# What is the problem in this solution to the Two Child Problem?

The Two Child Problem states

In a family with two children, what are the chances, if at least one of the children is a girl, that both children are girls?

It is well attested that the answer is 1 in 3. You can view the other question for an explanation why.

I thought up this solution to the same problem, but since the answer is not the same, there must be some part of my reasoning that is unsound, but I can't quite figure out where the problem is. This is my reasoning:

• If it is given that the older child is a girl, the probability that the younger child is also a girl, and thus that both children are girls, is 1 in 2.
• Similarly, if it is given that the younger child is a girl, the probability that the older child is also a girl, and thus that both children are girls, is also 1 in 2.
• If we know that one of the children is a girl, then it can easily be concluded that either the younger child or the older child is a girl.
• Either way, the probability is still 1 in 2.

Where have I gone wrong?

• Have you also seen this? – joriki Dec 24 '11 at 23:20
• Why do you believe this is "well attested"? – joriki Dec 24 '11 at 23:22

You play a lottery. One of the numbers 1,2,3 will be chosen uniformly at random, and you win if 1 is chosen. You have probability 1/3 of winning, but this argument allows you to conclude it's 1/2:

1. If 3 was not chosen, then it's 1 or 2, so you have 1/2 chance
2. If 2 was not chosen, then it's 1 or 3, so you have 1/2 chance
3. Since either 2 or 3 is not chosen, in both cases you have 1/2 chance

or:

You've got a five-sided dice with numbers 1,2,3,4,5. The probability of getting odd number is 3/5, but with this argument you can get wrong result - 2/3:

1. If you the number is at most 3, then possibilities are 1,2,3 so 2/3 chance of being odd
2. If you the number is at least 3, then possibilities are 3,4,5 so 2/3 chance of being odd
3. Either the number is at most 3 or at least 3, and in both cases the chance is 2/3

The error in both examples is that the cases are not disjoint. To reason about probability based on conditional probabilities, you're implicitly using law of total probability which requires that the space is split into disjoint sets.

The first example is exactly the same as boy-girl situation, where 1=GG, 2=BG, 3=GB. More dramatically, you could do the lottery with numbers 1-100 (only 1% chance of winning) and the argument would give you 50% chance of winning.

For a more intuitive reason than the formal answer:

Let's say I use your first two bulleted statements to guess how often both children are female.

When the older child is female, half the time I will guess that both children are female.

Similarly, when the younger child is female, half the time I will guess that both children are female.

Seems consistent, right? However, the logic fails when both children turn out to be girls! In that case, both of your probability statements fire at the same time, and the problem is that I might end up saying "Both children are girls" twice in the same universe. When that happens, we're overcounting the number of times that both children are girls, which is why your answer ended up too large.

Others have dealt with the specific error in your reasoning. I want to point out that the problem is not well-defined: without more information it is impossible to say what the probability is that both children are girls. In particular, it not true that the answer is definitely $1/3$.

Consider the following three scenarios. In each of them I choose one family at random from the pool of all families with exactly two children, in such a way that each such family is equally likely to be chosen, and then I make one of the following two statements to you:

Statement 1: At least one of the children is a girl.

Statement 2: At least one of the children is a boy.

Scenario A: If the older child is a girl, I make Statement 1; if the older child is a boy, I make statement 2.

Scenario B: If both children are girls, I make Statement 1; otherwise, I make Statement 2.

Scenario C: If both children are boys, I make Statement 2; otherwise, I make Statement 1.

Suppose that I’ve made Statement 1: I’ve told you that at least one of the children is a girl. What is the probability that both are girls?

Scenario A: The older child is definitely a girl, so the probability that both are girls is simply the probability that the younger child is a girl, which is $1/2$.

Scenario B: Both children are definitely girls, or I’d have made Statement 2, so the probability that both are girls is $1$.

Scenario C: The only possibility that is ruled out is that both are boys. Three possibilities remain: both are girls; the elder is a girl and the younger a boy; the elder is a boy and the younger a girl. These are equally likely, so the probability that both are girls is $1/3$.

In short, the probability depends on what I would have said had I chosen a different family. Using more complicated decision rules, I could arrange scenarios in which the probability that both children are girls has other values. Here’s an example:

Scenario D: I roll a fair die. If both children are girls, I make Statement 1. If both are boys, I make Statement 2. If one is a boy and the other a girl, I make Statement 1 if the die comes up $6$; otherwise, I make Statement 2.

There are $6\cdot 2\cdot 2=24$ equally likely events, one for each combination of die roll, sex of older child, and sex of younger child. I’ll represent a specific combination by a string like $3BG$, meaning that I rolled a $3$, the elder child is a boy, and the younger child is a girl. The fact that I made Statement 1 rules out six of these $24$ events: $1BB,2BB,3BB,4BB,5BB$, and $6BB$. It also rules out all five $nBG$ events in which $n\ne 6$ and all five $nGB$ events in which $n\ne 6$. The only possible events, therefore, are the six $nGG$ events, $6BG$, and $6GB$. In six of these eight equally probable cases both children are girls, so the probability that both are girls in this scenario is $6/8=3/4$.

Let me try and summarize your proof attempt, using $G_1$ and $G_2$ to denote the statements that younger and older child, respectively, is a girl:

(1) $P(G_1 | G_2) = 1/2$ the symbol | means "given"

(2) $P(G_2 | G_1) = 1/2$

(3) We are given $G_1 \vee G_2$ the symbol $\vee$ means "or"

(4) Therefore $P(G_1 \wedge G_2) = P(G_2 | G_1) = P(G_1 | G_2) = 1/2$ the symbol $\wedge$ means "and"

The problem is this last step. With probabilities, you cannot break the statement $G_1\vee G_2$ into two cases $G_1,G_2$. When you break the statements up, you are making the false assumption that $P(G_1 | G_2) = P(G_2 | G_1)\implies P(G_1\wedge G_2 | G_1\vee G_2)$. This certainly seems like it should be true, but if you try to prove it you will need to make use of values for $P(G_1),P(G_2)$, which you aren't given. All the information you have is that either $G_1$ or $G_2$ is true, which doesn't tell you how likely it is that $G_1$ is true or how likely it is that $G_2$ is true.

The probability of $Y$ given $X$,

where $X$=at least one is a girl and $Y$=both are girls,

is handled by the conditional probability $$P(X\cap Y)/P(X).$$

The four possibilities are BB, BG, GB, GG.

In this case $X\cap Y$=both are girls so $P(X\cap Y)=1/4$ corresponds to GG

However $P(X)$ is 3/4 since 3 of the four cases have at least one girl.

Hence the probability is 1/3.

The math actually adds up. Your original probability space is the space formed by the sample space {GG,BG,GB}, all points having a 1/3 chance. Now by the law of total probability, let G1 be the event the older one is a girl, and G2 that the younger is a girl. Then

$$Pr(GG) = Pr(GG|G1)*Pr(G1) + Pr(GG|G2)*Pr(G2) - Pr(GG|G1\cap G2)*Pr(G1\cap G2)= 1/2*2/3 + 1/2*2/3 - 1*1/3 = 1/3$$,

because G1 and G2 aren't mutually exclusive, hence the - part in the expression.

Notice that $$Pr(GG|G1)=1/2$$ which is the probability the younger one is a girl given the older one is a girl, and the opposite for $$Pr(GG|G2)=1/2$$