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.