Probability of second child being a girl

Question

In A family has two children. One child is a girl. What is the probability that the other child is a boy?, the question was

A family has two children. One child is a girl. What is the probability that the other child is a boy?

it was shown that the probability of the other child being a boy given that you know one child is a girl is $2/3$.

In another variant of this problem, the question was

I tell you that I have two children and that one of them is a girl (I say nothing about the other). You knock on my front door and you are greeted by a girl who you correctly deduce to be my daughter. What is the probability that I have two girls? Compare and contrast your answer to the answer to the previous question. Assume that boys and girls are equally likely to be born and that the gender of one child is independent of gender of another.

Apparently, the answer to this question is 0.5, but how is this any different from the previous problem? It seems like the same problem except we're asking what the probability of the second child being a girl is given that the first child is a girl rather than asking about the probability of the second child being a boy given that the first child is a girl. So shouldn't the answer just be $1/3$, the complement of the first question?

This is the solution to the second problem that I saw in the book:

I'm skeptical about the part of the solution where it says ". Having pinpointed the gender of Child #1 (we might as well call her that), however, the latter two outcomes are excluded." Getting rid of BG seems to rely on that you see the children in the order that they're listed, so if it's BG, it would have to be a boy that greets you at the door. But nothing in the problem suggests this, so I don't think this solution is accurate.

Answer 1

The issue is that in the original case talking about the "other child" doesn't make sense - the statement didn't specify a particular child, and if both the children are girls, there is no way to tell which is the "other child".

So in the original case we have to use some other criterion to distinguish the children - say order of birth - and then there are three possibilities: BG, GB, GG.

In the second case there is a well-defined "other child" - the one who didn't answer the door. You have absolutely no information about this other child. So there are two possibilities (putting the child who answered the door first): GB or GG.

Answer 2

The first question is: given that a family has two children and at least one is a girl what is the probability both are girls.

The second question is: given that a family has two children and one of them is a specific girl who answered the door. What is the probability that the child who didn't answer the door is a girl.

Do you see the difference?

o if it's BG, it would have to be a boy that greets you at the door.

Not really.

The couple has two children: DOORANSWERER and HERMIT. We are told DOORANSWERER is a girl. The question is what is the probability HERMIT is a girl.

ANd it's $\frac 12$. DOORANSWERER = GIRL and HERMIT = girl. Or DOORANSWERER = GIRL and HERMIT = boy are the only two equally likely outcomes.

Compare to the first question: The couple has two children C and D. We are told one of them is a girl but not which one is the girl. We are asked what is the probabily they are both girls.

And it is $\frac 13$. C= GIRL and D= girl. C=girl and D=boy. C=boy and D=girl are the three equally likely outcomes.

The second question distinguishes a distinction between the children that the first one didn't. It doesn't matter how they are distiguished and there is no requirement the oldest child or the shorter child is the one how answer the door. The only distinction is we say here THIS is a child, this one the one I'm poking with my fingers. If you distinguish that we are focusing on that one it changes everything from a case where we are not identifying any child specifically.

=======

Consider there are two children $C$ and $D$ and one of them answers the door.

There are eight possibilities.

$1:$ C is boy. C answers door. D is boy.

$2:$ C is boy. C answers door. D is girl.

$3:$ C is boy. D answers door. D is boy.

$4:$ C is boy. D answers door. D is girl.

$5:$ C is girl. C answer door. D is boy.

$6:$ C is girl. C answers door. D is girl.

$7:$ C is girl. D answer door. D is boy.

$8:$ C is girl. D answers door. D is girl.

......

Question 1: I told you that at least one is a girl. What is the probability they are both girls.

Well. 1 and 3 are out. Of the remaining six options, 6 and 8 are both girls while 2,4,5,7 have a boy and a girl. SO the probability is $\frac 26 = \frac 13$.

Question 2: We are told a girl answered the door. What is the probability they are both girls.

Well. 1,2,3, and 7 are out. Of the remaining four options, 6 and 8 are both girls while 4 and 5 have a boy and a girl. SO the probability is $\frac 24 = \frac 12$.

Answer 3

Maybe this analogy helps:

I flip two coins, and you know that at least one of them is tails. What is the probability that both are tails?

Well, the answer to this question, like the girl question, depends on how you know that at least one of the flips is tails. Consider the following two scenarios:

Scenario 1: I flip two coins, but but you don't get to see the outcome of either of them. I, however, do look at the outcome, and then I tell you: "At least one of them is tails". This is like your first girl scenario, and now the probability of both coins being tails is $\frac{1}{3}$

Scenario 2: I flip two coins, and you see one of them lands on tails on the table, but the second coin rolls of the table, and you don't see what it lands on. So, again you know that at least of them is tails, but it is pretty clear that in this scenario the probability of having two tails is $\frac{1}{2}$

Answer 4

Here is one more way to convince yourself, using Bayes' theorem.

$\displaystyle \small P(A|B) = \frac{P(A \cap B)}{P(B)}$

$B$ is the event of at least one of the two children being girl and a girl answering the door. $A$ is the event of one child being a girl, the other being a boy and the girl answering the door.

$ \displaystyle \small P(A \cap B) = \frac{1}{2} \cdot \frac{1}{2} \ $, as the probability of one child being girl and the other being boy is $ \displaystyle \small \frac{1}{2}$ and the probability that the girl answers the door is $ \displaystyle \small \frac{1}{2}$.

$ \displaystyle \small P(B) = \frac{1}{4} + \frac{1}{2} \cdot \frac{1}{2} \ $, as the probability of both being girls is $ \displaystyle \small \frac{1}{4}$ and in that case, it must be a girl answering the door. We add it to the probability that just one of them is girl and she answers the door.

So, $ \displaystyle \small P(A|B) = \frac{1}{2}$

Answer 5

In the first problem, you are considering the two children collectively or in summary; whereas in the second problem, they are an ordered/labelled pair, so you are considering each child separately/independently of the other.

In the first problem, you are using prior knowledge about the collective pair, so Bayesian/conditional probability applies.

In the second problem, it really doesn’t matter if the children are considered successively or concurrently (or even if the family has 13 children), since each child is distinguished from and independent of the other.