The Goat and The Car Puzzle: An Explanation

We must be careful to count the number of possible configurations of prizes and doors, and take all combinations of configurations with choices made, understanding the relative probabilities of each.

First, we begin by assigning prizes to doors. There are only three configurations. The doors are named A, B, and C. The dollar sign $ marks the shiny new car. (I hope this looks good in all web browser... I'm using Netscape, Netpositive and Opera. This won't work as well with Lynx or other text-only browsers.)

A B C
$    

A B C
  $

A B C
    $

These three cases of prize assigments occur with equal probabilies. (Actually, if they don't, it doesn't affect the result. I'll leave that as an "exercise for the student.") Only one of these cases will describe reality as we play the game, but we can imagine that the others describe alternate universes. If we imagine playing the game over and over, three thousand times total, then each case represents one thousand of those games. At least, on average, statistically, they should.

Now the contestant chooses a door. There are three, and at this point in the game, there's no reason to pick one instead of another. This decision is made based on pure whim, not depending in any way upon the assignment. List this choice vertically.

(A) B C

A (B) C

A B (C)

With three possible assignments, and three picks by the contestant, there are nine ways for things to be:

A B C
$    

A B C
  $

A B C
    $

(A) B C

(A) B C
 $    

(A) B C
    $

(A) B C
      $

A (B) C

A (B) C
$    

A (B) C
   $

A (B) C
      $

A B (C)

A B (C)
$    

A B (C)
  $

A B (C)
     $

Now the host reveals a goat. In every one of the nine cases, there exists at least one door, unpicked by the contestant, which hides a goat. That's the one the host will open, of course. If there are two such doors, as when the contestant has luckily picked the car's door, it doesn't matter which goat's door the host opens. Denoting the opened door with an 'x', the possible situations are

(A) B x
 $    

(A) B x
    $

(A) x C
      $

A (B) x
$    

x (B) C
   $

x (B) C
      $

A x (C)
$    

x B (C)
  $

A x (C)
     $

These nine cases all occur with equal probability. It is a simple matter to examine all of them and see that for three, the contestant has already picked a winner, but that for six cases, the contestant should switch to win.

Conclusion

Thus, by switching, the contest will win 6 out of 9 tries. By staying with the original choice, only 3 wins out of nine tries. It's twice as good to switch. This is what Marilyn Vos Savant said, and it's what Jeffrey Kooistra said, and it's what I say.

Of course, all this is not valid if the contestant enjoys goats and lacks a driver's license...