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

Question

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

I just dipped into a book, The Drunkard's Walk - How Randomness Rules Our Lives, by Leonard Mlodinow, Vintage Books, 2008. On p.107 Mlodinow says the chances are 1 in 3.

It seems obvious to me that the chances are 1 in 2. Am I correct? Is this not exactly analogous to having a bowl with an infinite number of marbles, half black and half red? Without looking I draw out a black marble. The probability of the second marble I draw being black is 1/2.

Answer 1

I think this question confuses a lot of people because there's a lack of intuitive context -- I'll try to supply that.

Suppose there is a birthday party to which all of the girls (and none of the boys) in a small town are invited. If you run into a mother who has dropped off a kid at this birthday party and who has two children, the chance that she has two girls is $1/3$. Why? $3/4$ of the mothers with two children will have a daughter at the birthday party, the ones with two girls ($1/4$ of the total mothers with two children) and the ones with one girl and one boy ($1/2$ of the total mothers with two children). Out of these $3/4$ of the mothers, $1/3$ have two girls.

On the other hand, if the birthday party is only for fifth-grade girls, you get a different answer. Assuming there are no siblings who are both in the fifth grade, the answer in this case is $1/2$. The child in fifth grade is a girl, but the other child has probability $1/2$ of being a girl. Situations of this kind arise in real life much more commonly than situations of the other kind, so the answer of $1/3$ is quite nonintuitive.

Answer 2

In a family with 2 children there are four possibilities:

1) the first child is a boy and the second child is a boy (bb)

2) the first child is a boy and the second child is a girl (bg)

3) the first child is a girl and the second child is a boy (gb)

4) the first child is a girl and the second child is a girl (gg)

Since we are given that at least one child is a girl there are three possibilities: bg, gb, or gg. Out of those three possibilities the only one with two girls is gg. Hence the probability is $\frac{1}{3}$.

Answer 3

Here's some idea.

Suppose you sample $4n$ families with two children, where $n$ is a very large integer. Then, with a very high probability, about $n$ of them have two boys, about $n$ have two girls, and about $2n$ have one girl and one boy. Since you ignore the families with two boys, the desired probability is given by $n/(3n) = 1/3$.

EDIT: The above solution corresponds to the following situation. You visit a large number of families. In each family, you check if there are two kids. If no, you ignore this family. If yes, you check if one of the kids is a girl. If no, you ignore this family. If yes, you check if both kids are girls.

EDIT: Among all the OP's examples, only one is relevant here. Namely, "I visit this family. I know they have 2 kids. One of them, a girl, comes into the room. The probability that the 2nd kid is also a girl is 1/2, no?". Well, in this situation the probability is indeed $1/2$, assuming the following interpretation. You visit a large number of families. In each family, you check if there are two kids. If no, you ignore this family. If yes, you wait until one of the kids comes into the room. If it is a boy, you ignore this family; otherwise you check if both kids are girls. It is readily understood that this leads to a probability of $1/2$.

Answer 4

It really depends on what knowledge you have to restrict the probability space. (The original question is a little ambiguous, but I agree that 1/3 is the best answer.)

The marble example and the example of the firstborn daughter are the same situation-- the second trial is independent, and the odds are 1/2. In these cases you know the first trial is independent, and you can ignore it.

The original question intends you to imagine a randomly chosen 2 child family from among all families with at least 1 daughter. For example, suppose there is a social science study on 2 child families with at least 1 daughter-- in this situation, about 1/3 of the families will be daughter-daughter, 1/3 will be daughter-son, and 1/3 will be son-daughter. You have to consider the full probability space of two trials (d-d,d-s,s-d,s-s) and eliminate the s-s possibility. This leaves a 1/3 chance of 2 daughters.

For other variations on this question, you have to decide the appropriate probability space before answering. (One common, confusing variation is to suppose the family has two children, one of which is a son born on a Tuesday-- what are the odds of two sons?)

In your example where the first child in the room is a daughter, if we suppose that the order the children enter has nothing to do with sex, then the probability the other child is a daughter is 1/2. However, maybe the family encourages their daughters to play inside and their sons outside-- then you could assume the other child is a son, since he hasn't come in yet. Maybe the culture has a tradition that the oldest daughter in the household (if there is a daughter) offers guests drinks-- then you know the family has at least one daughter, and the odds that they have another daughter is 1/3. Once you've chosen your assumptions, you can define the probability space and compute the answer.

Answer 5

One possible solution for this problem might be though using conditional probability.

Let $B_i$ and $G_i$ represent $i$th boy and $i$th girl respectively. Here the sample space could be expressed as $S =$ {$(B_1,B_2),(B_1,G_2),(G_1,B_2),(G_1,G_2)$}

Let $A$ = An event of having both girl chil. $B$ = An event of having at-least one girl child.

$A = ${$(G_1,G_2)$} and $B =${$(B_1,G_2),(G_1,B_2),(G_1,G_2)$}

So what you are asking is $$P(A / B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} $$

Answer 6

Please ignore the old comments - I have radically changed my answer. It's long, but bear with me - after reading this answer, you should feel intuitively-comfortable with almost all other probabilistic fallacies, not just this one.

The first and most important thing is to define what we mean by "probability." Let's define it to mean "the expected percentage of positive outcomes when repeating an observation over a large sample" (Go ahead, read that again, more slowly). Also, let's call the method we use to chose this large sample the "model."

This may sound trivial, but it has some important implications. For example, what is the probability you will die before age 40? According to our definition, this question has no meaning: we can't observe you multiple times before age forty and record how many times you die. Instead, we observe other people below age 40, and record how many of them die.

So let's say we observe all other people on earth below age 40 (our model), and find that 1/2 of them die before hitting the big four-oh. Does this mean the probability of you dying before 40 is 50%? Well, according to this model, it does! However, this is hardly a fair model. Perhaps you live in a first-world country - now when we revise our model to include only people below 40 in first-world countries, your chances become a much-less-grim 1/10^†. But you are also not a smoker, and don't live in the city, and ride your bike on Sundays with your wife and crazy mother-in-law, which makes your chances 123/4567! That's much better... however, our model still doesn't take into account that you are also an avid skydiver ;)
^† I am pulling these numbers out of nowhere, they are not real statistics.

So the point is, asking for a "probability" only makes sense in the context of a certain model - a way of repeating our observation many times. Without that, asking for a probability is meaningless.

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Now, back to the original question. Before we can assign a probability, we must choose a model; how are we choosing the families to sample from? I see two obvious choices, which will lead to different answers:

1. Consider only families which have two children, one of whom is a girl, and choose one randomly.
2. Consider only girls who have exactly one sibling, and choose one randomly.

Do you see the difference? In the first case, every family has the same probability of being chosen. However, in the second case, the families with two girls are more likely to be chosen than families with only one girl, because every girl has an equal chance of being chosen: the two-girl families have doubled their odds by having two girls. If children were raffle tickets, they would have bought two tickets while the one-girl families bought only one.

Thus, we should expect the probabilities in these two cases to be different. Let's calculate them more rigorously (writing BG to mean "boy was born, then girl):

1. There are three equally probable family-types: BG, GB, and GG (BB was removed from consideration, because they have no girls). Since only one of the three has two girls, our chances of having two girls are 1/3.
2. We have the same possibilities as above, but now GG is twice as likely as BG or GB. Thus, the probabilities are GG: 2/4, GB: 1/4, and BG: 1/4, meaning the probability of a girl-sibling is 2/4 = 1/2 (alternatively, we could have noted that there are only two equally-probable possibilities for the sibling: boy or girl).

Here lies the fallacy: the model our intuition assumes is the second one, but the way the problem is worded strongly implies the first one. When we think in terms of "randomly choosing a family (over a large number of families)," our intuition meshes perfectly with the result.

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Let's take a look at another similar problem

In a family of two children, where the oldest child is a girl, what is the probability they are both girls?

Once again, I can see two different, plausible models for observing our random sample:

1. Consider only families with two children, the oldest of whom is a girl, and choose one randomly.
2. Consider only girls who have exactly one (younger) sibling, and choose one at random.

So once again, the question strongly implies the first model, though arguments for either could be made. However, when we actually calculate the probability...

1. Same as before, but we've also eliminated BG, where the first child was a boy. This leaves only two equally-probable possibilities, GB and GG. Thus, the chances are 1/2.
2. We've also eliminated BG from this case, leaving GB and GG. However, unlike the original question, GG is no longer twice as likely, since the younger child can no longer be the one who was randomly chosen. Thus, GB and GG are equally likely, and we again have a probability of 1/2 (alternatively, we could have noted that there are only two equally-probable possibilities for the sibling: boy or girl).

...we find that the choice between these two models doesn't matter, because in this case both have the same probability! Among two-child families with an older daughter, it doesn't matter if we randomly choose the family or the older-daughter, because in both cases there is only one of each per family.

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Hopefully that all made sense. For bonus points, try applying this reasoning to the Monty Hall problem. What is our model - how are we making repeated observations? Why does it clash with our intuition?

For even more bonus points, try to figure out the following question; the math isn't too difficult, but it took me a long while to figure out why, intuitively, the answer should be correct:

In a family of two children, one of whom is a girl named Florida, what are the chances of two girls?

(If you have troubles, post a question and leave a link to it in the comments, and I'll try to answer it there as best I can :) )

Answer 7

This is really a continuation of BlueRaja's answer, where I'm changing the question slightly to give another illustration of the underlying point. Consider the following two procedures:

Procedure A: I flip two coins, a nickel and a penny. After the flip, I pick one at random, and report its value to you.

Procedure B: I flip two coins, a nickel and a penny. If exactly one lands heads, I report that one's value to you, i.e. I say either "the nickel landed heads" or "the penny landed heads" as appropriate. If both land heads, I report on one of them chosen at random. If neither lands heads, I say "both landed tails."

Now, if I follow Procedure A and say "the nickel landed heads", then the probability that the penny landed heads is 50/50. But if I say the very same thing after following Procedure B, the probability that the penny landed heads is 1 in 3.

Now take the following problem: I flip two coins and say to you "the nickel landed heads". What is the probability that the penny landed heads? Clearly there's no way to answer without knowing or assuming something about the procedure I'm following.

(All this having been said, I agree that the best answer to the original question is 1/3.)

Answer 8

I think that the reason that these puzzles are so often confusing is that they rely on the limitations of the English language rather than on any mathematical difficulties. Of course this is not unique to English, and I think you should be able to find similar similar puzzles in pretty much any natural language.

Here is an example that brings out the difficulty more clearly. First, consider the following similar puzzle:

• A family has two children, Robin and Lindsay. Lindsay is a girl. What is the probability that both children are girls?

In elementary probability class, they teach you to answer this by making a table of the four options

  Robin    Lindsay
  B        B
* B        G
  G        B
* G        G

The starred rows are the ones where Lindsay is a girl, and we compute from them that the chance that both children are girls is 1/2.

Now consider this puzzle

• A family has two children, Robin and Lindsay. At least one of them is a girl. What is the probability that both children are girls?

The elementary probability method gives the following table, and a probability of 1/3. The difference is that we gain one more row, compared to the previous puzzle.

  Robin    Lindsay
  B        B
* B        G
* G        B
* G        G

After looking at these, you can see that the difficulty of the original puzzle comes because, in English, "one of the children" can mean several different things:

• "one" can mean "a particular one". If you read the original puzzle like this, it becomes analogous to the first puzzle I wrote, and the answer will be 1/2.

• "one" can mean "at least one". If you read the original puzzle like this, it become analogous to the second puzzle I wrote, and the answer is 1/3.

• "one" can mean "exactly one". If you read the original puzzle like this, the answer is 0.

There is a common convention in mathematics that "one" usually means "at least one". For example, this is the sense intended in the following sentence, which is a typical example of mathematical English: "if a natural number $n$ is a multiple of a prime number $p$, and $n = ab$, then one of $a$ and $b$ is divisible by $p$." We would not read this as saying that exactly one of $a$ and $b$ is divisible by $p$.

I don't believe this convention is very common in non-mathematical English. If I say, "one of my children is a girl", in normal English this means that the other is a boy. Similar discrepancies between mathematical and non-mathematical English come up with our use of the word "or" and our use of the phrase "if/then". When we teach mathematics, we have to spend time explaining this mathematical argot to students, so they can use the same English conventions that we do.

Probability puzzles like the one you're asking about rely on these differences of English meaning, rather than on any logical or mathematical problem. In that sense, they aren't really puzzles, they're just tricks.