Frog on a 3 by 3 matrix Imagine a frog (f) on a 3 by 3 matrix, with the frog being able to jump to neighboring places from where it sits. So for example if the frog is at $s_{0,1}$, it can subsequently jump to $s_{0,0}, s_{0,2},s_{1,0},s_{1,1}$ and $s_{1,2}$. 
$$ \left( \begin{array}{ccc}
s_{0,0} & f & s_{0,2} \\
s_{1,0} & s_{1,1} & s_{1,2} \\
s_{2,0} & s_{2,1} & s_{2,2} \end{array} \right)$$
Overall there are $9$ configurations for the frog, what is the probability that the frog sits at $s_{0,1}$ after many random jumps? I thought of assigning weights to places on the matrix, by the fact that e.g. $s_{0,1}$ can be reached by $5$ other places, compared to $s_{1,1}$ that has $8$ ways to it. But I don't know how to include this information into the probability calculation. 
 A: The frog jump process is a random walk on this graph.

It is easy to show that the long run probability
at each node is proportional to its degree, i.e., the number of edges
 adjacent to that node. See gmath's answer here.
The degree of the nine nodes are either 8, 5, or 3 and simple
calculations show that the long run probability is 1/5 at the center,
3/40 at each of the four corners, and 1/8 at the other four edge nodes. 
A: let $p_{i,j}$ be the probability of frog being on $s_{i,j}$ after "many" jumps. One more jump added is still going to be "many" so you can use the law of total probability :
e.g.: $p_{0,0} = p_{0,1}/ 3 + p_{1,0} / 3$ (the only way for the frog to get to s00 is if it was on s10 or s01, if the frog was on s10 there would be a probability of 1/3 that it will jump onto s00 since s00 is one of three neighbours of s10, similarly with s01)
You do this with each box giving you 9 simultaneous equations with 9 unknowns, which you can do
edit: you also need the fact that they all add up to 1
