^ Continued ...

The last you’ll ever have to read about this

Dear Cecil:

To beat the dead horse of Monty Hall’s game-show problem: Marilyn was wrong, and you were right the first time …

— Eric Dynamic, Berkeley, California

Dear Cecil:

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

Dear Cecil:

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, Madison, Wisconsin; Stuart Silverman, Chicago; Frank Mirack, Arlington, Virginia; Dave Franklin, Boston; many others

Cecil replies:

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 non-prize 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 ticketholder to exchange tickets with you? (Answer: hey, after all this drill, you figure it out.)

So I lied — This is the last you’ll ever have to read about this

Dear Cecil:

The answers to the logic questions submitted by Jordan Drachman were illogical. In the first problem he says there is an equal chance the card placed in a hat is either an ace of spades or a king of spades. An ace of spades is then added. Now a card is drawn from the hat — an ace of spades. Drachman asks what the odds are that the original card was an ace. Drawing a card does not affect the odds for the original card. They remain 1 in 2 that it was an ace, not 2 in 3 as stated.

In the second problem we are told a couple has two children, one of them a girl. Drachman then asks what the odds are the other child is a boy, assuming the biological odds of having a male or female child are equal. His answer: 2 in 3. How can the gender of one child affect the gender of another? It can’t. The answer is 1 in 2.

— Adam Martin and Anna Davlantes, Evanston, Illinois

Dear Cecil:

In a recent column you asked, “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 ticketholder to exchange tickets with you?”

If you think the answer is “yes,” you are wrong. If you think the answer is “no,” then you are intentionally misleading your readers …

— Jim Balter, Los Angeles

Cecil replies:

Do you think I could possibly screw this up twice in a row? Of course I could. But not this time. Cecil is well aware the answer to the lottery question is “no” — if there are only two tickets left, they have equal odds of being the winner. The difference between this and the Monty Hall question is that we’re assuming Monty knows where the prize is, and uses that information to select a non-prize door to open; whereas in the lottery example the fact that the first 9,998 tickets are losers is a matter of chance. I put the question at the end of a line of dissimilar questions as a goof — not very sporting, but old habits die hard.

The answers given to Jordan Drachman’s questions — 2 in 3 in both cases — were correct. The odds of the original card being an ace were 1 in 2 before it was placed in the hat. We are now trying to determine what card was actually chosen based on subsequent events. Here are the possibilities:

(1) The original card in the hat was an ace. You threw in an ace and then picked the original ace.

(2) The original card in the hat was an ace. You threw in an ace and then picked the second ace.

(3) The original card was a king; you threw in an ace. You then picked the ace.

In 2 of 3 cases, the original card was an ace. QED.

The second question is much the same. The possible gender combinations for two children are:

(1) Child A is female and Child B is male.

(2) Child A is female and Child B is female.

(3) Child A is male and Child B is female.

(4) Child A is male and Child B is male.

We know one child is female, eliminating choice #4. In 2 of the remaining 3 cases, the female child’s sibling is male. QED.

Granted the question is subtle. Consider: we are to be visited by the two kids just described, at least one of which is a girl. It’s a matter of chance who arrives first. The first child enters — a girl. The second knocks. What are the odds it’s a boy? Answer: 1 in 2. Paradoxical but true. (Thanks to Len Ragozin of New York City.)

Cecil is happy to say he has heard from the originator of the Monty Hall question, Steve Selvin, a UCal-Berkeley prof (cf American Statistician, February 1975). Cecil is happy because he can now track Steve down and have him assassinated, as he richly deserves for all the grief he has caused. Hey, just kidding, doc. But next time you have a brainstorm, do us a favor and keep it to yourself.