= 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. Therefore
$$
\boxed{P=\frac{({}_{32}P_{16})^2}{{}_{64}P_{32}}
=\frac{(32!)^3}{(16!)^2\,64!}}.
$$
Here ${}_nP_r=n!/(n-r)!$. 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.
Back to article page
