 games when you switch doors when offered.

Anyway, I'll stop bothering you and we can move on to another riddle.

Which shows the results are wrong, because that's not the actual odds. The actual odds are

You will win roughly 1/3 games when you stick with your initial choice.
You will win roughly 1/3 games when you switch doors when offered since you can only switch to 1 of the 2 other options.

Don't take my word for it, take it to a math professor at your local college/university.

For anyone reading having a hard time understanding, here it is in plain English with ALL possibilities explored.

If you initially choose A:

And are offered the option to switch the ONLY possibilities are:

Possibility ONE: You stay with A (don't swap)

Then if A is correct, you win.
If B is correct, you lose.
If C is correct, you lose.

Possibility TWO: You switch to B

Then if A is correct, you lose.
If B is correct, you win.
If C is correct, you lose.

Possibility THREE: You switch to C

Then if A is correct, you lose.
If B is correct, you lose.
If C is correct, you win.

Those are the ONLY THREE possibilities and in each case, you have a one in three chance of winning. And the will be the EXACT same if you initially choose B or initially choose C. Write it all out if you don't believe me.

Like I said, you can't cheat math.

Are you taking into account the fact that someone who knows the contents of the doors is revealing a dummy prize? It doesn't seem like you are. And if you're not, then yes, you would be correct that any choice would be a 1/3 chance correct. However, the whole point of the Monty Hall puzzle is that someone with foreknowledge of the doors is revealing an incorrect door after you've chosen a door.