Mess in Chess...

There are $2*7*8 = 112$ interior sides on a chess board, and there is a one-to-one correspondence between these sides and the pairs of squares that have a side in common.

Now without restrictions there are $\binom{64}{2}$ pairs of squares that can be chosen at random, and so the desired probability is

$\dfrac{112}{\binom{64}{2}} = \dfrac{224}{64*63} = \dfrac{1}{18}$ .

Thus $m + n = 1 + 18 = \boxed{19}$ .

1) There total 64 squares on a chess board, arranged in 8x8 form

2) There are 4 corner squares where only 2 other squares share a side so total number of ways for such a square = (4x2)
3) There are 24 squares touching the boarder of the chess board (excluding 4 corner squares) where 3 other squares share a side.=> (24x3)
4) There are remaining 36 squares where 4 other squares share a side => (36x4)

So, total number of ways in which a square can be selected with one side shared is = (4x2) + (24x3) + (36x4) = 224

& total number of ways in which a square can be selected = 64x63

So, probability of selecting one square such that one of it sides is in common = 224/(64x63) = 1/18

On a chess board we have 64 squares,

And there are:

(1) 4 corner squares, with ONLY 2 squares with which they share a common side..

(2) 6*4 = 24 side squares, with 3 squares with which they share a common side..

(3) 6*6 = 36 inner squares, with 4 squares with which they share a common side..

So we have

P = (4/64) (2/63) + (24/64) (3/63)+ (36/64)*(4/63) P = 1/18