Two Children and a Boy
Kevin has two children. Each child is equally likely to be a boy or a girl, and the genders are independent.
You are told that at least one of Kevin's children is a boy. What is the probability that both children are boys?
Hints
- Write out the full sample space for two children -- there are four equally likely outcomes. Conditioning on "at least one boy" simply removes one of them.
- Use the definition of conditional probability: $P(BB \mid A) = P(BB \cap A) / P(A)$. Note that $BB$ is a subset of $A$, which simplifies the numerator.
- The complement gives $P(A) = 1 - P(GG) = 3/4$. There is only one outcome ($BB$) with both boys, so the numerator is $1/4$.
Worked Solution
How to Think About It: The heuristic is condition by counting the surviving equally-likely outcomes, and the trap is the phrasing. "At least one is a boy" is a statement about the *pair*, not about a specific child you inspected — so it kills only the all-girl outcome, not three of four. The intuitive $1/2$ comes from silently rewriting the problem as "a particular child is a boy, what's the other?" That is a different (and here, incorrect) conditioning. Whenever a gender/coin puzzle feels like it "should" be $1/2$, check whether the information is about one labeled item or about the collection.
Quick Estimate: List the four equally likely ordered pairs — $BB, BG, GB, GG$ — and cross out the one that violates the condition ($GG$). Three survive, each equally weighted, and exactly one ($BB$) is a win: $\tfrac{1}{3}$. You can do the whole thing on your fingers; no arithmetic beyond "1 out of 3."
Formal Solution:
Sample space (each probability $1/4$): $\{BB, BG, GB, GG\}$.
Let $A = \{\text{at least one boy}\} = \{BB, BG, GB\}$, so $P(A) = 3/4$. Since $\{BB\}\subset A$, we have $P(BB\cap A) = P(BB) = 1/4$. Then
$$P(BB \mid A) = \frac{P(BB\cap A)}{P(A)} = \frac{1/4}{3/4} = \boxed{\tfrac{1}{3}}.$$
Answer: $P(\text{both boys}\mid\text{at least one boy}) = 1/3$.
Intuition
The answer $1/3$ feels wrong because most people implicitly read the problem as: "I looked at one of the children and saw a boy -- what's the chance the other is also a boy?" That version gives $1/2$ -- knowing one specific child is a boy tells you nothing about the other. But the problem is asking something different: given that the pair of children contains at least one boy (without identifying which one), what is the probability both are boys? Conditioning on a set-level event (the pair has at least one boy) is weaker than conditioning on an individual (this specific child is a boy), and the weaker condition leaves more uncertainty -- hence the lower probability.
This is a classic example of how the precise framing of information matters enormously in probability. In market-making or signal processing, a similar subtlety arises constantly: did you observe a specific signal, or do you merely know that at least one of several possible signals fired? The distinction changes the posterior. Getting this right is the difference between correct Bayesian reasoning and a systematically biased intuition.