An ant does a randon walk on the complete graph on $n$ vertices (on each step, it jumps to one of the other $n-1$ vertices, regardless of the past). It starts from the vertex $1$.
There are few problems regarding this situation, I'm stuck on the last one. The only seemingly relevant prior problem is to find the expected time until the vertex $2$ is visited. Denoted $E$, it can be computed by the law of total expectation and by symmetry: $$ E=\frac{1}{n-1} \cdot 1 +\frac{n-2}{n-1}(E+1) $$ so $E=n-1$.
The last problem is to find the expected time (=number of steps) until $2$ is visited, given that $2$ was visited before $3$. I had 2 different approaches in mind:
- Given that $2$ was visited before $3$, we can "forget" about $3$ and act like we have the same problem as the one below, but with one vertex less, which yields $E=n-2.$
- $n-1=E(\text{time until 2})=\frac{1}{2}E(\text{time until 2 | 2 before 3})+\frac{1}{2}E(\text{time until 2 | 3 before 2})$ so we need to find the last expectation. The expectation of time until $3$, given "$3$ before $2$", should be $n-2$, and the expectation of the time from $3$ to $2$ should be $n-1$. So the last expectation is $2n-3$ which doesn't add up.
I believe there's a problem with the reasoning in the second approach. Could anyone help me find it? Also, other approaches will be welcomed.