Help me find the fallacy in these specific arguments re: the Boy-Girl Paradox [duplicate]

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).

Solution 1:

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

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