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I'm wrestling with the classic problem called the Boy-Girl Paradox in the following simple form: Suppose you know that a given family has at least one girl. What is the probability that the other child is a boy? (We're assuming boys and girls are equally likely, so boy-boy, girl-girl, boy-girl, and girl-boy each make up 1/4 of the population.)

I'm going to present the usual argument given for the answer being 2/3. Then I'll present a counter-argument which seems to me to show that the usual argument leads to contradiction.

Note: My goal here is not necessarily to just understand the problem in general. I would like someone to show me specifically where and how my counter-argument fails, because I find it equally convincing as the usual argument. I've been mulling this for a couple days and can't find the fatal flaw in either argument, and yet they lead to contradictory conclusions, so one of them must be rotten.

(For brevity/clarity, I'll refer to a family with one boy and one girl as "mixed". So the problem is: given that you know that at least one of the family's children is a girl, what is the probability that the family is mixed?)

Usual argument: Since you know one of the children is a girl, you can eliminate all boy-boy families, leaving only the girl-boy, boy-girl, and girl-girl families. So that's 3 equally likely possibilities for which family this girl could be from, 2 of which are mixed, so the probability that her family is mixed is 2/3.

Counter-argument: Take the usual argument and interchange "girl" and "boy". This leads you to conclude that if you know that at least one child is a boy, then the probability that the boy's family is mixed is 2/3. So whether you start out knowing "at least one girl" or "at least one boy", you get the same answer. But because every family has either at least one boy or at least one girl, this means that the answer must apply to the whole population. Because say you pick a child at random. That child will be either a boy or a girl. But we've seen that in either case, if the first argument is sound, we'll find that it's a 2/3 chance the family is mixed. But this is just a way of saying that the chance of mixed family is 2/3 over the whole population, i.e. no matter which child we pick. But we know this is false, because only 1/2 the families are mixed.

A restatement of the counter-argument: Suppose the usual argument is sound. Then given at least one girl, P(mixed|girl) = 2/3. But by symmetry under interchange of boy and girl, we have P(mixed|boy) = 2/3. But since we get the same P(mixed) in both cases (and those cases exhaust all possibilities), this means knowing the sex of the child actually gives us no new information about P(mixed), over and above what we already knew about the population as a whole, which is that P(mixed) = 1/2. Therefore learning the sex of the child can't possibly change our prior evaluation of P(mixed).

RobPratt
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nothingIsMere
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    These things often come down to "how do you know the given information?" In the usual statement, I think the best answer to that is "you asked some knowledgeable observer if that family had at least one daughter and received the answer Yes." Use this interpretation to sort out your computation. In particular, your "pick a child at random" methodology is radically different. – lulu Dec 24 '21 at 17:39
  • Side note: I don't understand why you see your argument as contradicting the usual solution. If you are told that the family has at least one daughter, that rules out $BB$ so it is certainly new information. Similarly, if you are told that it has at least one son, that rules out $GG$, so again new information. – lulu Dec 24 '21 at 17:42
  • That is new information in a sense. Just like learning the sex of the specific child is new information. But it is not information that is relevant to P(mixed). That is the thrust of my counter-argument. Yes, if you learn that one child is a girl (boy), you eliminate BB (GG), but this elimination leads you to the same answer either way. Therefore learning it can't change your mind from the P(mixed) = 1/2 that you had before you learned it. – nothingIsMere Dec 24 '21 at 17:49
  • Again, the "specific child" thing is a misunderstanding. If the situation was "of the two children we selected one uniformly at random and observed it to be a girl, what's the probability that the other child is also a girl?" the answer is $\frac 12$, of course. Each child has equal chance of being of either gender (for probability purposes). – lulu Dec 24 '21 at 17:54
  • I think it clarifies things enormously if you write out exactly how the information came into your possession. If you write that out clearly and completely, I believe all confusion should fade away. – lulu Dec 24 '21 at 17:56
  • I'll spend some time with this prompt, but I have to say I have no idea what you could mean by "writing out how the information came into my possession" or why that would be relevant. It seems quite cryptic. Can you just state clearly what you think the fallacy is in my argument? – nothingIsMere Dec 24 '21 at 18:02
  • The usual argument makes no appeal to 'how the information came into your possession." All it stipulates is that you know at least one of the children is a girl and gets its conclusion from there. So I'm not sure why you're pressing that issue. – nothingIsMere Dec 24 '21 at 18:04
  • I have done so already, and I have given a clear example of what I meant (in my first comment). The fallacy in your argument is that it is imprecise. If you define precisely the events you are conditioning on, there is no contradiction. – lulu Dec 24 '21 at 18:04
  • To stress: the usual statement of the "paradox" absolutely emphasizes the source of the information. If the source is "you asked if the family had at least one daughter and were told Yes" then the answer is $\frac 23$. If, alternatively, you specified one child and learned it was a girl, then the answer is $\frac 12$. The apparent, but not actual, equivalence of these two formulations is why people consider this a paradox. – lulu Dec 24 '21 at 18:06
  • Hm. That is a dimension of the problem I hadn't considered. I've only ever seen presentations that didn't address the issue of how you obtained the knowledge, they simply assumed you had it to start with. I'll explore that, thank you. – nothingIsMere Dec 24 '21 at 18:10
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    Does this answer your question? In a family with two children, what are the chances, if one of the children is a girl, that both children are girls? – Peter O. Dec 24 '21 at 20:33

1 Answers1

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The mistake in your arguement is assuming the event a family has a boy is the compliment/inverse of the event of a family having a girl.

If they were compliments, then you could write

$$P(mixed) = P(mixed\mid boy)P(boy) + P(mixed\mid girl)P(girl)$$

And with the intuition you described, you get $P(mixed)=2/3$. However, the events of having a boy and having a girl are not exclusive, and thus do not make a partition of the event space. That is why the above formula does not work.

Modifying the above formula to account for this using the inclusion-exclusion principle, we get

$$P(mixed) = P(mixed\mid boy)P(boy) + P(mixed\mid girl)P(girl) - P(mixed\mid boy\cap girl)P(boy\cap girl)$$ $$P(mixed)=(2/3)(3/4)+(2/3)(3/4)-(1)P(mixed)$$ $$2P(mixed)=1$$ $$P(mixed)=1/2$$

Angelica
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