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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!

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  • $\begingroup$ I don't understand.. wasn't the whole point of the original 4 possibilities GG, GB, BG, BB already sorted by youngest and oldest? Otherwise what is the real difference between GB and BG? $\endgroup$ – Albert Renshaw Jul 16 '14 at 15:07
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    $\begingroup$ kind of, it's just a different explanation - you could say i need more elements of concrete thinking, say pedagogical, because i'm not used to mathematics. $\endgroup$ – vic Jul 17 '14 at 18:43
  • $\begingroup$ Ahh well the main problem here is that it's not just GG, GB, BG, BB ... if order matter like the difference between GB & BG then you also have to have the difference between the girls... girl_1 then girl_2 or girl_2 then girl_1, so your real data set is: gg, gg, gb, bg, bb, bb — Which still equates to 50% chance it's a boy, not 66.666% $\endgroup$ – Albert Renshaw Jul 17 '14 at 18:48
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The problem is not really with your reasoning as much as with your interpretation of the question.

In the first interpretation, we know that there is a girl, and then we pick one of them, and then ask what is the probability that the other child is a boy. In this case, since we know that there is a girl, there are as you say exactly 3 equally probable possibilities, GG,GB,BG. And hence we can check that the answer is $\frac{2}{3}$.

In the second interpretation, we pick a child at random, and then it turns out to be a girl, and then we ask what is the probability that the other child is a boy. In this case, there are exactly 2 equally probable possibilities, GG and GB, where the first letter denotes the gender of the picked child, and the second letter denotes the gender of the other child. Clearly the answer is $\frac{1}{2}$. (Your added "reasoning" about the picked child being elder or younger isn't important.)

So the underlying issue is that when specifying a probability question, it is absolutely crucial to clearly state the order in which actions are performed.

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  • $\begingroup$ oooh, yeah, so i pick a child, regardless if he is elder or younger, it drives me out of the "at least one" territory. I didn't realize i picked a child... i built the reasoning around a sampled child... i thought that, by examining each "child picked is a girl" situation, it would automatically add up to all equiprobables situations, which it didn't. Thanks ! $\endgroup$ – vic Jun 22 '14 at 16:35
  • $\begingroup$ @vic: You're welcome! By the way, you should upvote all helpful answers, and accept the one that helped you the most so that this question doesn't remain unanswered. $\endgroup$ – user21820 Jun 23 '14 at 14:17
  • $\begingroup$ i would gladly upvote all the answers here, but it says i don't have enough reputation to upvote (need 15). (also i didn't see that i could select the most helpful answer or something) $\endgroup$ – vic Jun 24 '14 at 15:35
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It looks like you're double counting the two girls case. If you assign 1/8 chance to G(L)(y) G(e) and another 1/8 to G(y) G(L)(e), then you have:

B(y) G(L)(e) with chance .25 G(y) G(L)(e) with chance .125 G(L)(y) B(e) with chance .25 G(L)(y) G(e) with chance .125

We've obviously excluded the double boy case, but we should condition on no boys. So then we have 2/3 chance of a brother.

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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.

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How exactly do we know one of the two is a girl? We don't just abstractly know it. There is some story how we received the information. If we meet a person in hospital visiting a sick child, and that is how we gained the information, there is an equal chance that we would have received the information "Ms Michu has at least one boy" or "Ms Michu has at least one girl". If we meet Ms Michu picking up her child at a girls' school, we couldn't have received the information "Ms Michu has at least one boy" because it's a girls' school and she had to pick up a girl.

People with two boys turn up at hospitals visiting a sick child. People with two boys don't turn up at girls' schools to pick up their child. So how we got the information influences the result.

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