For a Markov chain with states $S=\{1,2,3,4,5,6\}$ and transition probabilities matrix $$P=\begin{pmatrix} 1/2 & 1/4 & 1/4 & 0 & 0 &0 \\ 0 & 1/3 & 0 & 0 & 2/3 & 0 \\0 & 0 & 1/2 & 1/2 & 0 & 0 \\ 0 & 0 & 1/3 & 2/3 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1/3 & 2/3 \\ 0 & 0 & 0 & 0 & 2/3 & 1/3 \end{pmatrix}$$ I want to find $\lim_{n \to \infty} p_{ij}^{(n)},i,j\in S$.
The classes $\{1\},\{2\}$ are transient and $\{3,4\},\{5,6\}$ are recurrent. For the recurrent states I have found stationary probabilities $(2/5,3/5)$ and $(1/2,1/2)$.
I am stuck with how to compute the probability of reaching the recurrent states $\{3,4\}$ and $\{5,6\}$. I know a theorem for reaching a set before another set using the total probability rule so I tried something similar. Let $$T=\min\{n\geq 0:X_n \in \{3,4\}\}$$ and $$h_i=P(T<\infty|X_0=i).$$ I want to find $P(T<\infty)$. I tried to use total probability: $h_3=1,h_4=1,h_1=p_{11}h_1+p_{12}h_2+1/4$ which gives me $1$ which I don't think is correct...
HINT
Why not just solve $x = xP$?
If you insist on doing this from first principles, note that you can absorb $\{3,4\}$ into one state, and $\{2,5,6\}$ into another -- now you have a 3-state chain with the first state transitive and the problem becomes much easier...
UPDATE
Following the suggestion of
NCh, here is an elaboration. When you collapse the states as I mentioned, you get 3 states $A=\{1\}, B = \{2,5,6\}$ and $C = \{3,4\}$ with transition matrix $$ \begin{pmatrix} 1/2 & 1/4 & 1/4\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} $$ so to solve $x = xP$ we can solve $x(P-I)=0$ which implies $x_A=0$ and has $x_B$ and $x_C$ unconstrained.Since from the statement of the problem, the states $B$ and $C$ are interchangeable, we have $x_B = x_C$ by symmetry, hence $x_B=x_C = 1/2$.
Now similarly you can solve the subsystem implied by $B$ and $C$ to get $x_2=0$ and $x_3=2/5, x_4=3/5, x_5=x_6=1/2$. Now when you recombine them with the conditional probability of ending up in $B$ and $C$ respectively you get $$ x_1=x_2=0 \quad \text{and} \quad x_3=2/10, x_4=3/10, x_5=x_6=1/4. $$