What is the problem in this solution to the Two Child Problem?

For a more intuitive reason than the formal answer:

Let's say I use your first two bulleted statements to guess how often both children are female.

When the older child is female, half the time I will guess that both children are female.

Similarly, when the younger child is female, half the time I will guess that both children are female.

Seems consistent, right? However, the logic fails when both children turn out to be girls! In that case, both of your probability statements fire at the same time, and the problem is that I might end up saying "Both children are girls" twice in the same universe. When that happens, we're overcounting the number of times that both children are girls, which is why your answer ended up too large.

You play a lottery. One of the numbers 1,2,3 will be chosen uniformly at random, and you win if 1 is chosen. You have probability 1/3 of winning, but this argument allows you to conclude it's 1/2:

1. If 3 was not chosen, then it's 1 or 2, so you have 1/2 chance
2. If 2 was not chosen, then it's 1 or 3, so you have 1/2 chance
3. Since either 2 or 3 is not chosen, in both cases you have 1/2 chance

or:

You've got a five-sided dice with numbers 1,2,3,4,5. The probability of getting odd number is 3/5, but with this argument you can get wrong result - 2/3:

1. If you the number is at most 3, then possibilities are 1,2,3 so 2/3 chance of being odd
2. If you the number is at least 3, then possibilities are 3,4,5 so 2/3 chance of being odd
3. Either the number is at most 3 or at least 3, and in both cases the chance is 2/3

The error in both examples is that the cases are not disjoint. To reason about probability based on conditional probabilities, you're implicitly using law of total probability which requires that the space is split into disjoint sets.

The first example is exactly the same as boy-girl situation, where 1=GG, 2=BG, 3=GB. More dramatically, you could do the lottery with numbers 1-100 (only 1% chance of winning) and the argument would give you 50% chance of winning.

Others have dealt with the specific error in your reasoning. I want to point out that the problem is not well-defined: without more information it is impossible to say what the probability is that both children are girls. In particular, it not true that the answer is definitely $1/3$.

Consider the following three scenarios. In each of them I choose one family at random from the pool of all families with exactly two children, in such a way that each such family is equally likely to be chosen, and then I make one of the following two statements to you:

Statement 1: At least one of the children is a girl.

Statement 2: At least one of the children is a boy.

Scenario A: If the older child is a girl, I make Statement 1; if the older child is a boy, I make statement 2.

Scenario B: If both children are girls, I make Statement 1; otherwise, I make Statement 2.

Scenario C: If both children are boys, I make Statement 2; otherwise, I make Statement 1.

Suppose that I’ve made Statement 1: I’ve told you that at least one of the children is a girl. What is the probability that both are girls?

Scenario A: The older child is definitely a girl, so the probability that both are girls is simply the probability that the younger child is a girl, which is $1/2$.

Scenario B: Both children are definitely girls, or I’d have made Statement 2, so the probability that both are girls is $1$.

Scenario C: The only possibility that is ruled out is that both are boys. Three possibilities remain: both are girls; the elder is a girl and the younger a boy; the elder is a boy and the younger a girl. These are equally likely, so the probability that both are girls is $1/3$.

In short, the probability depends on what I would have said had I chosen a different family. Using more complicated decision rules, I could arrange scenarios in which the probability that both children are girls has other values. Here’s an example:

Scenario D: I roll a fair die. If both children are girls, I make Statement 1. If both are boys, I make Statement 2. If one is a boy and the other a girl, I make Statement 1 if the die comes up $6$; otherwise, I make Statement 2.

There are $6\cdot 2\cdot 2=24$ equally likely events, one for each combination of die roll, sex of older child, and sex of younger child. I’ll represent a specific combination by a string like $3BG$, meaning that I rolled a $3$, the elder child is a boy, and the younger child is a girl. The fact that I made Statement 1 rules out six of these $24$ events: $1BB,2BB,3BB,4BB,5BB$, and $6BB$. It also rules out all five $nBG$ events in which $n\ne 6$ and all five $nGB$ events in which $n\ne 6$. The only possible events, therefore, are the six $nGG$ events, $6BG$, and $6GB$. In six of these eight equally probable cases both children are girls, so the probability that both are girls in this scenario is $6/8=3/4$.