A question about how to get the limiting probability. Suppose $p=\begin{bmatrix}
 0& 1\over 3 &0  &2\over 3 \\ 
 0.3&  0& 0.7 &0 \\ 
 0&  2\over 3&0  &1\over3 \\ 
 0.8&  0&  0.2& 0
\end{bmatrix}$is the transition probability matrix of a Markov
chain with state space {1, 2, 3, 4}. How to get the limiting probabilities of the Markov chain. I think if more different alternative answer and apporach would be better. In fact i know a little bit about limiting probability but not sure how to apply in this question. Clear step to illustrate how it works or explaination would be appreciated and i wanna learn how others interpret the concept of probability
 A: To find an invariant distribution for a Markov chain, which will give you information about the long term probability, you can use two methods. Solving the "left hand equations" for an irreducible, recurrent Markov chain: $$\pi_i = \displaystyle \sum_{j\in I} P_{ji} \pi_j$$ 
(Where $I$ is the state space)will give you an invariant measure, and the restriction 
$$\displaystyle \sum_{j\in I} \pi_j = 1$$
will give you an invariant distribution. These equations give the long term proportion of time spent in each state $i$ as $\pi_i$. This is invariant because it is saying "the probability of being in a state $i$ is the same as the sum of the probabilities being in any other state $j$ (which is $\pi_j$) and then moving into state $i$ (which is $P_{ji}$)
When you're more comfortable with the idea of invariant distributions, you can often save time (as you can in this case) by looking at the so-called "detailed balance equations", but they're more complicated, and I've run out of time! It's best to start with the basics though, and look up more info on detailed balance a bit later if you ask me.
