Probability returning to initial state 
Let $P=\begin{bmatrix}0&\frac{1}{2}&\frac{1}{2}\\\frac{1}{2}&0&\frac{1}{2}\\\frac{1}{2}&\frac{1}{2}&0\end{bmatrix}$ and $P^{(n+1)}=P^{(n)}P.$
I know that if you start in any vertex the probability of return after $n$ hops is the same $P_{11}^{(n+1)}=P_{22}^{(n+1)}=P_{33}^{(n+1)}$
then $$p_{11}^{(n+1)}=\frac{1}{2}p_{12}^{(n)}+\frac{1}{2}p_{13}^{(n)}$$
since $p_{12}^{(n)}+p_{13}^{(n)}=1$
$$p_{11}^{(n+1)}=\frac{1}{2}\left(1-\frac{1}{2}p_{12}^{(n)}\right)$$
but I do not know how to solve this recurrence relation or if is right.

Transition probability matrix (P) in this case will be
$$
P=\begin{bmatrix}0 & 0.5 & 0.5\\
0.5 & 0 & 0.5\\
0.5 & 0.5 & 0\end{bmatrix}.
$$
Though $p_{ii}=0$ for $i=0,1,2$, it can be observed that $p_{ii}^n\ne0$ for $n\ge 2$ (also intuitive). Rest is a straight forward calculation depending on $n$.
