Many new game shows have appeared in recent years, among them "Who Wants To Be a Millionaire," "Deal or No Deal," and "Show Me the Money." So far none has aroused the mathematical interest of the quiz show, "Let's Make a Deal." Having received so many emails over the years about the show's so-called Monty Hall problem, I thought I'd devote this holiday column to the famous problem, to a question about a new variant of it (with answer at end), and to a more general question about its applicability.

"Let's Make a Deal" was popular in the '60s and '70s and has been resurrected in various formats since then. In its original version the game show contestant was presented with three doors, behind one of which is a new car. The other two doors have booby prizes behind them.

"The Let's Make a Deal" host, Monty Hall, asks the contestant to pick one of the three doors. Once the contestant has done so, Monty opens one of the two remaining doors to reveal what's behind it, but is careful never to open the door hiding the car. After Monty has opened one of these other two doors, he offers the contestant the chance to switch doors. The question is: Should the contestant stay with his (or her) original choice and hope the car is behind it or switch to the remaining unopened door?

Many people reason that it doesn't make any difference since there are two possibilities, and thus the probability is 1/2 that the car is behind the original door. The correct strategy, however, calls for the contestant to switch. The probability he picked the correct door originally is 1/3, and the probability the car is behind one of the other two unopened doors is 2/3. Since the host is required to open a door behind which there's a booby prize, the 2/3 probability is now concentrated on the other unopened door. Switching to it will increase the contestant's chances of winning from 1/3 to 2/3.

One common way to make the decision to switch more intuitive is to imagine that there are 100 doors behind one of which is a new car, behind the other 99, booby prizes. The host again asks the contestant to pick one of the doors. Once the contestant has done so, the host opens 98 of the 99 remaining doors to reveal what's behind them, but is careful never to open the door hiding the car. After the host has opened 98 of these other 99 doors, he offers the contestant the chance to switch to the other unopened door. The question again is: Should the contestant switch?

That switching is the correct answer is clearer in this case. The probability the contestant picked the correct door originally is 1/100, and the probability the car is behind one of the other ninety-nine unopened doors is 99/100. Since the host is constrained to open 98 doors behind which there's a booby prize, the 99/100 probability is now concentrated on the other unopened door. Switching to it will increase the contestant's chances of winning from 1/100 to 99/100.

It's interesting that even in the latter case many people refuse to switch. One factor may be an extreme fear of the regret they'd feel if they switched away from their original pick, and it happened to be correct. The same fear of regret underlies people's reluctance to trade lottery tickets with friends. They imagine how bad they'd feel if their original ticket were to win.

If this is all clear, here's a variation of the problem to test your understanding of the probability involved. *(The answer is at the end of this column.)* Let's say that "Let's Make a Deal" were to attempt a comeback. The producers, fearing the audience would be small if the game were exactly the same, devise a variant game in which the contestant is presented with 10 doors. Again behind one of them is a car, behind the others booby prizes. After the contestant picks a door, Monty (or his avatar) opens just seven of the remaining nine unopened doors, but is careful never to open the door hiding the car. There are now three unopened doors -- the one that the contestant originally picked and two others. Which strategy works best, switching to one of the other two unopened doors or sticking with the original pick? Furthermore, what is the probability of winning by following these two strategies?

One last question: Can you think of any real-world situations -- crime mysteries, world politics, administrative deceptions -- which might be modeled on some close variant of the Monty Hall problem (and not simply by Bayes' theorem)? That is, are there situations in which the "contestant," say a reporter, must choose among various alternatives and the "host," say an official, knows the true answer, but is evasive about it and instead answers a question different from the one the contestant asks?

**Answer:** The chance the prize is behind the door originally chosen by the contestant is 1/10 and remains 1/10. The chance it's behind one of the nine other unopened doors is 9/10. Since the host opens seven of these nine other unopened doors, the 9/10 probability that it's behind one of them is divided between two of these nine doors. So the contestant should switch to one of these two. Doing so raises his probability of winning from 1/10 to one-half of 9/10 or 45 percent.

*Professor of mathematics at Temple University, John Allen Paulos is the author of best-selling books, including "Innumeracy" and "A Mathematician Plays the Stock Market." His "Who's Counting?" column on ABCNEWS.com appears the first weekend of every month. *