Markov chain state probability 
Given the above Markov chain diagram and letting $ b_k$ denote the probability that the process is in state $(k)$, the book derives this : 
$$b_k = \frac {W_{\min}-k}{W_{\min}} \cdot \frac{p}{1-p}b_0 \text{ where } 1\le k \le (W_{\min}-1)$$
If I understand it correctly, we have the probability $b_0$ of being in that state $0.$ Then $\frac p {W_{\min}}$ to reach the next, but not sure about $\frac 1 {1-p}$ and $W_{\min}-k$
Could you please explain the step by step process behind this derivation.
Thank you.
 A: I am assuming that the sequence $(b_k)_{k=0,\ldots,W_\min-1}$ denotes the invariant probability of the chain and that the down arrow "win" from state 0 to the outside does not play a role (that is, from state 0, the chain can only jump to states 0,1,.., $W_{\min}-1$).
In this case, $(b_k)$ must satisfy the global balance equations. In state 0, the global balance equation is
$$
b_0 \Big(1 - (1- p) - \frac{p}{W_{\min}}\Big) = b_1 (1-p).
$$
This gives:
$$
b_1 = \frac{p}{1-p} b_0 \frac{W_{\min} - 1}{W_{\min}} 
$$
For state $k>0$, the global balance equation is
$$
b_{k} ( 1 - p ) = b_0 \frac{p}{W_{\min}} + b_{k+1}(1-p).
$$
This gives
$$
b_{k+1} = b_{k}  - b_0 \frac{p}{1-p} \frac{1}{W_{\min}}.
$$
At this point, we substitute.  When $k=1$, we get
$$
b_{2} = b_{1}  - b_0 \frac{p}{1-p} \frac{1}{W_{\min}}
= \frac{p}{1-p} b_0 \frac{W_{\min} - 1}{W_{\min}}  - b_0 \frac{p}{1-p} \frac{1}{W_{\min}}
= \frac{p}{1-p} b_0 \frac{W_{\min} - 2}{W_{\min}}
$$
and iterating over all $k$ we get the formula you wrote.
