The two children paradox

Example — Standard version

A family has two children. You know that at least one is a boy. What is the probability that both are boys?

Temptation: $1/2$ (the other child is $50$–$50$). Wrong. The correct answer is $1/3$.

Property — Reading via the sample space

Sample space (assuming equiprobability $1/4$ for each pair): $\{MM, MF, FM, FF\}$. The information “at least one is a boy” excludes $FF$, leaving three equiprobable cases: $\{MM, MF, FM\}$. The condition “both boys” is realised only in $MM$. Hence $P(MM \mid \text{at least one M}) = \frac{P(MM)}{P(\text{at least one M})} = \frac{1/4}{3/4} = \frac{1}{3}.$

Warning — The conditioning trap

Change the wording, change the answer. If instead the statement were “the firstborn is a boy”, then “both boys” only means that the second is a boy: $P=1/2$. The information “at least one” is less specific than “the first”, and this difference translates into the conditional probability.

Links

Topics: Probabilita
Concepts: Paradosso dei due figli · Probabilita condizionata · Spazio campione
Skills: Calcolo probabilita · Ragionare per casi