Particle random walk on a graph Suppose that particle does a random walk on the following graph:

It starts from $A$ and at every moment of time goes to any adjacent vertex with the same probability. Find


*

*expected time returning to $A$

*how many times it visits $D$ before returning to $A$

*how many times it visits $C$ before returning to $A$


My attempts on solving first part of the problem:
Denote by $a$ expected time to return to $A$ starting from $A$, $b$ -- expected time to return to $A$ starting from $B$ and so on for $c$, $d$, $e$.
They are linked as
$\begin{cases}
a = \frac{1}{2}(c+1) + \frac{1}{2}(b+1)\\
b = \frac{1}{2} + \frac{1}{2}(c+1)\\
c = \frac{1}{2} + \frac{1}{2}(b+1)\\
d = \frac{1}{2}(c+1) + \frac{1}{2}(e+2)\\
e = \frac{1}{2}(d+2) + \frac{1}{2}(c+1)
\end{cases}$
and the solution of this system is $a=3, b=2, c=2, d=5, e=5$, so the answer is that expected time returning to $A$ is $3$.
But one of the comments to this question says that for connected undirected graph the expected returning time is given by $\frac{2n}{d}$, where $n$ is the number of edges and $d$ is the degree of the starting vertex. For this graph it gives us $a=\frac{2\cdot6}{2}=6$.
Where is my mistake? It seems that I wrote wrong system of equations. Can anyone fix it and also explain how second and third parts should be done? 

I have successfully solved it and now want to upload an answer because it may be helpful for someone in a future.
Denote the set of all vertices of a graph by $\{V_i\}$ and the degree of $i$-th vertex by $\mbox{deg}V_i$. Then $k$-th element of stable distribution vector is given by $\pi_k = \frac{\mbox{deg}V_k}{\sum_i\mbox{deg}V_i}$ since $\sum_k\pi_k=\frac{\sum_k\mbox{deg}V_k}{\sum_i\mbox{deg}V_i}=1$ and $(\pi P)_k=\sum_i \pi_iP_{ik}=\frac{\mbox{deg}V_k}{\sum_j\mbox{deg}V_j}$. Random walk on a graph can obviously be interpreted as a regular Markov chain, so $\pi_k=\frac{1}{\mu_k}$, where $\mu_k$ is the average time required for returning to $V_k$.
Denote by $\rho_k(s)$ an average number of visits $k$ before returning to $s$. In that way for a particular task we obtain:


*

*$\sum_i\mbox{deg}V_i=12$ and $\mu_A=\frac{1}{\pi_A}=\frac{12}{2}=6$

*$\rho_D(A)=\frac{\pi_D}{\pi_A}=1$

*$\rho_C(A)=\frac{\pi_C}{\pi_A}=\frac{1/3}{1/6}=2$

