Let a spider take a random walk on an $n$-vertexes complete graph, starting from vertex number $1$. (I guess it has no significance.). Given the spider visits vertex $2$ before he visits vertex $3$, what is the expected time it would take him to visit vertex $2$?
I got really confused during the computation: a previous question asked for the expected time without the the given condition. Symmetrically (or by Markov Chains), the expected time it would take the spider to return to $1$ is $n$, and noting $e$ as the expected time it will take the spider to visit vertex $1$ where starting at another vertex(which is symmetrically what I was looking for in that question), I get that $n=1+(n-1)\cdot{1\over n-1}\cdot e$ which gives $e=n-1$.
However, when I am given this condition, it starts to get a little confusing. One possibility is $n-2$ (it is a multiple choice question.)but I couldn't assume it is so simple. The computation also have me disputed. I could really use your help.
Michael Hardy has already given in an elegant solution in a comment. To solve the problem more mundanely, note that each step before reaching $2$ has the same probability of $1/(n-2)$ of reaching $2$ (given that it does not reach $3$). Thus the desired time is the expected time until the first success in a series of Bernoulli trials with $p=1/(n-2)$, which is $p^{-1}=n-2$.