### Look Before You Leap

Even the best of us get to make ourselves look silly once in a while, and today it appears to be Bob Somerby's turn. Bob writes the Daily Howler, which you'll find in my list of links, and I highly respect his work. Most of the time. But today he's come up with this, well, you know . . . howler:

Good God! Kaplan and Kaplan spent some time writing a book on “adventures in probability.” For their trouble, they get to read this, from William Grimes’ review in the Times:Sometimes harder than they look. For the record: Bob's entirely wrong.GRIMES (3/31/06): Before scoffing, chew on the now famous Monty Hall problem, named after the host of ''Let's Make a Deal.'' A contestant knows that concealed behind three doors there are two goats and one new car. The contestant chooses Door No. 1. The beaming host opens Door No. 3 to reveal a goat, and then asks the contestant if he would like to change his choice to Door No. 2. Two doors add up to a 50-50 proposition, obviously. So why bother? Because the odds have actually shifted.The chances are now two out of three that changing to Door No. 2 will obtain the car.Say what? We don’t know what the Kaplans wrote to provoke that highlighted sentence. But for the record: If the contestant changes to Door No. 2, he’ll obtain the car half the time—and “half the time” is not “two out of three.” . . .

Numbers, dear friends, are hard work.

To see why this is so, consider the initial choice of a door. There are three doors, each with an object behind it. Two of these objects are goats; the third is a car. If you pick randomly (and there's no reason to think you won't), you have a 1 in 3 chance of choosing the door concealing goat A, a 1 in 3 chance of choosing the door concealing goat B, and a 1 in 3 chance of choosing the door concealing the car.

So far, so clear. Now Monty shows you a goat. At this point, what happens if you stick? If you started out choosing a goat, there's no chance that you'll switch to the other goat; that goat has been revealed and is

*hors de combat*. You'll get the car. On the other hand, if you started out with the car, you will switch to the still-hidden goat. It looks like this:

- If you started with goat A, you'll get the car if you switch
- If you started with goat B, you'll get the car if you switch
- If you started with the car, you'll get the remaining goat if you switch

Bob's mistake is to assume that after Monty opens up door #3, the fact that two doors remain in play makes the odds of finding the car behind either one of them exactly 50:50. But this is magic. The probability that the car is behind door #1 (the one you've chosen)

*has not changed*; it's still 1 in 3. (Grimes is wrong when he says "the odds have shifted"). What

*has*changed is that you don't have to pick between the other two doors if you decide to abandon your initial door. Monty has effectively now given you the chance to exchange what's behind your door for what's behind

*both*other doors. Remember, he didn't choose to open door #3 at random; he chose a door that he knew harbored a goat. The aggregate probability of finding a car behind either door #2 or door #3 is still 2 in 3, but now you know which of these doors not to pick.

This knowledge yields a simple set of equations (where P(doorX) is the probability of finding a car behind door X):

- P(door1) = 1/3
- P(door1) + P(door2) + P(door3) = 1
*therefore*P(door2) + P(door3) = 1 - 1/3 = 2/3*but we now know*P(door3) = 0*therefore*P(door2) = 2/3 - 0 = 2/3

[For those of you with an aversion to equations, there's a simulation of this problem here. Just play it a few times and keep track of how often you win the car by switching (or win a goat by sticking).]

The only thing that could alter these odds would be some sort of unfairness on Monty's part. And in fact you wouldn't

*expect*Monty to play fair in real life. For example, he might show the goat and offer a switch only when the contestant has chosen the car, which would reduce the chance of getting the car by switching all the way to zero. You could imagine other behavior on Monty's part that would produce other probabilities, including Bob's answer of 50%. But that 50% doesn't follow necessarily from any of the premises of the problem; it would be an arbitrary importation by the host. And it wouldn't be, as Bob thinks,

**the**one and only answer to the problem.

In Bob's defense, he's by no means alone in this error. When the Monty Hall problem appeared years ago in Marilyn vos Savant's Sunday supplement column

*Ask Marilyn*, and Marilyn gave the correct answer, she was deluged with complaints. Some pointed out, quite reasonably, that the answer depends on whether Monty offers the same choice to all contestants. But there were others, including some with serious math credentials, who got it wrong in just the same way that Bob has, and made no secret of their scorn (sometimes sexist) at Marilyn for coming up with a different answer.

Yes, dear friends, numbers are hard work.

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