 Solution

The directed positive-transition graph gives the communicating classes $\{1,3,6\}$, $\{2,4\}$ and $\{5\}$. The first and last classes are closed; $\{2,4\}$ is open. In particular the PDF's fourth row has all six entries equal to $1/6$. The converted TeX incorrectly drops the last entry; that incomplete row is not a transition distribution.

The finite irreducible closed class $C=\{1,3,6\}$ reaches state six almost surely. For example, from every state in $C$ there is a uniformly positive probability of hitting six within a bounded number of steps, so the probability of avoiding it for successive blocks tends to zero. Put $h_i={\mathrm{Pr}}_i(T_6<\infty )$, with $h_1=h_3=h_6=1$ and $h_5=0$. First-step conditioning at states two and four givesSolving $4h_2-h_4=2$ and $5h_4-h_2=3$ yields $\boxed{h_2=13/19}$, with $h_4=14/19$.

Starting in state three, the chain remains in $C$. In the order $(1,3,6)$ its transition matrix isIts characteristic polynomial is $(\lambda -1)(\lambda +1/4)(\lambda +1/6)$. Thus, by the Cayley-Hamilton theorem, $q_n=(M^n{)}_{2,3}$ is a linear combination of $1,(-1/4{)}^n,(-1/6{)}^n$. The initial values are $q_0=0$, $q_1=1/3$, and $q_2=(1/3)(1/2+1/3+1/4)=13/36$. Solving for the three coefficients givesThe limiting value $12/35$ agrees with the state-six probability of the stationary distribution $(8/35,3/7,12/35)$ of this closed class.