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Good morning,

Here's an issue I was told some years ago by a maths professor, and I can't figure out what is actually going on, so I wanted to ask you if you can help me. Imagine there is family from which you know that it has two children, but you can't recall their gender. Both genders, male and female, shall be equally probable.

1) You are told that one of the children is female. What is the probability that the other is male? Answer: There are four possibilities: First child male, second child male (MM), first child male, second child female (MF), FM, FF. There first one is ruled out by the assumption that one is female, the others are equally probable. As two of them contain a male and one a female, the probability that the other is male is 2/3 (and not 1/2, as one might think).

2) Now it is getting pretty weird: You are told that one is female and born on a monday. What is the probability now that the other one is male? Answer: Now, there are 14*14 possibilities: First child female born on monday, second child female born on monday (F1F1); F1F2; F1F3;...; F1F7; F1M1;...; F1M7; F2F1;...; F2M7; F3F1;...; F7M7; M1F1;...; M7M7. Every possibility without a female child on a monday is ruled out, so there are 27 remaining; F1F1, ... F1M7,F2F1,...,M7F1. 14 of these contain a male child, 13 a female, and all are equally probable, so the final probability is 14/27 ???

3) (That one is by me). Imagine you are given that one of the children is female and the exact time of birth, e.g. 10am, 57 minutes, 12 seconds,...; up to some interval Dt. If you take Dt-> 0, the probability should get arbitrarily close to 1/2

I know thats all pretty messed up. Do you have any idea on how to fix it?

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  • $\begingroup$ What's your question exactly? The probabilities that you end up with in part 1) and 2) are correct, even though your reasoning at part 2) is a bit unclear. Regarding part 3); what makes you think that the probability should get arbitrarily close to $\frac{1}{2}$? (because it shouldn't) $\endgroup$ – titusAdam Nov 1 '17 at 11:32
  • $\begingroup$ A relevant problem that supports your answer to 3): math.stackexchange.com/questions/2497435/… $\endgroup$ – awkward Nov 1 '17 at 11:38
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There is a distinction here between being told one of the children is female and then being given some extra information (day/time of birth), and the issue your professor is talking about where you ask about both things at once.

Imagine you live in a town where every family has exactly two children (who are not twins). You go knocking on doors and asking the following.

Scenario 1

You ask "Do you have a daughter who was born on a Monday?"

Most people will say "no". In fact, most families won't even have any children born on Mondays. Of the families that have one child born on Monday and one not (this is $\frac{12}{49}$), half of those won't have a daughter born on a Monday, and of the half that do, their other child is equally likely to be a boy or a girl. The only factor that pushes the probability away from $\frac12$ is the small proportion of families where both children were born on a Monday. So the probability will be close to $\frac 12$.

(In fact, of $196$ families you'd expect $12$ with a girl born on a Monday and a girl born on another day, $12$ with a girl born on a Monday and a boy born on another day, $1$ with two girls born on Mondays, and $2$ with a girl born on a Monday and a boy born on a Monday. So the probability is $\frac{14}{27}$ as you say.)

Scenario 2

You ask "Do you have a daughter?" If the answer is "yes", you then ask what day she was born on.

In this scenario, most people ($\frac34$) will say "yes". You then get no extra (useful) information from the second question, since you don't eliminate any families. Since exactly the same families are involved as if you only asked your first question, the probability of the other child being a boy is the same as your option 1) - $\frac23$.


For your option 3), to do this like scenario 1 what you need to do is pick a time window (say 00:00-00:01) and go round asking everyone whether they have a daughter born in that time window. It will take you a very long time to get a "yes", since this is a very unlikely event (even assuming everyone knows the exact time of birth of each of their children). But if/when you do get a "yes", the chance of the other child being a boy is almost exactly $\frac12$, since the chance of that family having two children born in the time window you specify is extremely low.

However, if you get your information as in scenario 2 - by finding someone who has a daughter and then asking for a time interval - the extra information about the time is irrelevant and the probability is still $\frac23$.

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