Random rook movements

There are three types of square on the chessboard. Type $1$ squares are those from which the rook must take at least two moves to get to the top-right corner; these are squares in rows $1-7$ and in columns $1-7$ . Type $2$ squares are those from which the rook can reach the top-right corner in one move; these are squares in row $8$ and columns $1-7$ , or else in column $8$ and rows $1-7$ . The only Type $3$ square is the top-right target square in row $8$ and column $8$ . We are counting rows from the bottom, and columns from the left.

From a Type $1$ square, the rook will move to another Type $1$ square with probability $\tfrac67$ , and will move to a Type $2$ square with probability $\tfrac17$ . From a Type $2$ square, the rook will move to a Type $1$ square with probability $\tfrac12$ , will move to another Type $2$ square with probability $\tfrac37$ , and will move to the Type $3$ square, ending the game, with probability $\tfrac{1}{14}$ .

Given these observations, the expected number of moves until the game ends only depends on the Type of the starting square, and not more generally on the actual square. Let $E_1$ be the expected number of moves to end the game, starting from a Type $1$ square, and let $E_2$ be the expected number of moves to end the game, starting from a Type $2$ square. Conditional expectation arguments tell us that $\begin{array}{rcl} E_1 & = & \tfrac67\big(E_1 + 1\big) + \tfrac17\big(E_2 + 1\big) \\ E_2 & = & \tfrac12\big(E_1 + 1\big) + \tfrac37\big(E_2 + 1\big) + \tfrac{1}{14}\times1 \end{array}$ and hence $E_1 \; = \; E_2 + 7 \qquad \qquad 8E_2 \; = \; 7E_1 + 14$ Solving these simultaneous equations yields $E_2 = 63$ and $E_1 = \boxed{70}$ .