There is a principle which is a bar against all information, which is proof against all arguments and which cannot fail to keep a man in everlasting ignorance—that principle is contempt prior to investigation.” – Herbert Spencer

Our Question:

A woman and a man (unrelated) each have two children. At least one of the woman’s children is a boy, and the man’s older child is a boy. Do the chances that the woman has two boys equal the chances that the man has two boys?

Assumptions:

Boys and girls are equally likely.

Our problem statement is true.

Marilyn answers no, I answer yes. She is wrong; she is on the wrong side of the paradox - on the naive side of a counter-intuitive question - and refuses to know it. I told her I would send $1000 to her favorite charity if “my argument is wrong.” She said, “Eldon, I take the bet, you’re wrong, send the money to the American Heart Association.” To my knowledge she has not even looked at my argument. She offered the conventional argument, then has remained silent amid proofs of its falsity. Put her argument into logical steps; I’ll show you where the false statements are.

Marilyn’s argument: Marilyn says that there are four equally likely ways for a woman to have two children. (That is correct, prior to inspection, without prejudice toward boys, or girls) Then, Marilyn said that because we know that there is at least one boy, there are only three ways left. (That is also correct) Marilyn now thinks that the three remaining ways are still equally likely. (Obviously false)

Why obviously false? To have the three remaining ways be equally likely, boys must be selected over girls, prior to inspection. When boys are selected prior to inspection, we start with three equally likely. When we start with four equally likely, boys must have been selected after inspection.

Our statement, “at least one is a boy”, can’t be stated, as a true statement, prior to inspection. When boys are, or were selected, after inspection, then the remaining three ways are no longer equally likely.

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Eldon’s Argument: By the numbers

1.       There is a woman. All we know about the woman, we learned from the statement.

2.       The statement could not have been made, as a true statement, prior to some kind of investigation.

3.       Boys were chosen. Prior to the choosing, girls were equally likely to be chosen.

4.       The woman appeared, and then there was an investigation, then a choosing.

5.       When the choosing was after the investigation, the answer is, unambiguously, 1/2.

For years I have said that to disprove the above argument, you must disagree with 1. If you agree with 1, then the rest of the argument follows. Several have written and said, “I accept the first four, but disagree with 5. That just can’t be. Gemde is challenging 1. That’s honest, but wrong. 1 is a true statement. One of our assumptions is that the problem statement is true. Our problem statement talks of “the woman”. The woman exists. The statement was made about her.

Gemde also states that the problem statement says that the woman was chosen from amongst all the women who have two children and at least one boy. That’s false. It does not.

Consider a woman with a boy and a girl. At least one is a boy, and at least one is a girl are both true statements about her. If the “at least one is a boy” statement says that she was selected from bb, bg, gb, then “At least one is a girl” states that she was selected from gg, gb, bg. With one woman, that can’t be true.

Gemde isn’t thinking.

We arguing that on Marilyn’s website!!! I don’t know Gemde, she may not know better than that. I know that Marilyn would not make that simple mental error. It’s on her website. I wonder if she reads it. She has said nothing publicly on this question, to my knowledge, since the turn of the century. She may not care about the truth. She may speak up, or she may stay anonymously ignorant. It’s her website. It’s her reputation.

Consider the Conditional Probability Formula: P (A|B) = P(B|A)P(A) / P(B)

For problems of this type, there is a quarter (1/4) in the numerator. The devil resides in the denominator. B is defined as the event that has definitely happened; therefore P (B) is the probability that our event has happened. As the probability for A depends on ¼ divided by the probability for B, precisely defining what has happened, and what was the probability for that is imperative for answering our question. When B is precisely defined, the math is trivial.

There are four equally likely ways to have two children. Bb, bg, gb, and gg.

Therefore:

P (B) = P (B|bb) + P (B|bg) + P (B|gb) + P (B|gg)

The argument is always over what happened at bg and gb: and WHY!

When P (B) = (1/4|bb) + (1/4|bg) + (1/4|gb) + (0|gg) = ¾, then our answer one quarter divided by three quarter = one third.

When P (B) = (0|bb) + (0|bg) + (0|gb) + (1/4|gg) = 1/4, then our answer one quarter divided by one quarter = one.

When P (B) = (1/4|bb) + (1/4*1/2|bg) + (1/4|gb) + P (0|gg) = 1/2, then our answer one quarter divided by one half = one half

Those are three mathematical solutions. It is undeniable that the first family has at least one and ½ odds for two boys.

It is undeniable that the second family has at least one boy and probability for two boys of one.

The third family has at least one boy and chances for two boys of ½.

Those three families and their outcomes are undeniable. Understanding who they are, and why they got there seems a little tricky. Most one thirder’s don’t wish to go there. Most leave before they get this deep.

The first family had a success with probability one at bb, bg, and gb.

The second family had a success with probability one at bb.

The third family had a success with probability one at bb, probability ½ at bg, and gb.

Success? What is success? Who defined it and where? Understand these three families and it’ll go a long way toward understanding Conditional Probability.

Understand? Understand what happened, with each family, at bg and gb!!

How did the first family get a success with probability one? Why did the second family have no success? The third family had a success ratio of one at bb, ½ at bg and gb. How did this happen? Who, and or what defines success? Understand this, then you can answer Our Question.

Hint: With the first family, the maker of the question was looking for a boy and found one, with the second, looking for a girl and didn’t. With the third family the writer took what came, and reported on it.

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Another argument: September 10, 2006

There are four ways to have two children: bb, bg, gb, gg

A family has two children: That’s all we know.

Statement A: At least one of the children is a boy.

Statement B. At least one of the children is a girl.

Prior to investigation, the probability that A and B are both true is ½.
Neither statement can be stated, as a true statement, prior to investigation. Therefore, we know that when we hear one of those statements, there has been some kind of investigation.
When we hear one of those statements, does the probability that they are both true go up to 2/3? !!!Certainly not!! But this is what Marilyn is claiming.
When we hear one, then the probability for both to be true is still ½.
When the probability for both to be true is ½, the probability for two of a kind is ½. That’s the woman of our question.

Eldon’s $1000.00 challenge:

Prove my logic wrong and I’ll send the money to the American Heart Association in your honor. (Whoever you are but disprove the logic, typos don’t count). My argument is correct, or one of those five statements is false. To claim the challenge, show me a false statement. Which one of those five statements can you prove false? Email me Eldon Moritz . In the meantime, everyone, right or wrong, feel free to send money to the American Heart Association.

Niles said: Let's face it. I've got to make some restriction on the population of women with two children upon which to select randomly. If I don't make a restriction, I select randomly from the general population of women with two children, and I get boy-boy with probability 1/4. Nobody believes that is the answer is 1/4.

I called this a false statement. All we know is that the statement was made about the woman. Because of the statement we know that she has bb and the statement was made about her, or she has bg and the statement was made about her, or she has gb and the statement was made about her. She can have come from the general population.

Email me? Certainly. But first, understand that two coins flip four equally likely ways. To flip two coins three equally likely ways, the null must be selected, prior to the flip. As Martin Gardner said, “the flipper must agree in advance…”

Next understand that when the writer of a question thinks one question, but writes another, the written question prevails.

Political Question? This is not a political question, it is not a matter of opinion. This is a simple little logic question, it has an answer and Marilyn is wrong.

Marilyn vs Marilyn! A textbook example with Marilyn on the other side of the fence.

A Maths Solution! With help from Dr. Terry Moore of New Zealand 12/12/'03.

How to win!! Send me a working model that works. Do not send me Marilyn's silly little, incorrect, argument. (several people have done that already)

Moritz Gray Paper! In this paper Dr Gray put my argument into mathematical terms. This will show you that I'm not operating entirely without adult supervision.

The Experts can be wrong!!

Marilyn is wrong on this one but so far has refused to discuss it. She defers to Martin Gardner as her expert but on this one Martin agrees with me.

Martin Gardner said that he has spent too much time on this question and doesn’t wish to discuss it. I respected his wishes but he has written several times on said question and his writings agree with me. Martin said that when you select a family from the short group, then you get one third. I agree with this, whether it be the man’s or the woman’s family. In his book, Aha Gotcha! , he wrote that to get 1/3 when flipping two coins the flipper had to agree, in advance to always call heads when possible and to reflip on two tails. This agrees with my logic. This advance agreement is between the asker of the question and the answerer. As our question is a written question, this agreement must be in writing, in our statement. As it is not, then our answer is 1/2.

Eldon's homepage

For those of you who came from the wiskit, probability of boys and wish to return. Hello, thanks for coming, good bye and God Bless.

I appreciate Herb Weiner, writer of the The Wiskit. I disagree with much of his logic.

Herb specializes in finding Marilyn wrong. I don't care about Marilyn being wrong, she just stumbled in on the incorrect side of my question. My interest is in getting this question correct. Marilyn is smart enough to understand the logic. I have no evidence that she has looked at any of my arguments.

Herb sees that my answer is correct but then he sees an ambiguity. What he sees as an ambiguity is actually an incorrect assumption. He wrongly assumes that the gender decision was made. As per his homepage he bungee jumped off The Royal Gorge Bridge. Methinks he went A Bridge Too Far.

Return to the wiskit, Marilyn is Wrong page.

Nick Ratti of Bristol RI

Nick understands this question, maybe better than I do. He says that Marilyn has committed the "Fallacy of Equivocation", or "Sophismata Equivocatio" (Aristotle). He says that, "Most, if not all basic logic textbooks explain it, and at least one of them uses this very problem to illustrate it, clearly and indisputably enough for any Freshman to "get it".

My apologies to Nick Ratti

I quoted Nick in a way which he didn’t wish to be quoted. I’ll print any retraction which he sends. I think it’s a moot point as not very many people read this page.

Nick has written:

For your info, Marilyn’s trail of “authority” has led to yet another blind alley. She told me to get the proof of her answer from Gardner. He declined to do so. When I told him that that means he doesn’t have one, he told me to get it from a Stanford professor named Persi Diaconis. Diaconis has declined to give it to me. The blind alley has gone far enough to prove that her answer is based on no proof and false attributions. You may quote me on that.

Eldon Moritz sticks by his argument. It is correct.

References:

I reference these books:

St. Martin’s Press, 175 Fifth Avenue, New York, N.Y. 10010

Aha! Gotcha: paradoxes to puzzle and delight. By Martin Gardner

Twenty-first printing, 2000