HMMT-Nov 2019 guts, Problem 8

There are 36 students at the Multiples Obfuscation Program, including a singleton, a pair of identical twins, a set of identical triplets, a set of identical quadruplets, and so on, up to a set of identical octuplets. Two students look the same if and only if they are from the same identical multiple. Nithya the teaching assistant encounters a random student in the morning and a random student in the afternoon (both chosen uniformly and independently), and the two look the same. What is the probability that they are actually the same person?

Solution: Let $X$ and $Y$ be the students Nithya encounters during the day. The number of pairs $(X, Y)$ for which $X$ and $Y$ look the same is $1 \cdot 1+2 \cdot 2+\ldots+8 \cdot 8=204$, and these pairs include all the ones in which $X$ and $Y$ are identical. As $X$ and $Y$ are chosen uniformly and independently, all 204 pairs are equally likely to be chosen, thus the problem reduces to choosing one of the 36 pairs in 204, the probability for which is $\frac{3}{17}$. $\fbox{\frac{3}{17}}$.

Answer: \frac{3}{17}