 Solution

Temporarily label the individual pieces. A random placement is a uniform injection of thirty-two distinct pieces into sixty-four squares, giving ${}_{64}{P}_{32}$ equally likely outcomes. For a colour-matching placement, the black pieces have ${}_{32}{P}_{16}$ available injections, and independently the white pieces have the same number within their disjoint set of squares. ThereforeHere ${}_n{P}_r=n!/(n-r)!$. Treating same-colour pieces as indistinguishable gives the same probability, because the common within-colour labeling factors cancel. The two colour choices in the favourable count occupy disjoint square sets; no independence assumption about unrestricted placements is required.