The Woman With Two Children Hustle

A woman has two children, at least one of whom is a boy. What is the probability that the
other child also is a boy?

This is a classic puzzle that has been discussed by Martin Gardner, Marilyn Vos Savant, and
others. The classic solution is as follows:

If the woman's first child was a girl, the second must have been a boy. If the first child
was a boy, the second could have been either a girl or a boy; therefore there are three
equally likely possibilities, BB, BG, and GB. The probability that the second child is a boy
therefore is 1/3.

Having read this problem and its solution, a huckster thinks he can make some money from it.
He stops a man in the street, we'll call him the chump, and asks him to nod his head if he
has a brother or sister but not to tell him whether his sibling is male or female. If he
gets a positive response, he offers to bet the chump three dollars against his two if he can
guess which sex his sibling is.

The huckster reasons: If I guess female I am right with 2/3 probability and make $2 and I am
wrong I lose $3 with 1/3 probability so my average gain per bet is $0.33.

The chump reasons: As far as he knows, I am equally likely to have a brother as a sister, so
he has a 50-50 chance of being right. Therefore I stand to win $3 with probability 1/2 as
against losing $2 with probabilty 1/2, so I can expect to win $0.50 on this bet.

Who is right, the huckster or the chump? Of course the huckster can work this on women as
well, merely guessing the opposite sex.

The chump because all he has to do is lie!

The chump. The huckster improperly concludes that the probabilities are the same as the classic puzzle. In reality, by fixing one of the siblings (the chump, a male), one of the three possibilities has been eliminated: the chump's mother could not have had "at least one boy" in the combination "female chump + chump's brother." Ruling that out reduces the possibility to two: the chump (a male) has either a sister or a brother.

MathExtremist: I don't understand what a "female chump" or "chump's brother" are in this example. The chump is a male and the chump's brother can be paired only with the chump himself and not with his sister. What you must be saying is that the chump can have either an older or younger sister but not both. In the literature it is generally agreed that how one learns that at least one of the children is a boy affects whether the probability is 1/3 or 1/2.

Quote: puzzlenut

MathExtremist: I don't understand what a "female chump" or "chump's brother" are in this example. The chump is a male and the chump's brother can be paired only with the chump himself and not with his sister. What you must be saying is that the chump can have either an older or younger sister but not both. In the literature it is generally agreed that how one learns that at least one of the children is a boy affects whether the probability is 1/3 or 1/2.

I think we're assuming that the chump has exactly one sibling, so it's not possible for the chump to say "yes, I have a sister" and still also have a brother, or vice-versa.

Perhaps my explanation was inartfully stated, so let me try again. There is a difference in the amount of information you have between the two possible phrasings:
a) A woman has two children, at least one of whom is a boy
and
b) My mother has two children, at least one of whom is a boy (and I am male)

The initial puzzle was phrasing A, but the huckster/chump scenario is phrasing B. In B, one of the three arrangements is not possible; it is not possible for the chump's sibling to be male but not the chump because the chump actually is male. The relative ages of the children doesn't matter. What matters is that you've identified the gender of one particular child merely by talking to the chump.

The gender of the chump's sibling is irrelevant to the gender of the chump and the order of birth. The huckster is incorrect. If the chump did not exist then the chump's sibling could be any gender with the same probability. Knowing the chump's gender adds no information. 50/50

MathExtremist: Bravo! Much more understandable, but it might not convince everyone. I have a simpler explanation in mind. I think you are saying that it makes a difference whether you learn that at least one of the children is a boy from the boy himself or from some other source of information, even though the information is the same in either case. Marilyn never adopted this point of view but instead conducted a poll of her readers and found that among those families who had two children at least one of which was a boy the percentage of those with two boys was 35%. I would be interested to know whether or not you agree with Marilyn's solution to the original problem, that is Pr(bb) = 1/3.

I have a problem with your phrase "it is not possible for the chump's sibling to be male but not the chump because the chump actually is male." It sounds like you are saying that it is not possible for the chump to have a brother, nor is it possible for him to be male because that's what he is.

For those who like math, here is a more mathematical approach to the problem.

The probabilities of various family combinations are given by the binomial distribution. For two children:

(b + g)² = bb + 2bg + gg

If at least one of the children is a boy, then gg is eliminated and bg is twice as likely as bb, making Pr(bb) = 1/3.

We now are in a position to approach the following problem:

A woman has three children, at least one of which is a boy. What is the probability that at least one of them is a girl? I'm not sure that Marilyn could handle this variation, but we may proceed as follows:

(b + g)³ = bbb + 3bbg + 3bgg + ggg

ggg can be eliminated. Both bbg and bgg are favorable to at least one of them being a girl, giving a Pr of 6/7. This could be extended indefinitely.

"One of them is a boy."

"This one is a boy."

Different statements, different chances.

Quote: puzzlenut

MathExtremist: Bravo! Much more understandable, but it might not convince everyone. I have a simpler explanation in mind. I think you are saying that it makes a difference whether you learn that at least one of the children is a boy from the boy himself or from some other source of information, even though the information is the same in either case.

No, the information isn't the same in either case. In the first case, you're told that "at least one of the children is a boy." But you haven't actually been told that in the second case: on the street, by talking to the chump, you've learned that one specific child (the chump) is a boy. In the first problem, the possibilities are (BB, BG, GB). In the second, the possibilities are only (BB, BG).

That problem reminds me of this one:
I have two U.S. coins totaling thirty cents. One of them is not a nickel. What are the two coins?

Quote: puzzlenut

Who is right, the huckster or the chump?

The chump is right. From his perspective as far as all information revealed, he could either have a sister or a brother. Simple as that, the huckster will lose his bets.

On the contrairy, the huckster could offer an 1001:1000 bet on the gender, and could always pick the same gender. He could then be fishing for identical twins which occur at a rate fraction of about .3%. Although a real identical twin might simply decline the game....

Suppose I change the problem to say that the huckster knocks on doors and poses the question only if a man answers instead of accosting people in the street. Who has the advantage then?

The average neighborhood has households with equal numbers of bb, bg, gb, and gg siblings. If he passes up the gg households and guesses the opposite sex he will be right two thirds of the time. Why is this different from accosting people in the street? I am assuming that both siblings live at home and that one of them answers the door.

I predict that nearly everyone will say that the chump still has the advantage in such a case.

Its not.

Quote: puzzlenut

Suppose I change the problem to say that the huckster knocks on doors and poses the question only if a man answers instead of accosting people in the street. Who has the advantage then?

The average neighborhood has households with equal numbers of bb, bg, gb, and gg siblings. If he passes up the gg households and guesses the opposite sex he will be right two thirds of the time. Why is this different from accosting people in the street? I am assuming that both siblings live at home and that one of them answers the door.

I predict that nearly everyone will say that the chump still has the advantage in such a case.

I won't. To make the problem and discussion simpler, let's assume there is a law where every family must have two children.

If a male answers the door there are four types of men it could be. Let's label the possibilities: b1-b2, b3-g1, g2-b4, g3-g4.

If b1 or b2 answers the door, his sibling is male. If b3 or b4 answers the door his sibling is female. Thus, 50% chance the other sibling is male.

I think a more exploitable problem would the two coin problem.

Quote: puzzlenut

Suppose I change the problem to say that the huckster knocks on doors and poses the question only if a man answers instead of accosting people in the street. Who has the advantage then?

The average neighborhood has households with equal numbers of bb, bg, gb, and gg siblings. If he passes up the gg households and guesses the opposite sex he will be right two thirds of the time. Why is this different from accosting people in the street? I am assuming that both siblings live at home and that one of them answers the door.

I predict that nearly everyone will say that the chump still has the advantage in such a case.

First, I want to say that this is a pretty good puzzle. It almost tricked me! I had it right, then I changed my mind and had it wrong, then I realized why it was wrong, and got it right again. Please keep posting stuff like this; I love it!

The Wizard gave a good explanation of the correct answer, but let me explain it differently in a way that directly refutes your argument. By ignoring the times when a girl answers the door, he is not ONLY skipping the gg households. He is skipping the gg households, half the bg households, and half the gb households. So you have to give half-weight to bg and gb, which is why you get 1/2 -- it is 1 / (1 + 0.5 + 0.5)

Quote: AxiomOfChoice

First, I want to say that this is a pretty good puzzle. It almost tricked me! I had it right, then I changed my mind and had it wrong, then I realized why it was wrong, and got it right again. Please keep posting stuff like this; I love it!

Thanks for the encouragement! Please see the new puzzle "The Bridge Hands With an Ace" in this section.