Steve's son goes to an all-boys secondary school and Steve is going to collect him from school after his first day of school. Steve does not know any other parent yet and meets Richard at the gates, who also has a son in the same class. Richard tells Steve that he also has another child.

What is the probability that Richard's other child is also a boy?

Note: this is to try to better understand the Boy-girl paradox (https://en.wikipedia.org/wiki/Boy_or_Girl_paradox), which was pointed out to me in an anwser to one of my earlier questions

  • $\begingroup$ On an empirical note, people might also want to look into this. Although I think it is completely unrelated to the the answer that you are seeking. $\endgroup$ – Gumeo Sep 25 '15 at 16:42

With all of the reasonable necessary assumptions (equal probability of boy or girl, probability of sending one child to an all boy's school independent from gender to other child......) the probability that Richard's other child is also a boy is (drum-roll)

$$ \frac{1}{2}. $$

When you hear these problems a good thing to think is "can I label them in such a way that I know the gender of one of the labeled children?" In this case I will say the child in Steve's sons class is child A and the other child is child B. For simplicity later on I'll refer to boy as 1 and girl as 0. Our assumptions boil down to:

$$ P(A=x,B=y) = P(A=x) \cdot P(B=y) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}. $$

The question you are asking concerns $P(B=1 | A=1)$. By independence or direct computation we will easily get $1/2$.

The paradox focuses on the scenario where I know someone has two children, and I know they have at least one boy. No matter how I try to label the children, (youngest=A, oldest=B) (shortest=A, tallest=B) I don't know the gender of one of the labelled children. Notice I can not talk about the gender of "the other child" because I don't know of one specific child who is a boy. This is quite different than the other problem. Once i pick some arbitrary labelling, the quantity of interest mathematically is

$$ P(A+B=2 | A+B\geq 1) = \frac{1}{3}. $$

  • 1
    $\begingroup$ While the intuitive answer would look 1/2 to me as well, I can't find what's wrong in the below reasoning: The parents of an all-boys school have all at least one boy. If we suppose that Steve is only interested to speak with families with 2 children, speaking with a parent of that school does not add any new information to him. This is simply a family who has at least one boy. Hence if he would ask the same question to 10 different parents with 2 children, this would be a random sample of families with at least one boy, for which the probability ot having 2 boys is 1/3. $\endgroup$ – user90213 Sep 25 '15 at 19:01
  • $\begingroup$ @user90213 haha. I love how subtle this problem is. You said Richard has a son in Steve's son's class (I assume you meant exactly one). This allows me to label the students and I know A=1. If I walk up to a random two child parent there all I know is A+B is at least 1. $\endgroup$ – jlimahaverford Sep 25 '15 at 19:21
  • $\begingroup$ It would have been more tricky if you just saw a parent at the boy school gates, without knowing whether his son is in your son class or not. In that case I'd say the probability is somewhere in between 1/2 and 1/3, and it would depend on things like number of classes in the school, spacing distribution between childs etc $\endgroup$ – user90213 Sep 30 '15 at 11:23

It's 50% (or whatever the actual $p$ of a male vs female set is in the population drawn from, but 50% for simplicity.)

We know for a fact that this concrete child is a boy, so the only open situation is the gender (sex?) of the other child.

Another way to frame this particular question is:

"There is a child of unknown gender. What is the probability that the child is male?"

To illustrate the paradox, it would be better to examine a question like this:

From the families with exactly two children and at least one boy, what is the probability that one of these families chosen at random has two boys?

  • $\begingroup$ I'm woondering how the probabilty would change if Steve would have meet Richard in a boys clothing section of a department store. In that case I assume it would be more like 1/3? $\endgroup$ – user90213 Sep 30 '15 at 11:31

The probability is 1/2 cause P(girl|boy)=P(girl) cause they are independent and the fact that Richard has a boy doesn't give any useful information about the other child.

  • $\begingroup$ "Because they are independent" is precisely what needs to be shown. $\endgroup$ – whuber Oct 26 '15 at 22:52

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