**The Two Children Problem**

You meet a father and son, whose family was selected at random from among all two-child families where the eldest child is a boy. What are the chances that the man’s other child is also a boy?

The next day, you meet a different father and son, whose family was selected at random from among all two-child families with at least one boy. What are the chances that the man’s other child is also a boy?

For the purposes of both questions, assume that a child must be either a girl or a boy, that girls and boys exist in equal numbers in the population, and that the gender of a child is independent of the gender of the child’s sibling. If you think you have it figured out, read the answer and explanation below. Then scroll down for more boys and girls problems.

See Answer and ExplanationIt may seem like both questions should have the same answer, but that is not the case.

For the first question, the probability that the other child is also a boy is 1/2.

For the second question, the probability that the other child is also a boy is 1/3.

To understand why, begin with the following table, which shows the four gender-distribution possibilities for a two-child family, each of which has 25 percent probability:

Older Child | Younger Child | |

1 | Boy | Boy |

2 | Girl | Girl |

3 | Boy | Girl |

4 | Girl | Boy |

For the first question, with the information we are given, that the older child is a boy, we have narrowed down the possibilities to #1 or #3, so there is a 1/2 probability that the other child is also a boy.

For the second question, the information we are given, that one of the children is a boy, narrows down the possibilities only to #1, #3, or #4. Among those three choices, in two cases the other child is a girl and in one case the other child is a boy, so the probability that the other child is also a boy is 1/3.

**Another Boys and Girls Problem**

Assume that every pregnancy has an equal chance of producing a boy or a girl, and in a certain society, every family follows the policy of having as many children as they can until they have a boy, and then they stop having children. So a family that has a boy first will have only one child, but a family might have one, two, three, or more girls before having their first boy and then having no more children. In this society, after many generations, what will the ratio of boys to girls be?

See Answer

**The Four Children Problem**

Now suppose you hear of a family with four children, and you do not know the gender of any of them, but you wish to guess the most probable scenario. You reason that four boys or four girls is very unlikely. Three of one gender and one of another seems unlikely too. Since the probability for each child is 1/2, you decide the most likely breakdown is two girls and two boys. Are you correct?

See Answer

## Son Born on a Tuesday

Now suppose you encounter a mother and son, whose family was selected at random from among all two-child families with at least one boy born on a Tuesday. What is the probability that her other child is a boy?

Make the same assumptions as in the first problem, and also assume that births are equally likely on different days of the week.

See Answer

So while we might have been expecting the probability to be 14/28 or 1/2, we cannot count the “Tuesday boy – Tuesday boy” possibility twice. The real probability of two boys is therefore 13/27. This problem was devised by Gary Foshee.

Love your site! I disagree with your answer to the first question, though. Why is birth order relevant at all? In your chart, instead of “older child” and “younger child,” why not just “this child” and “that child?” Birth order is an arbitrary variable. You could just as easily have used height, weight, or any other distinguishing trait.

Glad you like the site, and thanks for the comment! I agree that using birth order as the distinguishing trait is somewhat arbitrary, and you could use another trait to achieve the same effect of narrowing the probability that the other child is a boy from 1/3 to 1/2. (Height or weight might be problematic since there are gender differences for those traits.)

The answer to both of the Two Children problems is 1/2. It does not matter how you identify the gender of one *specific* child. It can be “eldest”, “this one with me”, “named Aaron”, or even “sits to Mother’s right at the dinner table”. See https://philpapers.org/rec/BARSTC. (And order is “important” only because it provides an identification method.)

The point is that the child was chosen by that means, and not because he was a boy. In another two-child family, or on another day, that method could have chosen a girl. Even if she had a brother. So you can only count half of the boy+girl families – those days where the father would choose to take a walk with his son, and not his daughter. So fully 1/2 of what you should count have two boys.

In fact, this is the same reason why the answer to the Monty Hall Problem is 2/3. Those who say 1/2 are counting all possible games where Monty Hall can open Door #3, just like you are counting all possible families where the man has a son. You should only count half of the games, or families, where a different observation could be made by the implied method.

The Tuesday Boy Problem is the same. Unless there is a reason that, say, a father of a Tuesday Boy and a Thursday Girl cannot tell you that he has a Girl, you can only count half of every cell in your table EXCEPT the one with two Tuesday Boys. The answer is (1+12/2)/(1+26/2)=(7)/(14)=1/2.

Thank you for your comment! I think the original wording was ambiguous, and could allow for the answer to the second question to be 1/3 or 1/2, depending on interpretation. I have changed the wording to make the question more precise, which I hope makes the answer more clear, while still presenting an interesting puzzle.

The mechanisms you added do indeed make the original answers correct. But the original wordings of the “at least one boy” and “Tuesday boy” problems cannot be interpreted to have these mechanisms, so they were not ambiguous. If they were, the Monty Hall Problem would have the same ambiguity.

And the point is that your incorrect interpretation has been taught so often, that people who should know better believe it, and teach it in turn.