To derive a nice formula for the strength of a Rook on a board of Y height and Z width, here is what you do.

First, examine the geometry of the board. You can see that from the 4 corners, the King can be safely checked a total of (Y - 2) + (Z - 2) times. Recall is a Rook is next to the king while delivering check (like Y -1 or Z - 1) then the king can just capture it.

So, we have 4((Y-2)+(Z-2)) so far.

Basically, in the corner, the king can recapture 2 of the rook checks.

Now move over one square for the king (b8) or move down one square (a7). The king would be able to recapture against 3 rook checks. In the case of the a7 king, it could capture a Rook checking on a8,a6, or b7.

You notice on any rectangular board, there are 4 pairs of squares where this is true. On the 8x8 chessboard these are b8/a7, g8/h7, b2/a1, and g2/h1. So, we have 4 instances of (Y-2) + (Z-1) and 4 instances of (Y-1) + (Z-2).

To 4((Y-2)+(Z-2)) from the 1st calculation we add 4((Y-2) + (Z-1)) + 4((Y-1) + (Z-2)).

Recall a probability is a quotient, that being these total squares of safe check divided by the entire population of arrangements. After placing one piece on the board, there are ZY - 1 slots remaining for the next piece. But the first piece can occupy any one of those ZY squares, so you get fractions in terms of ZY-1 and ZY(ZY-1) when you compute the probabilities.

When you collect all such terms for the rook, you get:

P(safe check) = Z + Y - 6/(ZY - 1) + 2(Z + Y)/[ZY(ZY-1)] for the probability that a Rook can safely check a king on any such size board.

Plug in Y = Z = 8, and you get:

8+8-6/63 + 2(8+8)/64(63) =
10/63 + 32/64x63 =
640/64x63 + 32/64x63 =
672/4032 = 1/6

So my results match Taylor, I just computed it a different way since Gothic Chess has a rectangular board.