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Probability of two boys in a family with at least one

Sentient Potato • July 14, 2022

Recently Twitter user @RandomSprint tweeted

Mr. Smith has two children. At least one of them is a boy. What is the probability both are boys?

The correct answer is 1/3, but 1/2 won the attached poll. Why is 1/3 the correct answer? I’ll show you three ways: two simple analytical approaches and simulation evidence.

First, assuming that every time a child is born there is a 50% chance the child is a boy and a 50% chance the child is a girl, there are four possible outcomes for a family with two children, all of which are equally likely (25% chance of each outcome):

  • two boys
  • a boy born, then a girl (one boy)
  • a girl born, then a boy (one boy)
  • two girls

Since we know one of the children is a boy, we are now reduced to three outcomes, each of which would be equally likely (now a 33 1/3% chance of each outcome):

  • two boys
  • a boy born, then a girl (one boy)
  • a girl born, then a boy (one boy)

So, 1/3 is the correct answer. Some people’s confusion stems from not being able to distinguish between the two possible outcomes that result in the family having one boy. For me, the thing that helps is remembering that there’s a difference between having an oldest son and younger daughter than having an eldest daughter and younger son. Since the question doesn’t tell us which situation we’re dealing with, we have to treat them distinctly.

We can also consider the mathematical formula for conditional probability, Bayes’ rule. Bayes’ rule tells us what the probability of an event B happening, given that we already know event A has happened:

\Pr(A \mid B) = {{\Pr(B \mid A) \Pr(A)} \over {\Pr(B)}}

Here, A is the probability of two boys being born and B is the probability of at least one being born. We know the probability of at least one boy being born is 3/4, and the probability of two boys being born is 1/4; the probability of at least one boy being born given that we know two are born is 1. So plugging these numbers in,

\Pr(\text{two boys} \mid \text{one boy}) = {{1 \times {1 \over 4}} \over {3 \over 4}} = {1/4 \over 3/4} = {1 \over 3}.

In case the analytical solutions don’t convince you, let’s take a look at some simulation evidence:

set.seed(123)
n = 100000
g = c("M", "F")
families = replicate(n = n, expr = sample(x = g, size = 2, replace = TRUE))
n_boys = apply(families, 2, function(x) sum(x == "M"))
full_table = table(n_boys)
at_least_1 = which(n_boys > 0)
n_at_least_1 = length(at_least_1)
at_least_1_table = table(n_boys[at_least_1])
percent_0_boys = round(100 * (full_table / n)["0"])
percent_1_boy  = round(100 * (full_table / n)["1"])
percent_2_boys = round(100 * (full_table / n)["2"])
percent_2_boys_given_1 = round(100 * (at_least_1_table / n_at_least_1)["2"])

I simulated one hundred thousand sets of two children where each child had a 50% chance of being a boy (M) and a 50% chance of being a girl (F). As we would expect from the analysis above, 25% of the families had no boys, 25% had two, and 50% had one girl and one boy. Of the families that had at least one boy, what percent had two boys? It turns out 33%.

All of these avenues of showing the correct answer were previously presented by other twitter users:

  • The implicit Bayesian approach is given by Lucas and Roxie
    • Lucas’ tweet is here
    • Roxie’s tweet is here
  • The explicit Bayesian approach is given by Riley here
  • Simulation evidence is provided by Erin here