Consider a pond with one central lilypad and 9 outer lilypads surrounding the central pad in a circle.
A frog hops from one lilypad to another, determining its destination each time independently of where it has been before. When it is on the central lilypad it is equally likely to jump next to any of the outer lilypads. When on an outer one, it has a probability p (0 < p < 1) of jumping to the central one, or q/2 of jumping to each of the outer lilypads to either side of the current location (q = 1 - p)
Let $X_n$ be the lilypad occupied by the frog after the nth jump
In the long run, what fraction of time does the frog occupy each lilypad?
My working so far. Let the states be {0, 1, 2, .., 9} where 0 is the central pad and 1 is the outer pad. This means states 2 & 9 are adjacent pads to 1.
$$\begin{pmatrix} 0 & \frac{1}{9} & \frac{1}{9} & \frac{1}{9} & \frac{1}{9} & \frac{1}{9}& \frac{1}{9} & \frac{1}{9} & \frac{1}{9} & \frac{1}{9}\\ p & 0 & \frac{q}{2} &0 &0&0&0&0&0&\frac{q}{2}\\ p &\frac{q}{2} & 0& \frac{q}{2} &0 &0&0&0&0&0\\ p & 0 &\frac{q}{2} & 0& \frac{q}{2} &0 &0&0&0&0\\ p &0& 0 &\frac{q}{2} & 0& \frac{q}{2} &0 &0&0&0\\ ...\\ p & 0 &0 & 0& 0 &0 &0&\frac{q}{2}&0&\frac{q}{2} \end{pmatrix}.$$
So $\pi_j = \sum_i \pi_i P_{ij}$. Bascially its tedious to right out the limiting probabilities and not sure how to solve them either when written out. Am i doing it right and is there was to find the answer faster.
You are doing it right. But without "thinking" a bit more the solution would be tedious, indeed.
Let $a$ be the stationary probability that the frog sits on the central lilypad. By intuition, the frog will sit on one of the outer things will be of the same probability: $\frac{1-a}9$.
In order to prove that this is a solution, we will have to see if
$$\begin{pmatrix}a\ \frac{1-a}9\cdots\frac{1-a}9\end{pmatrix} \begin{pmatrix} 0 & \frac{1}{9} & \frac{1}{9} & \frac{1}{9} & \frac{1}{9} & \frac{1}{9}& \frac{1}{9} & \frac{1}{9} & \frac{1}{9} & \frac{1}{9}\\ p & 0 & \frac{q}{2} &0 &0&0&0&0&0&\frac{q}{2}\\ p &\frac{q}{2} & 0& \frac{q}{2} &0 &0&0&0&0&0\\ p & 0 &\frac{q}{2} & 0& \frac{q}{2} &0 &0&0&0&0\\ p &0& 0 &\frac{q}{2} & 0& \frac{q}{2} &0 &0&0&0\\ ...\\ p & 0 &0 & 0& 0 &0 &0&\frac{q}{2}&0&\frac{q}{2}\\ \end{pmatrix}=\begin{pmatrix}a\ \frac{1-a}9\cdots\frac{1-a}9\end{pmatrix}.$$
When we calculate the first element of the RHS we will multiply $a$ by $0$ then we will add $p\frac{1-a}9$ nine times. That is, we have the following equation
$$9\frac{1-a}9p=a.$$
So, $$a=\frac p{1+p}.$$
When we calculate the $i$th element then we will have the following equation
$$a\frac19+2\frac q2\frac{1-a}9=\frac{1-a}9.$$
The solution is again $$a=\frac p{1+p}.$$