Last visited vertex of a simple random walk on a cycle ($\mathbb{P}(L=j)=1/(N-1)(j\ne 0)$)

A simple symmetric random walk on a cycle of $N\ge 2$ vertices, started at vertex zero, has its last newly visited vertex uniformly distributed on the other $N-1$ vertices. To prove this for a specified vertex $j\ne 0$, cut the cycle at $j$, representing it by a path with absorbing endpoints $0,N$ both corresponding to $j$. All other vertices are visited before $j$ exactly when both interior endpoints $1,N-1$ are visited before absorption. After the first hit on either interior endpoint, the probability of reaching the other before the adjacent absorbing endpoint is $1/(N-1)$ by gambler's ruin. The Strong Markov property makes this the desired probability, independent of the original start position. For $N=2$ the sole unvisited vertex has probability one.