Marilyn Ignores the Obvious Regarding Probability of Boys

Marilyn is Wrong Copyright © 1996-1998 Herb Weiner. All rights reserved.

Ask Marilyn ® by Marilyn vos Savant is a column in Parade Magazine, published by PARADE, 711 Third Avenue, New York, NY 10017, USA. According to Parade, Marilyn vos Savant is listed in the "Guinness Book of World Records Hall of Fame" for "Highest IQ."

In her Parade Magazine column of May 26, 1996, Marilyn claimed that if a woman has exactly two children, at least one of whom is a boy, there is only a 1/3 chance that both of her children are boys. Marilyn has discussed this same problem in four additional columns, on December 1, 1996, March 30, 1997, July 27, 1997, and October 19, 1997.

In her column of July 27, 1997, Marilyn accepted this challenge from Eldon Moritz <> of Arlington, Texas: "I will send $1000 to your favorite charity if you can prove me wrong. The chances of both the woman and the man having two boys are equal." But rather than acknowledge that the fault might be in the interpretation of the problem rather than her math, Marilyn invited readers to participate in an experiment that simply demonstrated that her math was correct. On October 19, 1997, Marilyn declared herself the winner, gave Eldon no opportunity to respond, and told him he owed $1000 to the American Heart Association.

Marilyn has ignored my letter defending Eldon.

The readers disagree!

Readers have raised several objections to Marilyn's answer:

The Two Interpretations

Harry Eaton <> and Patrick Maupin <> both wrote to point out that Marilyn is assuming that the woman was randomly selected from the collection of all women having two children, at least one of whom is a boy. They believe that this assumption is not necessarily valid.

Consider the collection of all women with two children. Now, consider the following two experiments:

Since the original question states only that we know the sex of one child, Harry and Patrick believe that the first experiment described above is the intended question. They believe that this corresponds to the real world question that might be asked by someone who knows one of the two children but not the other. (If we knew the sex of both children, why ask the question in the first place?)

Harry predicted that this column would generate a lot of mail, just as the goat-in-a-game-show column did. Based upon the mail I've received so far, I belive Harry's prediction is correct.

Here's a copy of Harry's Letter to Marilyn.

In Defense of Marilyn's Math

Many readers are asking for a better explanation than the one Marilyn offered. Perhaps this will clear things up.

Begin with a group of 100 families, each with two children, distributed as follows:

Of this group, there are 50 families in which the oldest child is a boy. Of those 50 families, there are 25 families in which the youngest child is also a boy. In other words, out of the group of families in which the oldest child is a boy, 50% have two boys.

From the same group, there are 75 families in which at least one child is a boy. Of those 75 families, there are 25 families in which the other child is also a boy. In other words, out of the group of families in which at least one child is a boy, only 33% have two boys.

Expectation vs. Proportion

Thanks to Dr. Steven M. Carr <> for clarifying the differences between expectation and proportion.

What if we Reword the Question?

We know that the local university is conducting a study of two child families with at least one boy. In other words, a two child family must have either one or two boys in order to be accepted into the study. We are talking to a woman whose family has just been accepted into this study. What is the expectation that both of her children are boys?

Is the answer one third? Is this different from the answer to the original question? Why?

Marilyn Makes a Minor Math Error

Charlie Kluepfel <> wrote to report the following error.
Marilyn states "Given that there are about 106 boys born for every 100 girls, the actual percentage in the population would be closer to 33.9%." Based on my understanding of colloquial speech, I would take this to mean that doing the computations based on 106 boys for every 206 births, you would get 33.9% of families with at least one boy having two boys. But doing the calculations with 106/206 births' being boys does not result in this answer. That fraction, 106/206, is .51456. Out of all families with 2 children, (.51456)^2 would have two boys, or 26.4775%. The fraction that would have one boy and one girl is 2(.51456)(1 - .51456), or 49.958%. Thus the families with both boys out of all families with at least one boy is 26.4775/(26.4775+49.958) = 34.641%, not 33.9%.

Where did the 33.9% figure come from? It's certainly not "Given that there are about 106 boys born for every 100 girls," as we have just seen. Possibly the later reportage from the census bureau that "11,334 of the 33,355 ('at least one boy' group) had two boys (34%)" influenced this. These figures actually represent 33.98%, but maybe Marilyn was incorrectly rounding it to 33.9%. But in any case, her notation of "closer to 33.9%" came before she introduced these census-bureau figures as something to explain.

As an aside: Given the correct 34.641% based on 106/206 births being boys, one would have expected about 6216 responses, give or take about 80 (that is about the square root of 6216, the standard deviation of the corresponding Poisson distribution) would be two-boy responses, yet she got about 6443 such responses (based on 35.9% of 17,946 responses). Maybe some people have the urge to prove Marilyn wrong and those with two boys were thereby more likely to respond. On the other hand, the census bureau got only 11,334 both-boy families, where 35.9% would have been 11,979 give or take 110. I can only conjecture that this is the result of higher mortality rates for boys lowering the ratio from the 106/206 that applies at age 0.

Social Factors Affect the Results

Robert M Beals <> wrote to point out that social factors may affect the probability that the woman has two boys. In particular, there is a strong desire in many families to have a son, and couples whose first child is a girl are somewhat more likely to have additional children than couples whose first child is a boy. This social factor decreases the probability that the older child in two child families is a boy, and therefore decreases the probability that the woman has two sons. However, the fact that there are about 106 boys born for every 100 girls reduces the effect of this social factor.

How Old is this Problem

Dennis Rakestraw <> is searching for the oldest publication of this problem. Dennis recalls usenet posting reporting that problem appeared in a 1980 book from Life Science Library, Mathematics, Time-Life Books, Alexandria, Virginia; however, Dennis believes the problem is much older than twenty to thirty years. Any assistance would be much appreciated.

Letter to Marilyn

Subject: Thank you for the flawless analysis and effective experiment
Date: Thu, 05 Feb 1998 18:46:38 PST
From: Herb Weiner <>

Dear Marilyn,

Your mathematical analysis of the probability of the woman having two boys was flawless. And the experiment you conducted was certainly an effective way to prove it.

But consider this: You meet a woman and her son. You ask her how many children she has, and she replies "two." So now you know that this woman has exactly two children, at least one of whom is a boy. What is the probability that both of her children are boys? If the answer is different from your previous answer, why? How would you conduct an experiment to test your predictions?

In email correspondence with Eldon Moritz of Arlington, Texas, he argues that you have not proven him wrong, and that he does not owe $1000 to the American Heart Association. How do you respond to his claim?

-- Herb Weiner, Portland, Oregon (

For anyone wondering, the probability that this woman has two sons is 50%, and Eldon is correct.

Eldon's Statement

Eldon Moritz <> requested that I include the following statement:
God bless the American Heart Association, but Marilyn is wrong. There is a group from which you can pick a lady and make the problem statement whereas the probability for the other child would be one-third. Marilyn has very aptly defined this group. To assume that the lady was picked from this group because of the problem statement is an invalid assumption, a mistake in elementary logic. Obviously she has spent very little time on this question.

Many in the scientific community seem willing to make this same invalid assumption. When someone buys her argument there seems to be no more intellectual curiousity.

Scientific American Declared this Problem Ambiguous in the 1950's

Nick Ratti Jr. of Bristol, RI, sent me a photocopy of a clipping from "Scientific American, circa 1950's." (There is no page number, date, or other information on the clipping itself.)

Another example of ambiguity arising from a failure to specify the randomizing procedure appeared in this department last May. Readers were told that Mr. Smith had two children, at least one of whom was a boy, and were asked to calculate the probability that both were boys. Many readers correctly pointed out that the answer depends on the procedure by which the information "at least one is a boy" is obtained. If from all families with two children, at least one of whom is a boy, a family is chosen at random, then the answer is 1/3. But there is another procedure that leads to exactly the same statement of the problem. From families with two children, one family is selected at random. If both children are boys, the informant says "at least one is a boy." If both are girls, he says "at least one is a girl." And if both sexes are represented, he picks a child at random and says "at least one is a ..." naming the child picked. When this procedure is followed, the probability that both children are of the same sex is clearly 1/2. (This is easy to see because the informant makes a statement in each of the four cases -- BB, BG, GB, GG -- and in half of these case both children are of the same sex.) That the best of mathematicians can overlook such ambiguities is indicated by the fact that this problem, in unanswerable form, appeared in one of the best of recent college textbooks on modern mathematics.

Eldon Moritz <> reports that he also received a copy of this column, and was told that appeared in the October 1959 issue, and was written by Martin Gardner.

Ambiguous, or Marilyn Unambiguously Wrong?

My position is that the problem is ambiguous, because there are two valid interpretations. Marilyn's interpretation is valid, but as I point out in my letter to Marilyn, there is a second, equally valid interpretation.

Eldon Moritz <> and Nick Ratti Jr. of Bristol, RI both take the position that the problem is not ambiguous at all; that there is only one valid interpretation, and that Marilyn's interpretation is invalid. They argue that (in the Scientific American excerpt above) since we are talking about a specific man (Mr. Smith), the first randomizing procedure (from all families with two children, at least one of whom is a boy, a family is chosen at random) violates the original problem statement. Choosing a family at random violates the problem statement because the Smith family has already been chosen. Therefore, they argue that only the second randomizing procedure (telling us either that "at least one is a boy" or "at least one is a girl") is a valid interpretation of the problem.

Nick also points out that "In 1949, my Logic 101 Professor used the 1/3 solution to this problem as an example of bad logic which would be unacceptable in his class." So the problem is nearly 50 years old, and perhaps older, which answers Dennis Rakestraw's question above.

With More than One Interpretation, it's Ambiguous

I continue to get frequent letters claiming that one interpretation, or the other, is the best, or the only, interpretation of the question. In my mind, this is the very essence of ambiguity. In summary, here are the two most common interpretations of the original question:

Because the question does not state how we obtained the information, the question is ambiguous. And since Eldon challenged Marilyn to prove him wrong, which Marilyn has failed to do, Eldon does not owe $1000 to the American Heart Association.

Eldon's Web Page

Please visit Eldon's Web Page for further discussion on this topic. last updated August 15, 1999 by