Door Number Two

A few years back Marilyn vos Savant, who writes the “Ask Marilyn” column in Parade Magazine, caused quite a stir with her answer to the following reader’s question:
“Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No.1, and the host, who knows what’s behind the other doors, opens another door, say No. 3, which has a goat behind it. He then says to you, ‘Do you want to switch to door No.2?’ Is it to your advantage to make the switch?”

Her answer — that the contestant should always switch doors — has been analyzed by mathematicians at the Massachusetts Institute of Technology and computer programmers at Los Alamos National Laboratory in New Mexico. It has been tested in classrooms from second grade to graduate level at thousands of schools around the world. And it still stirs debate among people who see it for the first time.

Let’s restate the problem. Monty Hall, a fine, upstanding game-show host, has randomly placed a car behind one of three closed doors. There is a goat behind each of the other two doors. “First you point to a door,” he explains. “Then I’ll open one of the other doors to reveal a goat. After I’ve shown you, the goat, you make your final choice, and you win whatever is behind that door.”

You really want that new car. But you know you only have a one-in-three chance of getting it right. You point to Door Number One. Mr. Hall opens Door Number Three and shows you a goat. Now there are two doors remaining. It seems like your odds just improved to 50/50. Do you stick with Door One or do you switch to Door Number Two? Or doesn’t it matter?

To see which strategy works best you can try playing the game over and over to see which wins most often. The results contradict most people’s intuition that, when there are only two unopened doors left, the odds on each one must be one-in-two, or 50/50. But the fact that Mr. Hall opened another door did not affect the odds on Door Number One. You had a one-in-three chance of being right to begin with, and you still have a one-in three chance after he opens Door Number Three and shows you the goat. You knew he was going to open another door and reveal a goat regardless of what was behind Door Number one, so his action provides no new information about Door Number One. Therefore, since the odds on Door 1 are still one-in-three, and the only other place the car could be is behind Door Number Two, the odds of Door Number Two being the correct choice are now two-in-three.
Another way to explain this is as follows. The probability of picking the wrong door in the initial stage of the game is two-out-of-three. If the contestant picks the wrong door initially, the host must reveal the remaining empty door in the second stage of the game. Thus, if the contestant switches after picking the wrong door initially, the contestant will win the prize. The probability of winning by switching then reduces to the probability of picking the wrong door in the initial stage, which is clearly two-out-of-three.

Still don’t get it? There are literally hundreds of on-line math-game sites where you can test the theory. Here’s a link to one that works nicely:

http://math.ucsd.edu/~crypto/Monty/Montytitle.html

Have fun with it.