The Two-Children Problem
At least one is a girl. What's the probability both are? Not 1/2 — the classic conditioning trap.
also known as: boy-girl paradox · Mrs. Smith's children
A family has two children. You're told that at least one of them is a girl. What is the probability that both are girls?
ANSWER
1/3
CANONICAL SOLUTION
Enumerate the sample space of two-child families (by gender and birth order, each child independently boy or girl with equal probability): {BB, BG, GB, GG}. Each has probability 1/4.
Conditional on 'at least one girl', we exclude BB. The remaining outcomes are {BG, GB, GG}, each with probability 1/3 (they're equally likely). Only one of these three outcomes (GG) has both children as girls. So:
AI ALTERNATIVE FRAMING
The trap is intuition saying 1/2 because 'well, the other child is 50/50 boy-or-girl'. But conditioning on 'at least one girl' is not the same as conditioning on a specific child. Imagine you meet one of the two children randomly and she's a girl — then the remaining sibling is indeed 50/50, and P = 1/2. But the problem didn't tell you about a specific child; it told you something about the family as a whole (that BB is excluded). Those two conditions create different posteriors. The 1/3 answer respects the information given exactly.
On QuantPrep, every practice problem gets an AI-generated alternative explanation like this — tailored to the canonical solution, streaming live.
COMMON VARIATIONS
At least one child is a boy born on a Tuesday. What's the probability both are boys?
Answer ≈ 13/27, not 1/3. The extra day-of-week conditioning changes the posterior. This variant is a Jane Street favourite for testing conditioning rigor.
You randomly meet one of the two children, and she's a girl. What's the probability the other is a girl?
1/2. Because the conditioning is on a specific random child (not on the set), the outcomes where you could have met either child become distinguishable.
The eldest child is a girl. What's the probability both are girls?
1/2. Conditioning on a specific child (the eldest) collapses the asymmetry.
FIRMS THAT ASK THIS
TECHNIQUE
"Given X happened, what's the probability of Y?" — the second-most-common framing in quant interviews.
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In the classical secretary problem, as n → ∞, what fraction should you reject outright before accepting the next best-so-far candidate?
FAQ
Because 'at least one' describes a property of the set of children; 'the eldest' identifies a specific child. The first eliminates only {BB}; the second eliminates {BB, BG} (since BG means the eldest is a boy). Different conditioning events, different posteriors.
Some versions are deliberately ambiguous to probe how carefully you read. If the problem says 'I have two children and I tell you one is a girl', that's unambiguous — P = 1/3. If it says 'you meet one of my children and she's a girl', that's P = 1/2. Always clarify which phrasing you're working with.
Both involve subtle conditioning. In Monty Hall, the trap is ignoring that the host's choice is constrained; here, the trap is treating 'at least one girl' as if it were 'a specific random girl'. Both problems punish the same careless instinct: conflating set-level information with individual-level information.
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