Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 2 / 11F / b / ii / Solution

For each $j\ne 0$, cut the cycle at $j$. The event that $j$ is last is the event that the lifted walk covers the other $n-1$ residues before crossing that cut. Gambler’s ruin (or cyclic symmetry of the two expanding endpoints) gives $\mathbb{P}(Z=j)=1/(n-1)$.