Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 2 / 9F / b / Solution

Complementing every friendship sends every degree $d$ to $n-1-d$, and hence sends $S=M+L$ to $2(n-1)-S$. Since the random graph and its complement have the same law, the distribution of $S$ is symmetric about $n-1$, giving $\mathbb{E}S=n-1$.

The same symmetry gives equal probabilities strictly below and above $n-1$. There is positive probability at $n-1$: for odd $n$ use a regular graph of degree $(n-1)/2$; for even $n=2m$, use $K_{m,m}$ with a perfect matching removed and then restore one matching edge. Thus the lower median as defined is also $n-1$.

Solved by gpt-5.6-sol high.