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Let $L$ be the last vertex visited for the first time, and assume $N\ge 2$. Every vertex is eventually painted with probability one: from any position, $N-1$ consecutive clockwise steps visit the entire cycle, and the probability of such a block is $2^{-(N-1)}$. Thus the probability that coverage has not occurred after $m$ successive blocks is at most $(1-2^{-(N-1)}{)}^m$, which tends to zero.

Fix $j\ne 0$ and cut the cycle at $j$. Represent it as the path with positions $0,1,\dots ,N$, where the two endpoints both represent $j$ and the original starting vertex is at position $N-j$. Up to the first visit to $j$, the walk is a simple symmetric random walk on this path, killed at either endpoint. The event $L=j$ is precisely that it visits both $1$ and $N-1$ before being killed: travelling between these two positions visits all intermediate vertices, while visiting all non-$j$ vertices necessarily includes these two.

For $N\ge 3$, first wait until the walk reaches one of $1,N-1$. This happens almost surely before it can reach an absorbing endpoint. If it reaches $1$ first, it must next reach $N-1$ before zero. The gambler's ruin probability is $1/(N-1)$: the hitting probability $h(i)$ on $0\le i\le N-1$ satisfiesso its successive differences are constant and $h(i)=i/(N-1)$. Starting at $1$ gives the stated probability. If the first endpoint reached is $N-1$, reflection of the path gives the same probability of reaching $1$ before $N$. The Strong Markov property at the first such hit therefore gives $\mathbb{P}(L=j)=1/(N-1)$, independently of which endpoint was hit first and of the start position.

Consequently the last visited vertex of a simple random walk on a cycle has distributionFor $N=2$ the same formula follows directly since the only other post must be last. For $N=1$, painting the initial post already completes the task, so there is no subsequent last unpainted post.