If there are two siblings and you meet one of them and he is male, what is the probability that the other sibling is also male?

Question

To me it seems like the obvious answer, ½, is correct, and that this may have been intended to be an example of the boy/girl paradox but is not. My AI professor asked the class this today, and claims that the answer is ⅓. His logic is as follows:

Of the four possibilities, MM, MF, FM, FF, only FF can be eliminated. Considering that you know at least one sibling is male, of the remaining possibilities there is a ⅔ chance that the other sibling is female.

If you were to know that the older of the two siblings is male, you could cross out MF and FF, and the remaining probabilities are in fact 50/50 since in one case the younger sibling is male, and in the other she is female. Meeting one sibling instead of the other in an arbitrary way should not behave any differently, unless the question were to specifically imply you were more likely to meet a sibling of a certain gender. When I asked after class, the professor seemed to disagree on this point, so perhaps I am missing something here.

To me it seems like treating MF and FM as both valid possibilities after meeting a male sibling introduces knowledge monty-hall style. If this is the case, how can I state this formally?

Answer 1

I agree with $\frac{1}{2}$, and I'll present one intuitive justification and two formal arguments (each of which makes an assumption which I believe to be justified based on your question).

Intuitive justification: Suppose there are four families living on the block, MM,MF,FM,FF. Note that of the eight children there are four males, and two of the males ($\frac{1}{2}$ of them) have a male sibling. The case of randomly meeting one of these eight children seems equivalent to your case.

Argument #1: Here we assume that you are equally likely to have met the older child as you are two have met the younger child. Let MMO denote the event where both children are male and you have met the older child, and let FMY denote the event where the children are female and male (in that order) and you have met the younger child, etc. The events MMO, MMY, MFO, and FMY are the events which involve your meeting a male, and they are all equally likely. Two of them have the sibling male.

Argument #2: Here we assume that, given that the family has one male and one female child, you are a priori equally likely to have met the male as you are to have met the female, and we apply Bayes' Theorem. Let $M$ be the event that the family is MM and $F$ be the event that the family is FM. Let $A$ be the event that you meet a male; then the assumption stipulates $P(A|F) = \frac{1}{2}$, and we are looking for $P(M|A)$. We compute $$P(M|A) = \dfrac{P(A|M)P(M)}{P(A)} = \dfrac{P(A|M)P(M)}{P(A|M)P(M) + P(A|F)P(F)} = \dfrac{1\cdot\frac{1}{4}}{1\cdot\frac{1}{4} + \frac{1}{2}\cdot\frac{1}{2}} = \dfrac{1}{2} $$

Of course, if the circumstances of the meeting were such that you were a priori more likely to meet a male than a female child, these assumptions don't hold, and the answer may well not be $\frac{1}{2}$.

Answer 2

Formal settings

Since a formal solution is asked, I use a few notations in axiomatic probability in Kolmogorov's Foundations of the Theory of Probability to make things clear and concise.

• sample space: $\Omega = \{(C,c,s) \mid C \in \{F,M\}, c \in \{f,m\}, s \in \{0,1\}\}$
  • set of a priori outcomes
  • capital letter means older; small letter means younger
  • $M/m$ means male; $F/f$ means female
  • $1$ means younger child sampled (already met); $0$ means younger child unsampled (unseen)
  • each element is called elementary event in Kolmogorov's language
• $\sigma$-algebra: $|\Omega|<\infty \implies {\cal A} = {\cal P}(\Omega)$
  • we have finite sample space, so the $\sigma$-algebra has to its power set
  • due the this unique choice of $\sigma$-algebra, it doesn't have much meaning in finite $\Omega$, so we won't use this in the following
  • under the finite case, it's the same as the field of sets in the book
• probability: $P$ is the uniform distribution over $\Omega$, so for each $\omega \in \Omega$, $P(\omega) = \frac18$.

Wrong interpretation

\begin{array}{|c|c|c|} \hline & m & f \\ \hline M & \rlap{\checkmark}\square & \square \\ \hline F & \square & \times \\ \hline \end{array}

A wrong choice of sample space $$ \Omega' = \{(M,m),(M,f),(F,m),(F,f)\}$$ gives rise to the wrong probability $\frac13$ since $M/m$ of the sample is undistinguished. (We mistreat $(M,m,0)$ and $(M,m,1)$ as one case.)

Right table

\begin{array}{|c|c|c|c|c|} \hline {\small\text{sampled} \setminus \text{unsampled}} & M & m & F & f \\ \hline M & \times & \rlap{\checkmark}\square & \times & \square \\ \hline m & \rlap{\checkmark}\square & \times & \square & \times \\ \hline F & \times & & \times & \\ \hline f & & \times & & \times \\ \hline \end{array}

• Target event: $E = \{(M,m,s) \mid s \in \{0,1\}\}$ (both children are male)
• A posteriori condition: $G = \{(M,c,0), (C,m,1) \mid C \in \{F,M\}, c \in \{f,m\} \}$ (sample is male)

Required probability \begin{align} & P(E \mid G) \\ =& \frac{P(E \cap G)}{P(G)} \\ =& \frac{2}{4} \\ =& \frac12 \end{align}

Conclusion

It's very important to choose a correct $\Omega$.

Answer 3

I believe I found the correct way to cross out possibilities, among the 4 initial cases MM, MF, FM, FF.

Let us first bring some identification to distinguish the two 'M's in case of MM, because a major source of error can be the interchangeability of the two brothers. So let us assume that one child is older that other (even twins have minutes between their births). The first letter denotes the older child, while the second letter the small child.

"one of them is a male" is equivalent to the reunion of the cases "the older one is a male" and "the smallest one is a male", with equal chances for these two cases to happen, unless no other information is given. In the first case, we can cross out FM and FF, remaining with MM and MF, thus 50% chances for the other kid to be a male. In the second case, we can cross out MF and FF, remaining with MM and FM, thus again 50% for the other kid to be a male. Overall, the chances are 50%, even with this 'cross out possibilities' method.

I feel that when crossing out the FF option, because 'one of them is a male', as your professor did, is the same error as when saying that flipping a coin twice can bring only 3 possible results: two heads, two tails, a head and a tail - thus 33% chances for each of these possibilities. What is omitted here is that 'one head and one tail' case is a joint situation of two atomic situations. 'one head and one tail' - it is important to establish which flip is what - the order is important. Similarly, 'one of them is male' can lead to confusions unless we specify exactly which one is a male.

For the same reason i decided that some order relationship must be introduced in the notation 'MM', and I picked age as this order criteria. It could have been weight or height, as well. Unless we specify a clear identity to the letters, their interchangeability can lead to logic mistakes, as shown above.