# An unintuitive probability question

Suppose you meet a stranger in the Street walking with a boy. He tells you that the boy is his son, and that he has another child. Assuming equal probability for boy and girl, and equal probability for Monty to walk with either child, what is the probability that the second child is a male?

The way I see it, we get to know that the stranger has 2 children, meaning each of these choices are of equal probability:

(M,M) (M,F) (F,M) (F,F)

And we are given the additional information that one of his children is a boy, thus removing option 4 (F,F). Which means that the probability of the other child to be a boy is 1/3. Am I correct?

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The question in your title and the question you end off with are different. Which are you asking: whether it's a variant of the Monty Hall problem, or whether the probability is $1/3$? –  msh210 Nov 27 '11 at 18:37
Thanks, fixed title –  Tomer Nov 27 '11 at 18:42
See the Boy or Girl Paradox. –  r.e.s. Nov 27 '11 at 18:56
I am pretty sure that what is unintuitive is not the question but its answer, no? –  Mariano Suárez-Alvarez Nov 27 '11 at 19:07

I don't think so. The probability of this is $\frac 1 2$ which seems clear since all it depends on is the probability that the child not walking with Monty is a boy. The probability is not $\frac 1 3$! The fact that he is actually walking with a son is different than the fact that he has at least one son! Let $A$ be the event "Monty has $2$ sons" and $B$ be the event "Monty is walking with a son." Then the probability of $A$ given $B$ is $P(A | B) = \frac {P(A)} {P(B)}$ since $B$ happens whenever $A$ happens. $P(A) = \frac 1 4$ assuming the genders of the children are equally likely and the genders of the two children are independent. On the other hand, $P(B) = \frac 1 2$ by symmetry; if the genders are equally likely then Monty is just as likely to be walking with a boy as a girl. This particular problem isn't even counterintuitive; most people would guess $\frac 1 2$.
Now, on the other hand, suppose $C$ is the event that Monty has at least one son. This event has probability $\frac 3 4$ and so if you happen to run into Monty on the street without a child, but you ask him if he has at least one son and he says "yes," it turns out that Monty has two sons $\frac 1 3$ of the time using the same calculation. This seems a little paradoxical, but you actually get more information from event $B$ than you do from event $C$; you know more than just that he has a son, but you can pin down which one it is: the one walking with Monty.