2
$\begingroup$

Q1.) I tell you that I have two children and that one of them is a girl. What is the probability that I have two girls? Assume girls and boys equally likely to be born and that the gender of one child is independent of the other.

The answer is 1/3 which I understand and agree with.

It can be seen that of the possible combinations {BB,GB,BG,GG} the first combination is not possible. In the remaining three cases, there is only one in which there are two girls which gives the correct answer.

Q2.) I tell you that I have two children and that one of them is a girl. You knock on the front door and are greeted by a girl who you correctly deduce is my daughter. What is the probability that I have two girls? Assume girls and boys equally likely to be born and that the gender off one child is independent of the other. Compare and contrast the answers.

The claimed answer here is 0.5. That is on the basis that the girl at the door is taken to be #1 and therefore the only randomness remaining is the gender of the second child which is given in the question is 50%.

However this doesn't seem correct to me, how can you know that the girl is #1 and not #2? Seeing the girl does not seem to preclude {GB,BG,GG} so the answer should be the same as above?

Baz

$\endgroup$
  • 2
    $\begingroup$ You say for the first scenario that "the probability that I have two girls" yet the probability of $\frac{2}{3}$ seems to be for "the elder child is a girl." Shouldn't the probability of two girls in that scenario instead be $\frac{1}{3}$? $\endgroup$ – JMoravitz Oct 16 '15 at 18:21
  • 1
    $\begingroup$ As for defining the difference between child #1 and child #2, this could be done based on height, name, age, or in particular for the current situation first met. Each of which are perfectly valid. We would need an additional assumption that a boy is equally likely to answer the door as a girl in the case that both are present, (and to have height as the order, that boys and girls in this family are equally likely to be taller than one another). $\endgroup$ – JMoravitz Oct 16 '15 at 18:23
  • $\begingroup$ Sorry yes! I'm not sure this is a duplicate the questions are very similar but the wording is crucial? $\endgroup$ – Bazman Oct 16 '15 at 18:24
  • $\begingroup$ But the answer makes no mention of this it just says that the child which is seen is taken to be #1 $\endgroup$ – Bazman Oct 16 '15 at 18:25
  • $\begingroup$ Sorry it is a dupe: math.stackexchange.com/questions/15055/… the answer provided by blueraja seems to makes sense $\endgroup$ – Bazman Oct 16 '15 at 18:32
2
$\begingroup$

Assuming that boys and girls in this household are equally likely to answer the door and that only one person ever answers the door at a time, we may choose the order of the children to be based on first seen.

In this sense, we can indeed claim that the girl that we saw at the door is "child #1" and the result follows.

However, this assumes that boys and girls are equally likely to answer the door.

Consider an opposite extreme where you know that our entire family is home at the time and our family follows some very strict religious customs that only a boy may answer the door unless no boys are present. In this scenario, the fact that a girl answered the door directly implies that there were no boys available to answer the door, which in turn implies that I must have two daughters and no sons.

I would say then that the problem is poorly worded and there is not yet enough information to conclude that the probability is indeed $\frac{1}{2}$ without additional assumptions (even if the assumption seems valid).

$\endgroup$
  • $\begingroup$ Thanks, have to agree that the question is too loosely worded but I understand the difference better now. $\endgroup$ – Bazman Oct 16 '15 at 18:34
3
$\begingroup$

In the first case, you are quite correct. By asserting that one of your children is a girl, what you effectively said is that the older child is a girl, or the younger child is a girl, and you haven't ruled out the possibility that both children could be girls. Hence, the answer is, indeed, $\frac13.$

In the second case, however, exactly one of the children has answered the door, and that child is a daughter. The only question is whether the child who didn't answer the door is a girl or a boy. You don't know if it's the older child or the younger child who has answered the door, but exactly one of them has. If the older child answered the door, then it is impossible that the older child is a boy, so there are only two possibilities remaining. Likewise, if it was the younger child who answered the door, then it is impossible that the younger child is a boy, and so there are only two possibilities remaining. Regardless of which child answered the door, the answer is then $\frac12.$

$\endgroup$
  • $\begingroup$ OK but this is where I was/am confused. If a girl answers the door and you are not making any assumptions that she is the eldest (or indeed youngest) then surely you can still have {GB, BG, GG}? The observation that a girl answered the door is consistent with any of those three combinations? $\endgroup$ – Bazman Oct 16 '15 at 18:40
  • 2
    $\begingroup$ @Bazman You have $4$ times the letter $G$ in $\{GB,BG,GG\}$ and exactly $2$ of these letters $G$ are accompanied by another letter $G$. $\endgroup$ – drhab Oct 16 '15 at 18:44
  • $\begingroup$ True, we are not making any assumptions about whether the child who answered the door is the youngest or the oldest, but one of them did, and by answering the door, precluded one possibility. We don't know which possibility was ruled out, but we know that one of them was. After all, if the daughter who answered the door is the elder child, then the possibilities are GB and GG. On the other hand, if the daughter who answered the door is the younger child, then the possibilities are BG and GG. We don't know which of these two distinct events we're dealing with, but that doesn't change things. $\endgroup$ – Cameron Buie Oct 16 '15 at 18:54
  • $\begingroup$ We are dealing with exactly one of those two events, and the event we're dealing with (whichever it may be) has probability $\frac12.$ $\endgroup$ – Cameron Buie Oct 16 '15 at 18:56
  • $\begingroup$ @Bazman: Apologies! I made an error in my comments. I should have said that our sample space is either {GB,GG} or {BG,GG}, though we don't know which. However, regardless of which of these is our sample space, the event {GG} has probability $\frac12.$ Does that make more sense? $\endgroup$ – Cameron Buie Oct 16 '15 at 19:12
0
$\begingroup$

I'll expand two of the answers that are already given. They are based on two different models: in one, the child answering the door is chosen based on age, and in the other by sex. And the answer depends on which model is chosen.

  1. Suppose that the elder child opens the door with probability $p$ and the younger with probability $1-p$. Let G@D denote the event that you see a girl at the door. We have: $$ P\left(GG|G@D\right)=\frac{P\left(G@D|GG\right)P(GG)}{P\left(G@D\right)}=\frac{1/3}{1/3+p/3+\left(1-p\right)/3}=\frac{1}{2}. $$

  2. Suppose that, if the family has a boy and a girl, the girl opens the door with probability $q$. Now, $$ P\left(GG|G@D\right)=\frac{P\left(G@D|GG\right)P(GG)}{P\left(G@D\right)}=\frac{1/3}{1/3+q/3+q/3}=\frac{1}{1+2q}. $$

So the answer is $1/2$ only if we pick the first model or if $q=1/2$ in the second model.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.