Two Children, One Girl, and Why the Answer Is 1/3 Until She Opens the Door

Told that one of two children is a girl, the chance both are girls is 1/3. Watch a girl open the door instead and it is 1/2, from the same four families and the same prior. One likelihood separates them: a mixed family always satisfies the statement, but sends the girl to the door only half the time. Push the identifying detail to a girl born on a Tuesday and the answer slides to 13/27.

I have two children. One of them is a girl. What is the probability that both are girls?

Now the same family, asked a second way. I have two children. I knock, and a girl opens the door. What is the probability that both are girls?

The two questions look like the same question wearing different clothes. They are not. The first answer is $1/3$ , the second is $1/2$ , and a candidate who gives the same number twice has just told the interviewer something about how they handle conditioning.

This one shows up in first-round quant interviews, usually on the phone, and most often for trading and quantitative research seats rather than development ones. It is cheap to ask and it does a lot of work: the wrong answer is not a calculation error, it is a candidate conditioning on the wrong event.

Four families, and the one that gets removed

Take the sex of each child as an independent fair coin, older child first. Four sequences, each with probability $1/4$ :

\{\,GG,\; GB,\; BG,\; BB\,\}

The statement “at least one is a girl” is a fact about the pair. It is true of $GG$ , $GB$ and $BG$ , and false only of $BB$ . So it deletes one cell and leaves the other three at equal weight:

P(GG \mid \text{at least one girl}) \;=\; \frac{P(GG)}{P(\text{not } BB)} \;=\; \frac{1/4}{3/4} \;=\; \frac{1}{3}.

(1)

The reflex answer is $1/2$ , and it comes from reading the sentence as though it named a child. If you hear “one of them is a girl” as “this particular child is a girl,” then only the other child is still random and the answer is a coin flip. The sentence does not name a child. It reports a property of the set.

The asymmetry is easy to miss because $GB$ and $BG$ feel like one outcome, “one of each.” They are two. Mixed families are twice as common as two-girl families, which is the whole reason the answer lands below a half.

Why a girl at the door is a different event

Now change how the information arrives. Nobody makes a statement about the pair. One child walks to the door, and I see who it is. Model that the obvious way: the child who answers is picked uniformly from the two.

The evidence is now $E = \{\text{the child I see is a girl}\}$ , and its probability depends on the family:

P(E \mid GG) = 1, \qquad P(E \mid GB) = P(E \mid BG) = \tfrac{1}{2}, \qquad P(E \mid BB) = 0.

A two-girl family sends a girl to the door every time. A mixed family sends one half the time. That factor is the entire difference between the two puzzles. Averaging over the four equally likely families,

P(E) \;=\; \tfrac{1}{4}\!\cdot\! 1 \;+\; \tfrac{1}{4}\!\cdot\!\tfrac{1}{2} \;+\; \tfrac{1}{4}\!\cdot\!\tfrac{1}{2} \;+\; \tfrac{1}{4}\!\cdot\! 0 \;=\; \tfrac{1}{2},

and Bayes gives

P(GG \mid E) \;=\; \frac{P(E \mid GG)\,P(GG)}{P(E)} \;=\; \frac{1 \cdot \tfrac{1}{4}}{\tfrac{1}{2}} \;=\; \frac{1}{2}.

(2)

Fig. 1 — Same four families, two filters. The statement keeps the mixed cells whole; the sighting keeps them at half weight.

The difference is one likelihood, not one interpretation

It is tempting to file this under “ambiguous wording” and move on. That is the wrong lesson, and interviewers push on it. Both problems are perfectly well posed once you write down how the evidence was generated. They differ in a single number.

Where the two problems part company

In the statement version, $P(\text{evidence} \mid GB) = 1$ : a mixed family always satisfies “at least one girl.” In the door version, $P(\text{evidence} \mid GB) = 1/2$ : a mixed family only sometimes shows you the girl. Everything else in the two calculations is identical.

Under-weighting the mixed families relative to $GG$ is what drags the posterior up from $1/3$ to $1/2$ . This is the same machinery that decides the Monty Hall problem, where the host’s rule for choosing a door is what makes switching worth $2/3$ , and the same machinery that resolves Bertrand’s box paradox. In each case the puzzle is not about the outcome you observed. It is about which other outcomes could have produced the same observation.

The practical version of that, on a desk: a number that reaches you because someone chose to send it carries different information than the same number sampled at random. Selection is part of the likelihood.

Born on a Tuesday, and the slide toward one half

There is a follow-up that interviewers keep in reserve, and it looks absurd until the mechanism above makes it obvious. Suppose I say: one of my two children is a girl born on a Tuesday. Now what?

Give each child a sex and a weekday, so $14$ equally likely labels each and $196$ ordered pairs. Families with no Tuesday girl number $13^2 = 169$ , so $196 - 169 = 27$ families qualify. Among two-girl families, $7^2 = 49$ pairs exist and $6^2 = 36$ avoid Tuesday, leaving $49 - 36 = 13$ . So

P(GG \mid \text{a girl born on a Tuesday}) \;=\; \frac{13}{27} \;\approx\; 0.4815.

(3)

A detail with nothing to do with sex moved the answer from $0.333$ to $0.481$ . The reason is that the detail is rare. The more specific the identifying description, the less likely two children both match it, so the fewer double-counted families remain, and the closer the statement gets to naming an individual. Take the description to a unique one and the value reaches exactly $1/2$ , which is the door problem again. In general, if the marker has probability $p$ per child,

P(GG \mid \text{a girl with the marker}) \;=\; \frac{2p - p^{2}}{4p - p^{2}} \;\xrightarrow[\;p \to 0\;]{}\; \frac{1}{2},

which returns $1/3$ at $p = 1$ and $13/27$ at $p = 1/7$ . The two headline answers are the endpoints of one curve, not two rival opinions.

If you get asked the first version cold, the answer that scores is $1/3$ , followed immediately by the sentence that earns the rest of the marks: it is $1/3$ because the information is about the pair, and it becomes $1/2$ the moment the information is about one identified child. Say which model you are using and the interviewer has nothing left to trap you with.

Sources and further reading

Boy or girl paradox — the history of the problem and the Tuesday variant.
Bertrand’s box paradox — the same likelihood argument with coins in drawers.
Monty Hall problem — where the protocol behind the evidence decides the answer.