The Girl at the Door

I tell you I have two children and that one of them is a girl. You knock on my door and a girl greets you, whom you correctly take to be my daughter. What is the probability that I have two girls? Compare it with the version where you are only told that at least one child is a girl.

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#The four equally likely families

With two children the genders are $BB, BG, GB, GG$ , each with probability $\tfrac14$ .

#Only being told at least one is a girl

This bare fact rules out $BB$ and leaves $BG, GB, GG$ equally likely, so

\PP(GG \mid \text{at least one girl}) = \tfrac13. \tag{1}

#Meeting a girl at the door

Now you see one particular child, effectively drawn at random, and she is a girl. The chance of that is $0$ from $BB$ , $\tfrac12$ from $BG$ and from $GB$ , and $1$ from $GG$ . By Bayes,

\PP(GG \mid \text{met a girl}) = \frac{\tfrac14 \cdot 1}{\tfrac14 \cdot 0 + \tfrac14 \cdot \tfrac12 + \tfrac14 \cdot \tfrac12 + \tfrac14 \cdot 1} = \frac{\tfrac14}{\tfrac12} = \tfrac12. \tag{2}

All four families are equally likely, but they are not equally likely to send a girl to the door. Weighting each by that chance, zero up to one, makes the two-girl family worth a half of the girl-answered cases, not the third you get from the bare fact alone.

#Why they differ

Meeting a specific daughter carries more information than the bare existence of a girl. A two-girl family is twice as likely to send a girl to the door as a one-girl family, and that extra weight lifts the answer from $\tfrac13$ to $\tfrac12$ .

#Result

The girl at the door makes it $\tfrac12$ , against $\tfrac13$ when you are merely told at least one child is a girl. Same headline fact, different conditioning, different answer.