Invariant probability starting in transient class I am asked to find the invariant probability for the Markov chain given that the Markov chain starts in the transient state $(1,1)$. I already know the invariant probability of the recurrent class on the right.
I have a Markov chain with the following state space and transition probabilities:

My solution is a triple sum, because we have to include the possibilities that the Markov chain transitions $k$ times through the $(0,0)-(0,1)$ loop but 0 times through the $(0,1)-(2,1)$ loop, etc.
So my answer is
$$\frac{1}{3}\frac{1}{2}\sum_{k=0}^\infty 6^{-k} (\sum_{n=0}^\infty 3^{-n} (\sum_{m=0}^\infty 6^{-m}))=\frac{1}{3}\frac{1}{2} \cdot \frac{6}{5}\frac{6}{5}\frac{3}{2}=\frac{9}{25}$$
The fraction in front of the sums is because the chain has to finally move from state $(0,2)$ to state $(0,3)$.
I have calculated that the invariant probability measure for state $(1,2)$ with initial state inside the recurrent class is $\frac{1}{7}$, giving me an invariant probability of $\frac{9}{25} \cdot \frac{1}{7}=\frac{9}{175}$.
Which is a very ugly number compared to the rest of the assignment. So I am very skeptical. However it is not too improbable because $\frac{1}{7}\approx 0.14$ and $\frac{9}{175}\approx 0.05$ since there is a reasonable chance of ending up in the small recurrent class on the left.
What am I doing wrong?
 A: The triple summation is unnecessary: symmetry considerations tell you that from $(1,1)$ you have a $\frac12$ chance of eventually reaching $(1,0)$ and a $\frac12$ chance of eventually reaching $(0,3)$. 
(For each of them, you have to pass through $(0,1)$, then with probability $\frac13$ continue to an intermediate state (either $(0,0)$ or $(0,2)$), then with probability $\frac12$ continue on to $(1,0)$ or $(0,3)$ instead of returning to $(1,0)$.)
Even if the cases were not symmetric, the triple summation would not be the way to go. Write $p_{00}$, $p_{01}$, $p_{02}$ for the probability of reaching the recurrent class on the left from states $(0,0)$, $(0,1)$, and $(0,2)$. Then we have
$$
   \begin{cases}
      p_{00} = \frac12 + \frac12 p_{01} \\
      p_{01} = \frac13 p_{00} + \frac13 p_{01} + \frac13 p_{02} \\
      p_{02} = \frac12 p_{01}
   \end{cases}
$$ 
which we solve to get $p_{00} = \frac34$, $p_{01} = \frac12$, $p_{02} = \frac14$. This could be done even if we change the transition probabilities from $(0,1)$.
Your calculation of $\frac17$ for the invariant probability of $(1,2)$ given that you're in the right recurrence class matches mine. The overall probability, then, is $\frac1{14}$.

As for why your triple summation gives the wrong answer: to be correct, it needs to include a $\binom{k+n+m}{k,n,m} = \frac{(k+n+m)!}{k!\,n!\,m!}$ term, to account for the fact that the loops can happen in any order. The summation
$$
    \frac16 \sum_{k\ge0}\sum_{n\ge0}\sum_{m\ge 0}\binom{k+n+m}{k,n,m} 6^{-k} 3^{-n} 6^{-m}
$$
gives $\frac12$, as expected. One way to evaluate this sum is to write it as
$$
    \frac16 \sum_{t \ge 0} \sum_{\substack{(k,n,m)\\k+n+m=t}}\binom{k+n+m}{k,n,m} 6^{-k} 3^{-n} 6^{-m} = \frac16 \sum_{t \ge 0} \left(\frac16 + \frac13 + \frac16\right)^t = \frac{\frac16}{1 - (\frac16 + \frac13 + \frac16)} = \frac12
$$
where the sum over $k,n,m$ satisfying $k+n+m=t$ is done by the multinomial theorem, and the sum over $t$ is a geometric series.
