Oh, What A Knight!

Every square on the chessboard has a scope , equal to the number of squares to which a knight on that square can move. The knight's journey is a described by a Markov chain, where the transition probability $p_{ij}$ of a knight moving from square $i$ to square $j$ is $p_{ij} \; = \; \left\{ \begin{array}{lll} s_i^{-1} & \hspace{1cm} & \mbox{if } j \mbox{ is a knight's move away from } i, \\ 0 & & \mbox{otherwise} \end{array} \right.$ where $s_i$ is the scope of the square $i$ . Note that $s_ip_{ij} \; = \; s_jp_{ji}$ for all squares $i,j$ , since the above expression is equal to $1$ if squares $i$ and $j$ are a knight's move apart, and equal to $0$ otherwise. This means that the Markov chain is time-reversible, and in particular that $\sum_i s_i p_{ij} \; = \; \sum_i s_j p_{ji} \; = \; s_j$ for all $j$ , and hence this Markov chain has the (unique) invariant distribution $\pi_i \; = \; \frac{s_i}{s}$ where $s \; =\; \sum_i s_i$ is the sum of the scopes of all the squares. By standard theory (for a non-null recurrent irreducible Markov chain), the expected return time to a square $i$ is equal to $\pi_i^{-1}$ , and so is equal to $\frac{s}{s_i}$ .

The $4$ corner squares have scope $2$ , while the eight squares adjacent to the four corners have scope $3$ . There are $20$ squares with scope $4$ (the remaining $16$ edge squares and the four squares "one diagonal in" from the corner squares). There are then $16$ squares with scope $6$ , while the $16$ central squares have scope $8$ . Thus $s \; = \; 4\times2 + 8\times3 + 20\times4 + 16\times6 + 16\times8 \; = \; 336$ and hence the expected return time of the knight to a corner square is $\frac{336}{2} = \boxed{168}$ .