Two children paradox: where is my reasoning wrong? I hope here is the good place to be asking this. Apologies otherwise.
The statement is as follows:
"Ms Michu has two children. We know one of the two is a girl, we call that girl Ludivine. What is the probability that Ludivine has a brother, rather than a sister?"
This is with the usual assumptions that there is an equal chance for one person to be a boy or a girl, no cis or gemels, etc.
I read that the answer should be 2/3. Demonstration is :
For any given set of 2 children, there are 4 different equiprobable combinations:
GG BB BG GB
We get rid of the bb combination because there is no girl. That leaves us with 3 possibilities, of which 2 match the criteria of the statement, hence the probability of Ludivine having a brother is of 2/3.
But, if I follow the following reasoning, I obtain a probability of 1/2:
Ludivine(L) is either the elder(e) or the youngest(y), with an equal probability of 1/2.
If she is the elder(e), it gives 2 combinations, with again a probability of 1/2 for each:
B(y) G(L)(e)
G(y) G(L)(e)
If she is the youngest we again have 2 equiprobable combination :
G(L)(y) B(e)
G(L)(y) G(e)
By combining those probabilities, we have 2 cases out of 4 where she has brother. Therefore she has a probability of 2/4 or 1/2 to have a brother.
Now I feel like my reasoning is wrong somewhere, but I can't see where or why, even though i went through quite a few topics on the subject; hence why I'm re-discussing that old topic...
Thanks in advance!

The probability of the situation you have described as "B(y) G(L)(e)" is $\frac{1}{4}$, as you have said.  But the probability of "G(y) G(L)(e)" is $\frac{1}{8}$: probability that elder child is a girl, $\frac{1}{2}$; probability that younger child is a girl, $\frac{1}{2}$; probability that you give the name Ludivine to the elder of the two girls, $\frac{1}{2}$.  Therefore these two events are not equiprobable.
