I have a Markov Chain with transition matrix
\begin{equation} P= \begin{pmatrix} 0 & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4}\\ \frac{1}{5} & 0 & 0 & 0 & \frac{4}{5} \\ 0 & 0 & 1 & 0 & 0\\ \frac{5}{6} & 0 & \frac{1}{6} & 0 & 0\\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} \end{equation} The states are labelled $0,1,2,3,4$. States $2$ and $4$ are absorbing. I am interested in calculating the probabilities that I reach state $2$ and $4$ respectively. Initially, I did the following. $$$$ Define \begin{equation} f=\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\left(\frac{1}{4}\frac{1}{5}\right)^{i}\left(\frac{1}{4}\frac{5}{6}\right)^{j}=\frac{1}{1-\frac{1}{20}}\frac{1}{1-\frac{5}{24}}=\frac{480}{360} \end{equation} $f$ is the sum of probabilities that the chain loops between states $0$ and $1$ $i$ times and $0$ and $3$ $j$ times for all possible values of $i$ and $j$. Then \begin{equation} P(2)=\frac{1}{4}f+\frac{1}{4}\frac{1}{6}f=\frac{140}{361} \end{equation} Where the first term is the probability of accessing state $2$ from state $0$ and the second term is the probability of accessing state $2$ from state $3$. \begin{equation} P(4)=\frac{1}{4}f+\frac{1}{4}\frac{4}{5}f=\frac{216}{361} \end{equation} Where the first term is the probability of accessing state $4$ from state $0$ and the second term is the probability of accessing state $4$ from state $1$. The problem with this is that these 2 probabilities do not sum to 1. Does anyone know why?
