The rook and the bishop are moving independently on the chessboard starting at the same corner The rook and the bishop are moving independently on the chessboard starting at the same corner. What is the average number of steps until they meet again in the same corner, if we know that the bishop moves only on one quarter of the chessboard?
I suppose that I should use Markov chains / processes. I think that I should consider two transition matrices: one 8x8 and one 4x4, and then model a proper graph. We need to find the probability that those 2 figures meet in the same time and in the same spot, not only the probability of getting back to the corner.  I'm new to this subject, so please be indulgent. Any help or tips will be much appreciated.
 A: The PBS infinite series episode “Can a Chess Piece Explain Markov Chains?” can give you an insight into solving this problem.
I realize I’m several years late to the party here but the question is new to me.
To calculate how long it will take a piece to return to its starting position you take every square that piece can legally occupy and count the legal number of moves that reach that square. This is really easy for a rook since it can occupy every square and at every square there are 14 ways to reach that square.
Now sum all of the legal square moves for a piece. With 14 moves and 64 possible squares this is 896 for a rook. So the probability of being at any square is 14/896 (or 1/64, which makes sense because the rook can move anywhere and in our random walk all moves from a given position are equally probable. The probability of returning to the square is the reciprocal of p(occupy square). For a rook it takes on average 64 moves to return to any given starting point.
The Bishop can be a little trickier since the diagonals are of different lengths, but I count 7 legal moves to land in a deep corner and 280 moves overall. This means p(bishop in deep corner) is 7/280 which simplifies to 1/40. Taking the reciprocal we get that the average number of turns it takes a randomly walking bishop to return to a deep corner is 40.
To get the average number of turns it takes for two randomly-walking pieces to return to their shared starting square you take the Least Common Multiple of their average number of turns to return.
The least common multiple of rook return and bishop return is 320.
