Probability a Markov chain is absorbed at a specific state without initial distribution On a problem sheet I have been given (non-assessed) we are given the following transition matrix for a Markov chain $(X_0,X_1,\dots)$ with state space $S=\{1,2,3,4,5\}$:
$$\begin{pmatrix}
0 & 0 & 1 &0&0\\
0 & 0 & 4/5 &1/5 &0\\
0&1/6&2/3&0&1/6\\
0&0&0&1&0\\
0&0&0&0&1
\end{pmatrix} $$
We are asked to work out the probability that we are absorbed at state 4, i.e., let $T$ be the time of absorption we wish to find $\mathbb{P}(X_T=4)$.
My first thought is to use law of total-probability, conditioning on starting in state $i$ for $i=1,2,3,4,5$, but without the initial distribution this seems impossible. How can I proceed?
Edit: Answer is 1/6, for any curious.
 A: Clearly, the initial distribution has to be provided. To make the best out of the situation, notice (or recall) that the $n$-step transition probability matrix for the $1$-step transition probability matrix $W$ is $W^n$ (Chapman-Kolmogorov equation for time-homogeneous Markov chains). This means that independent of the initial distribution we can look at the transition probabilities from each state to each other state taking $n$ steps at once. This power converges fairly quickly ($n=20$ or so), I took the liberty to compute
$$W_\infty=\begin{pmatrix}
0 & 0 & 0 & \frac{1}{6} & \frac{5}{6}\\
0 & 0 & 0 & \frac{1}{3} & \frac{2}{3}\\
0 & 0 & 0 & \frac{1}{6} & \frac{5}{6}\\
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 1\\
\end{pmatrix}.$$
This means that after "infinitely" many steps we go from $1$ to $4$ with probability $1/6$ (!), to $5$ with probability $5/6$. From state $2$ to $4$ we make it with probability $1/3$, to $5$ with probability $2/3$. The interpretation for states $3$, $4$ and $5$ is analogous. My guess would be that the initial distribution was supposed to be $p_1=1$, but any initial distribution $(p_1,\dots,p_5)$ with
$$\frac{1}{6}p_1+\frac{1}{3}p_2+\frac{1}{6}p_3+p_4=\frac{1}{6}$$
would also be possible.
