Originally Posted by

**SlipEternal** If the rook may only move up and to the right, then this changes things. It starts off as being $(7,7)$ squares away (meaning it needs to move up 7 and right 7 to get to its target).

The first move has a uniform distribution of odds for which move the rook will make (a $\dfrac{1}{14}$ chance for each move).

The second move will not have a uniform distribution of odds for which move the rook will make.

Suppose the rook moves exactly one square on the first move. Then it is either at $(6,7)$ or $(7,6)$. By symmetry, we the probabilities will be the same, so $E_{(6,7)} = E_{(7,6)}$ and similarly for $E_{(a,b)} = E_{(b,a)}$. Now, there is a $\dfrac{1}{12}$ chance for any of the six positions in the same direction it has already moved, but a $\dfrac{1}{14}$ chance for any position in the direction it has not moved.

So, $E_{(7,7)} = \dfrac{2}{14}\left(7+E_{(6,7)}+E_{(5,7)}+E_{(4,7)} +E_{(3,7)}+E_{(2,7)}+E_{(1,7)}+E_{(0,7)}\right)$

$E_{(6,7)} = \dfrac{1}{12}\left(6+E_{(5,7)}+E_{(4,7)}+E_{(3,7)} +E_{(2,7)}+E_{(1,7)}+E_{(0,7)}\right)+\dfrac{1}{14 }\left(7+E_{(6,6)}+E_{(5,6)}+E_{(4,6)}+E_{(3,6)}+E _{(2,6)}+E_{(1,6)}+E_{(0,6)}\right)$

$\displaystyle E_{(a,b)} = \dfrac{1}{2\cdot a}\sum_{i=0}^{a-1}\left(1+E_{(i,b)}\right) + \dfrac{1}{2\cdot b}\sum_{j=0}^{b-1}\left(1+E_{(a,j)}\right)$

To calculate $E_{(7,7)}$, again I used Excel.

In cell A1, I put 0.

In cell B1, I put the formula:

=1+SUM($\$$A1:A1)/(COLUMN()-1)

I copied that formula to B1:H1

In cell A2, I put the formula:

=1+SUM(A$\$$1:A1)/(ROW()-1)

I copied that formula to A2:A8

In cell B2, I put the formula:

=1+SUM(B$\$$1:B1)/(2*(ROW()-1))+SUM($\$$A2:A2)/(2*(COLUMN()-1))

I copied that formula to B2:H8

In cell H8, I should have $E_{(7,7)} = 5.185714$.

If you do it in Mathematica, you can get an exact result instead of Excel's approximation.