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SO I LIED--THIS IS THE LAST YOU'LL EVER HAVE TO READ ABOUT THIS (I HOPE)

The answers to the logic questions submitted by Jordan Drachman [January 25] 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 one in two that it was an ace, not two in three 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: two in three. How can the gender of one child affect the gender of another? It can't. The answer is one in two. --Adam Martin and Anna Davlantes, Evanston, Illinois

In your column of January 25, 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

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 nonprize 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--two out of three in both cases--were correct. The odds of the original card being an ace were one in two 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 two of three 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 the fourth choice. In two of the remaining three 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: one in two. It seems paradoxical, but it's not. (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 prof at the University of California at Berkeley (see 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.

Art accompanying story in printed newspaper (not available in this archive): illustration/Slug Signorino.

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