Expected return time of a knight to a corner

Question

A knight starts on a corner square of a standard $8 \times 8$ chessboard. Each move it chooses uniformly at random among its currently legal knight moves (from a corner there are only $2$ ; from a central square there can be up to $8$ ).

What is the expected number of moves until the knight first returns to its starting corner?

Accepted Answer

The graph is irregular here, corner squares have degree 2, central squares degree 8, so lumping by symmetry is hopeless. But the return-time formula still applies to any graph with a uniform-random-edge walk: E[	ext{return to } v] = \frac{2E}{\deg(v)}. You only need two numbers: the total number of knight-move edges on the board, and the degree of a corner. Count the edges. Summing the number of knight moves available from every square of an 8 	imes 8 board gives 336 (this is a standard tally: 8 corners contribute few, the central squares many). Since that sum counts each edge from both end…

Expected return time of a knight to a corner

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