Expected number of steps to absorbtion - Markov chain I want to calculate the expected number of steps (transitions) needed for absorbtion. So the transition probability matrix $P$ has exactly one (lets say it is the first one) column with a $1$ and the rest of that column $0$ as entries.
$P = \begin{bmatrix} 1 & * & \cdots & * \\ 0 & \vdots & \ddots & \vdots \\ \vdots & & & \\ 0 & & & \end{bmatrix} \qquad s_0 = \begin{bmatrix} 0 \\ \vdots \\ 0 \\ 1 \\0 \\ \vdots \\  \\ \end{bmatrix}$
How can I now find the expected (mean) number of steps needed for the absorbtion for a given initial state $s_0$?
EDIT: An explicit example here:
$P = \begin{bmatrix} 1 & 0.1 & 0.8 \\ 0 & 0.7 & 0.2  \\ 0 & 0.2 & 0 \end{bmatrix} 
\qquad s_0 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \implies s_1 = \begin{bmatrix} 0.8 \\ 0.2 \\ 0 \end{bmatrix} \implies s_2 = \begin{bmatrix} 0.82 \\ 0.14 \\ 0.04 \end{bmatrix} \ldots $
 A: We conventionally read row by row, but it appears that in your matrix, the columns add to 1 instead. 
I'll write your matrix like this and proceed:
\begin{align*}
  P = 
\begin{bmatrix} 
0 & 0.2 & 0.8 \\ 
0.2 & 0.7 & 0.1  \\ 
0 & 0 & 1
\end{bmatrix}
\end{align*}
which is of the form 
$$P=\left[
\begin{array}{c|c}
Q & R \\ \hline
0 & 1
\end{array}\right]
$$  
and 
\begin{align*}
  s_0 = \begin{bmatrix} 1 & 0 & 0  \end{bmatrix}
\end{align*}
For the expected time to absorption, we need to calculate
$\left(I-Q\right)^{-1} \cdot \begin{pmatrix}
1 \\ 
1 
\end{pmatrix} \approx \left(\begin{array}{r}
1.92307692307692 \\
4.61538461538461
\end{array}\right)
$
A: Let $t_i$ denote the mean time to absorption at state $0$ starting from state $i$ then $t_0=0$, one is after $t_2$, and the usual one-step Markov property yields 
$$t_2=1+\frac15t_1,\qquad t_1=1+\frac7{10}t_1+\frac15t_2,$$ which is solved by
$$
t_2=\frac{25}{13}=1.923\ldots
$$
For $n+1$ states, one gets an $n\times n$ affine system whose solution $(t_i)$ describes the mean times to absorption at state $0$ starting from each state $i\ne0$.
