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For a finite Markov chain with transition matrix given by: $P=\left[ \matrix{1\over 3 & 2\over 3&0&0 &0 &0 \\ 2\over 3&0&1\over 3&0 &0&0\\0&1\over 3&0&2 \over 3&0&0\\0&0&2\over 3 &0 &1\over 3&0\\0&0&0&1\over 3 &0&2\over 3\\0&0&0&0&2 \over 3&1\over 3}\right]$

I need to find:

  1. the probability of visiting state 6 before visiting state 1, given that my initial state is 2 (so that my initial distribution vector is $\mu_{0}=\left[ \matrix{0 & 1 &0 &0&0&0}\right]$).
  2. The mean number of visits at state 1, until returning to state 6, and assuming $X_0=6$.

My attempts didn't help so much.

As for part one: I tried solving this according to a similar question, by denoting $p_i$ as the probability of visiting state 6 if I'm currently at state $i\in\lbrace1,\dots ,6\rbrace$, but I got a system of linear equations where all $p_i=1$... and generally I had trouble formalizing this question in terms of the random variables of the states.

As for part two: Denoting $t^*_6=\mathbb E\left[T_6|X_0=6\right]$ as the mean recurrence time for state 6, I tried solving the system given by $t_i=1+\sum_{i\neq 6}p_{ij}t_{i}$ but it became a mess. Also since the stationary distribution is $\pi=\left[ \matrix{1\over 6 & \dots & 1\over 6}\right]$, and the chain is aperiodic and irreducable, it seems that $t_6^{8}=6$, but it still doesn't answer the question, as for how many times in average state 1 is visited.

Any help would be appreciated.

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Let $p_i$ denote the probability that you reach state $6$ before state $1$ starting from state $i$. Then $p_2=\frac13p_3$ and $p_3=\frac13p_2+\frac23p_4$. By the symmetry of the problem, $p_4=1-p_3$. Substituting the first and third equations into the second yields $p_3=\frac19p_3+\frac23(1-p_3)$, with solution $p_3=\frac37$ and thus $p_2=\frac17$.

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