Stochastic Process and Probability A particle moves among $n + 1$ vertices that are situated on a circle in the following manner. At each step it moves one step either in the clockwise direction with probability $p$ or the counterclockwise direction with probability $q = 1 − p$. Starting at a specified state, call it state $0$, let $T$ be the time of the first return to state $0$. Find the probability that all states have been visited by time $T$.
I am quite confused as of to how I should start this question. Please guide me along if you know!
Thank you!
update: so "someone" (whose uni I shall not name) tried to defame me and that person (whom I shall not name either) can't even comment now
FINALLY COMPLETED THIS QUESTION! THANKS ALL!
CHEERS TO A BETTER COMMUNITY (:
 A: Expanding a bit on what Nate Elderedge suggested, since it's definitely the right idea:


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*Suppose the vertices are labeled from the starting point in a clockwise direction as $\{0, 1, \dots, n\}$. Consider two cases in the problem, based on where the walk's first step occurs. Specifically, if $A$ denotes the event that the walk visits all steps before returning to $0$, then $$\mathbb P_0[A] = \mathbb P_0[A \mid X_1 = 1] \cdot \mathbb P_0 [X_1 = 1] + \mathbb P_0[A \mid X_1 = n] \cdot \mathbb P_0[X_1 = n].$$
Note that $\mathbb P_0[X_1 = 1] = p$ and $\mathbb P_0[X_1 = n] = q$, so all that remains is to compute those two conditional probabilities.

*Use the Markov property to rewrite the conditional probabilities. Specifically, $$\mathbb P_0[A \mid X_1 = 1] = \mathbb P_1 [T_n < T_0]$$
where $T_n$ and $T_0$ are the times of the first visits to $n$ and $0$, respectively. This expression is now a classic "gambler's ruin" problem with an asymmetric random walk. There are many solutions to this problem: see this and this for two examples. (My favorite proof is the martingale approach, outlined in the second link -- but I'm not sure what you're already comfortable with.) You can then turn the other conditional probability into a gambler's ruin problem in the same fashion.

