Consider $n\geq 4$ points arranged clockwise on a circle. Label them $0,\ldots, n-1$. Let $X_1,X_2,\ldots$ be iid Rademacher and define $S_0=0$, $S_p=(\sum_{k=1}^p X_k) \text{ mod } n$. What is the probability that $(S_p)_{p\geq 1}$ visits all of $\{0,2,3,\ldots, n-1\}$ before it ever visits $1$ ?
Simulations (confirmed by this other question) suggest the answer is $\frac 1{2(n-1)}$. I've been trying to carry out an analysis similar to this answer. Note that my question is different from this because I'm not excluding the origin since I consider the walk $(S_p)_{p\geq 1}$, instead of $(S_p)_{p\geq 0}$.
Consider the symmetric random walk $(R_p)_{p\geq 0}$ on $\mathbb Z$, starting at $0$. For any $a\in \mathbb Z$, let $T_a = \inf\{p\geq 1: R_p = a\}$. The probability we're looking for should be $$P(T_0< T_{-(n-2)}<T_1) + P(T_{-(n-2)} < T_0 < T_{-(n-1)}).$$
Next, I think (but am not sure) we should have $$P(T_0< T_{-(n-2)}<T_1) = P(T_0< T_{-(n-2)})P(T_{-(n-2)}<T_1).$$
The probability $P(T_0< T_{-(n-2)})$ of returning to $0$ before ever reaching $-(n-2)$ is $1-\frac 1{2(n-2)}$.
The probability $P(T_{-(n-2)}<T_1)$ is $\frac{1}{(n-2)+1} = \frac 1{n-1}$.
Therefore $$P(T_0< T_{-(n-2)}<T_1) = (1-\frac 1{2(n-2)}) \frac 1{n-1},$$
however $(1-\frac 1{2(n-2)}) \frac 1{n-1} > \frac 1{2(n-1)}$, hence what I've found is incorrect.
I cannot find where I'm mistaken and this has been quite frustrating...
