You call a family that has two children. A boy picks up. What is the probability that the other is a boy?

The Prompt

Lots of Disagreement

The most liked comment answers 1/2, but the user got worn down by others until he recanted and agreed that the answer is 1/3. Many other comments give lengthy explanations as to why the answer is 1/3.

The actual answer is 1/2.

Framing the Question

Let's flesh out this hypothetical, to make it easier to reason about.

• You're a salesman who sells bunkbeds, specifically bunkbeds that only work for two boys.
• You have a phonebook of leads, containing the phone number of all households in the country that have exactly two kids.
• When you make a call, you anxiously wait to find out whether you have called the right kind of home. If the home has two girls or a girl and a boy, you've wasted your time.
• When you call, there's an equal probability that either child answers.
• When a boy answers, you're happy. The household has at least one boy.
• But how happy should you be? How likely is it, given that the kid answering the phone is a boy, that the other kid is also a boy?

Explanation

First, imagine there are 24 households with kids. The distribution of genders in those households is as expected, with a quarter of them (6) having two girls, half of them (12) having a girl and boy, and a quarter of them (6) having two boys.

GG	GG	GG	GG	GG	GG
BG	BG	BG	BG	BG	BG
BG	BG	BG	BG	BG	BG
BB	BB	BB	BB	BB	BB

An intuitive explanation is: look at all the Bs on the board. How many have a brother? Half. A longer explanation is as follows.

We want to solve P(called row 4 given that a boy picked up).

• P(called row 4) = 1/4
• P(a boy picked up) = 1/2, because:
  • half the children are boys, each family is equally likely to have been called, and each child in the family is equally likely to pick up.
• To re-emphasize, P(a boy picked up) =
  • P(a boy picks up from row 1) + P(a boy picks up from row 2) + P(a boy picks up from row 3) + P(a boy picks up from row 4) =
  • ((1/4) * 0) + ((1/4) * (1/2)) + ((1/4) * (1/2)) + ((1/4) * 1) =
  • 1/2.
• Therefore, P(called row 4 given that a boy picked up) =
  • (P(a boy picked up given that row 4 was called) * P(row 4 was called)) / P(a boy was called) =
  • (1 * (1/4)) / (1/2) =
  • 1/2.
  • The link between the first bullet in this nested list and the second is given by Baye's Theorem.

The answer is 1/2.

Why do so many people think the answer is 1/3?

The line of reasoning getting people to 1/3 is something like the following:

• After calling and discovering that at least one boy is in the family, row 1 is eliminated.
• Of the 18 remaining families, 6 have two boys.
• Therefore, you have a 6/18 = 1/3 chance of having called a home with two boys.

The flaw there is that you're ignoring the fact that for row 4, there's a 100% chance a boy answers the phone, whereas for rows 2 and 3, there's a 50% chance a boy answers the phone.

Explicitly, P(called row 2 column 1 given that a boy answered) =

• (P(a boy answered given row 2 column 1 was called) * P(row 2 column 1 was called)) / P(a boy was called) =
• ((1/2) * (1/24)) / (1/2) =
• 1/24

Similar logic can be used to conclude P(called row 4 column 1 given that a boy answered) = 1/12.

You know as soon as the boy answers that you're twice as likely to be talking to row 4 column 1 as row 2 column 1. But there are twice as many BG families as BB families, so that evens out, and you have an equal probability of talking to a BG family as a BB family, once you know a boy answered.

In what situation would the answer be 1/3?

If the question was phrased as "You are calling a family with 2 children with at least one boy. What's the probability that the family has 2 boys?" then 1/3 would be the answer.

You could imagine that our salesman, instead of having a list of all families with 2 kids, has a list of all families with 2 kids with at least one boy. When he picks up the phone, it's ringing, and he's anxiously waiting to hear the voice on the other line. He knows it's one of the families from rows 2 to 4, and it's equally likely to be each of them. Therefore, there's a 6 / 18 chance that it's row 4.

The key there is that there was no event, a boy picking up, which indicated a weighting toward one family versus another. All families in the set are equally likely from the salesman's perspective.

Conclusion

I think it's pretty funny that people got so dug in on the 1/3 thing. It feels like they were so primed for expecting a trick question that it felt like it had to be the unexpected 1/3 rather than the obvious 1/2.

The subtleties and confusion around this question are captured by Wikipedia's Boy or Girl Paradox. Looking at the "Second Question" section, you can see that our question is equivalent to one of their examples: "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."