There are a total of 5 nodes at the edge of a pentagram 
At each node, you have a 4 choices which will lead you to either a destination node or non-destination node. Assume the decision of path is indepedent of your previous decision, and all decision has to be made random. What is the expected number of node you have to visit, in order to reach to your destination node?
Please give me hint instead of answer
You can proceed with irreducible HMC since the graph given is irreducible. From the property of local balance of we can calculate the value of stationary distribution $\Gamma $ & the entries in it are reciprocal of the expected time to reach the corresponding state back. Here time is synonymous to number of nodes we visit in our journey. A brief hint & workout will be as follows:
Denote the time as : $ T_0 $ = inf{$n \ge 0 : X_n = 0 $ } [Denoting my destination node as 0 - just numbering - you can choose whatever you want]. What we want is then $E_0[T_0]$ i.e. expectation value of the time it takes to reach back my starting node. From the theory of invariants we can calculate $\Gamma$ as $\Gamma(i) = x_i/\Sigma_i x_i $. From the condition of local balance you can calculate the terms $\Gamma[i]$ and hence $\Gamma[0] = x_0/\Sigma_i x_i$. Basically it can be shown easily that $\Sigma_i x_i=E_0[T_0]$ and since we started with state 0, $x_o=1$. Hence, the required quantity is reciprocal of $\Gamma[0]$.
Answer that I got : $E_0[T_0]=5$.