Let’s the possibilities:
Alternative 1, first choice A, C shows a goat, A has the car.
Door A / Door B / Door C
Car / Goat / Goat n=1 (50%)
Goat / Car / Goat n=2 (50%)
Goat / Goat / Car n=3 (0%, we know C doesn’t have a car)
Not switching wins. Switching looses.
Alternative 2, first choice A, C shows a goat, B has the car.
Door A / Door B / Door C
Car / Goat / Goat n=1 (50%)
Goat / Car / Goat n=2 (50%)
Goat / Goat / Car n=3 (0%, we know C doesn’t have a car)
Switching wins. Not switching looses.
Alternative 3, first choice A, B shows a goat, A has the car
Door A / Door B / Door C
Car / Goat / Goat n=1 (50%)
Goat / Car / Goat n=2 (0%, we know B doesn’t have a car)
Goat / Goat / Car n=3 (50%)
Not switching wins. Switching looses.
Alternative 4, first choice A, B shows a goat, C has the car
Door A / Door B / Door C
Car / Goat / Goat n=1 (50%)
Goat / Car / Goat n=2 (0%, we know B doesn’t have a car)
Goat / Goat / Car n=3 (50%)
Switching wins. Not switching looses.
Alternative 5, first choice B, C shows a goat, B has the car.
Door A / Door B / Door C
Car / Goat / Goat n=1 (50%)
Goat / Car / Goat n=2 (50%)
Goat / Goat / Car n=3 (0%, we know C doesn’t have a car)
Not switching wins. Switching looses.
Alternative 6, first choice B, C shows a goat, A has the car.
Door A / Door B / Door C
Car / Goat / Goat n=1 (50%)
Goat / Car / Goat n=2 (50%)
Goat / Goat / Car n=3 (0%, we know C doesn’t have a car)
Switching wins. Not switching looses.
Alternative 7, first choice B, A shows a goat, B has the car
Door A / Door B / Door C
Car / Goat / Goat n=1 (0%, we know A doesn’t have a car)
Goat / Car / Goat n=2 (50%)
Goat / Goat / Car n=3 (50%)
Not switching wins. Switching looses.
Alternative 8, first choice B, A shows a goat, C has the car
Door A / Door B / Door C
Car / Goat / Goat n=1 (0%, we know A doesn’t have a car)
Goat / Car / Goat n=2 (50%)
Goat / Goat / Car n=3 (50%)
Switching wins. Not switching looses.
Alternative 9, first choice C, B shows a goat, C has the car.
Door A / Door B / Door C
Car / Goat / Goat n=1 (50%)
Goat / Car / Goat n=2 (0%, we know B doesn’t have a car)
Goat / Goat / Car n=3 (50%)
Not switching wins. Switching looses.
Alternative 10, first choice C, B shows a goat, A has the car.
Door A / Door B / Door C
Car / Goat / Goat n=1 (50%)
Goat / Car / Goat n=2 (0%, we know B doesn’t have a car)
Goat / Goat / Car n=3 (50%)
Switching wins. Not switching looses.
Alternative 11, first choice C, A shows a goat, C has the car
Door A / Door B / Door C
Car / Goat / Goat n=1 (0%, we know A doesn’t have a car)
Goat / Car / Goat n=2 (50%)
Goat / Goat / Car n=3 (50%)
Not switching wins. Switching looses.
Alternative 12, first choice C, A shows a goat, B has the car
Door A / Door B / Door C
Car / Goat / Goat n=1 (0%, we know A doesn’t have a car)
Goat / Car / Goat n=2 (50%)
Goat / Goat / Car n=3 (50%)
Switching wins. Not witching looses.
Not Switching wins alternatives 1, 3, 5, 7, 9, and 11 = 6 out of 12. Not Switching looses alternatives 2, 4, 6, 8, 10, and 12 = 6 out of 12.
50/50
The result is the same for any number of door with only one car (see simulator)… Can you find more alternatives? If the percentages I left showing throw you off, forget them, and look at the possible configurations, there aren’t others, so there are only twelve alternatives and not switching wins in 6 of them. In four of them the car is in A, in another four it is in B, and in the remaining four you find the car in C. Conversely, you chose A 4 times, B 4 times, and C four times. You will find that this mathematics cannot be blown by any bullet you might try.
Not switching has 6 wins of 12 alternatives, making it 50/50.
Now, why would you believe that not switching would only win in 4 alternatives and loose in the other eight? Belief bias, that’s why.

When there are two players, they both have a 50/50 chance right from the start. They pick, Monty reveals a goat, and someone wins.
Perhaps what makes all this so hard to fathom is the fact that your odds don’t really “change”. Your odds of winning are based on the rules of the game, not so much your initial guess, which means nothing, since you have to choose again after seeing a goat. What I mean is that your odds of winning on the first try are zero. You may pick the right door, but you haven’t won the prize yet, because the game is not over.

I have to admit that it was very hard for me to accept this at the three door level, but unless you picked the right door (only once every three attempts), Monty basically tells you where it is! So this is how I inferred that Monty tells you where the new car is two out of three times, giving legitimacy to Dr. Shermer’s Arument.

Clearly, at the beginning the contestent has a 1/3 chance of picking the right door when he picks #1. And when Monty Hall shows him that a goat is behind door #2 he has in effect changed the probability space from 3 to 2 possibilities. So when presented a chance to change his choice the contestent has in effect started over, i.e. it’s as though there was never a door #2. He now has 2 possible choices so he clearly has improved his chances of picking the right door from 1/3 to 1/2.

As is often the case according to Occam’s razor, the simplest solution/explanation is usually the right one. Many things in life and mathematics are counterintutive but this is not one of them.

The probability of picking the right door at the beginning is 1/3. In other words, most of the time, you’d lose this game if it ended here. Keep that in mind — most likely, you picked a loser. Well, if you most likely picked a lsoer, why not start with a different door? — because no matter which door we choose, it’s most likely a loser. Remember: YOUR ORIGINAL CHOICE IS MOST LIKELY WRONG. Now, once one of the other doors is shown to be a loser, and you know YOUR ORIGINAL CHOICE IS MOST LIKELY WRONG, the other remaining door is most likely the winner.

Thanks to Chris for providing a better explanation. I think I’ll go play a game and amuse myself with the results anyway. I’ve always struggled with assigning probabilities to anything other than a coin toss.

Since a goat was shown behind door 2, then the prize must be behind (1) or (3). Why should I change my mind about which door I chose? If the host always shows one of the losing doors and asks the player if they want to switch, the player may have the prize door and lose by switching or have the goat and lose by switching. (But hey, personally I wouldn’t mind having a goat to milk or cook.) So with the evidence presented, I just can’t see how the odds of winning are actually improved by changing the decision.

The way I see it:
- a door was chosen with a 1-in-3 chance of winning.
- a losing door was shown – had you known that door was a loser beforehand, you would have made your odds at 50/50.
- so you now know one losing door – big deal – you may or may not have the other losing door – *how* does knowing where one losing door is improve your chances of winning over the original 1-in-3?

I can see how the scheme may work if you had 10 doors and the host always showed you all but two doors (the one you chose plus one other). In that case it would make sense to switch doors. However, with a mere 3 doors I just can’t figure out the advantage in the switch.

Has anyone written a small computer program and played the game a few hundred times to verify this claim that the overall probability goes from 1/3 to 2/3 if you switch?

If one picks door #1 and thinks of doors 2&3 as a group, then he knows the odds of the car being behind 2 or 3 are 2/3s, while the odds of the car being behind door 1 are 1/3.
This remains true, even after the goat is revealed behind door 3.

Basically Monty is simply offering a chance to switch from your initial pick (1/3 chance) to the other group consisting of both doors (2/3s chance).

Of course you can’t actually trade for both doors, but if you think of it that way -as trading door 1 for doors 2&3, then you’re always being given a chance to change your odds for the better.