What algorithm to find end state probabilities
Take your matrix, and normalize the rows to sum to 1. Denote the result by $A$; in your example,
$$
A = 
\begin{bmatrix}
0&amp; 1/2&amp; 0&amp; 0&amp; 0&amp; 1/2 \\
4/9&amp; 0&amp; 1/3&amp; 2/9&amp; 0&amp; 0 \\
0&amp; 0&amp; 0&amp; 0&amp; 0&amp; 0 \\
0&amp; 0&amp; 0&amp; 0&amp; 0&amp; 0 \\
0&amp; 0&amp; 0&amp; 0&amp; 0&amp; 0 \\
0&amp; 0&amp; 0&amp; 0&amp; 0&amp; 0
\end{bmatrix}
$$
Let $v$ be a row vector with $v_0 = 1$ and $v_i = 0$ otherwise. In your example,
$$
v = \begin{bmatrix} 1&amp;0&amp;0&amp;0&amp;0&amp;0 \end{bmatrix}
$$
Then $vA^t$ gives the probabilities of being in each state after $t$ steps, and so $w := v \sum_{t=0}^\infty A^t$ gives the expected number of times in each state. Since a final state terminates the process, for each final state $i$, $w_i$ is in fact the probability of the process ending at $i$. The vector $w$ can be calculated using the formula
$$
w = v (I-A)^{-1}.
$$
In your case, we get
$$
w = \begin{bmatrix} 9/7 &amp; 9/14 &amp; 3/14 &amp; 1/7 &amp; 0 &amp; 9/14 \end{bmatrix}
$$
In particular, the probabilities of terminating at the final states $2,3,4,5$ are (respectively) $3/14,1/7,0,9/14$.

Regarding algorithms, we can rephrase the problem as follows: given $v,A$, find the vector $w$ such that
$$
w(I-A) = v.
$$
This requires solving a system of linear equations, for which there are many algorithms.