HMMT-Nov 2023 guts, Problem 6

There are five people in a room. They each simultaneously pick two of the other people in the room independently and uniformly at random and point at them. Compute the probability that there exists a group of three people such that each of them is pointing at the other two in the group.

Solution: Solution: The desired probability is the number of ways to pick the two isolated people times the probability that the remaining three point at each other. So, \[ P=\left(\begin{array}{l} 5 \\ 2 \end{array}\right) \cdot\left(\frac{\left(\begin{array}{l} 2 \\ 2 \end{array}\right)}{\left(\begin{array}{l} 4 \\ 2 \end{array}\right)}\right)^{3}=10 \cdot\left(\frac{1}{6}\right)^{3}=\frac{5}{108} \] is the desired probability. $\fbox{\frac{5}{108}}$.

Answer: \frac{5}{108}