THE LAST YOU'LL EVER HAVE TO READ ABOUT THIS
To beat the dead horse of Monty Hall's game-show problem [November 2 and 23]: Marilyn was wrong, and you were right the first time ... --Eric Dynamic, Berkeley, California
You really blew it. As any fool can plainly see, when the game-show host opens a door you did not pick and then gives you a chance to change your pick, he is starting a new game. It makes no difference whether you stay or switch, the odds are 50-50. --Emerson Kamarose, San Jose, California
Suppose our task is to pick the ace of spades from a deck of cards. We select one card. The chance we got the ace of spades is 1 in 52. Now the dealer takes the remaining 51 cards. At this point his odds are 51 in 52. If he turns over 1 card which is not the ace of spades our odds are now 1 in 51, his are 50 in 51. After 50 wrong cards our odds are 1 in 2, his are 1 in 2. The idea that his odds remain 51 in 52 as more and more cards in his hand prove wrong is just plain crazy. --John Ratnaswamy, Chicago; similarly from Greg in Madison, Wisconsin, Stuart Silverman in Chicago, Frank Mirack in Arlington, Virginia, Dave Franklin in Boston, and many others
Give it up, gang. It was bad enough that I screwed this up. But you guys have had the benefit of my miraculously lucid explanation of the correct answer! Since you won't listen to reason, all I can tell you is to play the game and see what happens. One writer says he played his buddy using the faulty logic in my first column and got skunked out of the price of dinner. Several other doubters wrote computer programs that, to their surprise, showed they were wrong and Marilyn vos Savant was right.
A friend of mine did suggest another way of thinking about the problem that may help clarify things. Suppose we have the three doors again, one concealing the prize. You pick Door #1. Now you're offered this choice: open Door #1, or open Door #2 and Door #3. In the latter case you keep the prize if it's behind either door. You'd rather have a two-in-three shot at the prize than one-in-three, wouldn't you? If you think about it, the original problem offers you basically the same choice. Monty is saying in effect: you can keep your one door or you can have the other two doors, one of which (a nonprize door) I'll open for you. Still don't get it? Then at least have the sense to keep quiet about it.
Other correspondents have passed along some interesting variations on the problem. Here's a couple from Jordan Drachman of Stanford, California:
There is a card in a hat. It is either the ace of spades or the king of spades, with equal probability. You take another identical ace of spades and throw it into the hat. You then choose a card at random from the hat. You see it is an ace. What are the odds the original card in the hat was an ace? (Answer: 2/3.)
There is a family with two children. You have been told this family has a daughter. What are the odds they also have a son, assuming the biological odds of having a male or female child are equal? (Answer: 2/3.)
Finally, this one from a friend. Suppose we have a lottery with 10,000 "scratch-off-the-dot" tickets. The prize: a car. Ten thousand people buy the tickets, including you. 9,998 scratch off the dots on their tickets and find the message "YOU LOSE." Should you offer big money to the remaining ticket holder to exchange tickets with you? (Answer: hey, after all this drilling, you figure it out.)
Art accompanying story in printed newspaper (not available in this archive): illustration/Slug Signorino.