What is the stationary distribution of a bishop's legal moves in chess? I have been asked to consider the following questions and I'm not sure what to do. Can anyone help out?
Consider the random movement of a knight on a chessboard. At each time step, we pick one of the bishop’s legal moves at random.
(1) What is the stationary distribution? 
(2) What is the expected number of moves to return to the corner (1,1) when we start there?
Any help would be much appreciated!
 A: The stationary distribution is not difficult. The expected return time then follows. 
Assuming we are only considering one colour, e.g. including $(1,1)$ but not $(1,2)$, then there are $32$ positions and from each each you can move to $7$, $9$, $11$ or $13$ other positions depending on where you start: $7$ from $14$ positions, $9$ from $10$ positions, $11$ from $6$ positions, and $13$ from $2$ positions. By symmetry the number of legal moves out of a position is equal to the number of moves into that position.  Multiply these up and there are $280$ legal moves in total.  
Since the underlying positions are connected and aperiodic, the stationary distribution makes all $280$ legal moves equally likely, so for example a position with $m$ legal moves has a probability of $\frac{m}{280}$ in the stationary distribution. 
Since $(1,1)$ has $7$ legal moves, its probability in the stationary distribution is $\frac{7}{280}=\frac1{40}$.  So the proportion of time spent at $(1,1)$ is also $\frac1{40}$.  There is no legal move directly from $(1,1)$ to itself, so the expected return time from $(1,1)$ is ${40}$.
