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The question itself is a very easy one:

Somebody has got two kids, one of whom is a girl. Then what's the probability that he's got at least one boy?

My answer is that, since he's already got a girl, then "he's got at least one boy" amounts to "the other kid is a boy", whose probability is apparently $\frac{1}{2}$.
But my friends argue that the probability should be $\frac23$: they say this is a binomial distribution, all the possible cases are (girl,girl),(girl,boy),(boy,girl) which yields that the probability is two cases out of three and is thus $\frac23$.
But I think this is totally unacceptable. I don't think it is a binomial distribution at all, at least not what my friends explained to me. However, I just can't disuade them of their opinion, nor can I prove that I am wrong.
So what on earth is the probability? and why? Any help is appreciated. Thanks in advance.

Esp. Can anybody show why my explanation is wrong? Isn't it that whether the other kid is a boy or a girl a 50/50 event?
EDIT:
Thanks for all the help you provided for me, and special thanks will go to @HammyTheGreek and @KSmarts, who have made it clear to me that there is in fact some ambiguity in my statement in this problem.
As is pointed in this link ,two distinct interpretation of the statement "one of whom is a girl" that gives rise to ambiguity:

From all families with two children, at least one of whom is a boy, a family is chosen at random. This would yield the answer of 1/3.
From all families with two children, one child is selected at random, and the sex of that child is specified to be a boy. This would yield an answer of 1/2.

Vim
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6 Answers6

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I'll try to explain without mentioning conditional probabilities or binomial distributions explicitly:

Having two kids, you can have two boys, two girls, or one girl and a boy. However those 3 possibilities don't have equal probabilities. Two get two boys, both your first and second child have to be boys, chance $\frac 1 4$. To get two girls, both your first and second child have to be girls, chance $\frac 1 4$. Now to get a boy and a girl, you can either first get a boy, then a girl, or first get a girl, then a boy, chance $\frac 1 2$.

Now assume we know that somebody has two kids, and we know one of them is a girl. We are left with the possibilietes of two girls or a boy and a girl, the second still having twice the probability. Hence the probability of that somebody also having a boy is $\frac 2 3$.

Christoph
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All of the answers so far explain why you are wrong. However, strictly speaking, you are not wrong. You are simply making different assumptions and interpreting the question differently than other people.

The question tells us that someone has (exactly) two children, (at least) one of whom is a girl. There is unstated information here that is necessary to answer the question, and thus the question is ambiguous. How did we select this "someone" and how did we find out that one child is a girl?

The existing answers treat it like this: we gather all people who have two children and at least one girl, and select one of them. In this situation, we consider the three situations of older boy, younger girl; older girl, younger boy; and two girls. This gives us a probability of $2/3$ that the other child is a boy.

Consider this alternate presentation, though. We select one person with two children (sex unknown). Then we determine that one of the children, randomly selected, is a girl. We still have the same three possible cases as before. However, in each of the two cases where there is one boy and one girl, there is only a $1/2$ chance that the child selected would be a girl, whereas if there are two girls, we will always select a girl. So, in this situation, there is a $1/2$ chance that the other child is a girl.

This question (or some form of it) is known as the Two-Child Problem, or the Boy or Girl Paradox. If you think this is frustrating or confusing, you should probably avoid the similar Sleeping Beauty problem.

KSmarts
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Imagine all the people in the world with two kids separated into 4 groups. Those with (B,B), (B,G), (G,B), (G,G).

How many of them could say one of their kids is a girl? (75%) And of those, what is the probability that the other child is a boy? You see that it is $\frac{2}{3}$

ETA: Note your line of thinking would be correct had the speaker specified that the OLDER or YOUNGER child was a girl.

turkeyhundt
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  • How can you show that the four cases are equally possible under the problem's context? – Vim Feb 03 '15 at 09:06
  • You can think of a probability tree. There is a 50/50 split on the gender of the first child and then from these two branches, there is a 50/50 split on the gender again. Making 25% BB, 25% GG, and 50% BG(in any order). That's probably where the confusion comes in. Not accounting for the fact that the world has twice as many 2 child families that are mixed than there are GG (assuming all births are 50/50) – turkeyhundt Feb 03 '15 at 09:09
  • @TheArtist If you choose to lump BG and GB together as one group, then you have to concede that that group is twice as big as the GG group. That is another way to look at it. – turkeyhundt Feb 03 '15 at 09:10
  • Your explanation sounds reasonable. But I can't prove why mine is wrong. Isn't it that the other kid's gender is completely independent from Jane's? – Vim Feb 03 '15 at 09:14
  • Because there is a 50% chance the speaker is referring to his older child and a 50% chance he is referring to his younger child. That's where your method falls. – turkeyhundt Feb 03 '15 at 09:16
  • If you ask "How many of them could say one of their kids is a girl named Jane?", the percentage is definitely waaaaaay lower than $;75$% , as there surely are millions of people with two kids and one being a girl but not named "Jane" . I think you meant to ask "how many of them can say they have at least one girl?" Then yes: $;75$% of them. – Timbuc Feb 03 '15 at 09:16
  • Well, the name is not important... Gender is what matters. – Vim Feb 03 '15 at 09:19
  • @Vim Of course, yet it is important to be as clear as possible when dealing with mathematics. – Timbuc Feb 03 '15 at 09:20
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I agree with your friend, and the reason follows using conditional probability. Define B=the event of having a boy, G= the event of having a girl, and we're in the space defined by "having two kids". Then we want the probability $\;P(B\backslash G)=$ the probability of having a boy knowing that there's already a girl:

$$P(B\backslash G)=\frac{P(B\cap G)}{P(G)}=\frac{\frac12}{\frac34}=\frac23$$

Timbuc
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The a priori probabilities indeed follow a binomial distribution, and all pairs are equiprobable $$P(BB)=P(BG)=P(GB)=P(GG)=\dfrac14.$$ The distribution of the number of boys follows $(\dfrac14,\dfrac12,\dfrac14)$.

Now you are told that $BB$ is excluded, then the a posteriori probabilities turn to $$P(BB|\lnot BB)=0,P(BG|\lnot BB)=P(GB|\lnot BB)=P(GG|\lnot BB)=\dfrac{1/4}{3/4}=\frac13$$ (all pairs but $BB$ remain equiprobable). The distribution of the number of boys follows $(\dfrac13,\dfrac23,0)$.

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A lot of great explanations about the correct answer. Your answer is wrong because it ignores that fact that people are distinguishable, but the information you got doesn't tell you which child is the boy. You are ignoring this and treating the problem as if you knew that, for example, Child 1 is a boy. As others here have stated, this is only half the number of cases that are consistent with the information you got.