I recently started playing chess (damn quarantine) and came up with this problem.
Supposing we have a blank 8x8 chessboard. We pick a spot at random and place a special black piece which cannot move. Then pick two another spots at random and place two white kings. (the kings move according to normal chess rules)
Q: What are the chances that either king could capture the black piece? If instead of kings we had white rooks, what would the answer be?
Additionaly, if we repeat the same problem with two black pieces which both have to be captured and two white kings what would the answer be?
For one black piece and two white kings:
The black piece has $8$ neighbouring squares with probability $\frac{6^2}{8^2}=\frac9{16}$, $3$ neighbouring squares with probability $\frac4{8^2}=\frac1{16}$ and $5$ neighbouring squares with the remaining probability $1-\frac9{16}-\frac1{16}=\frac38$. If the black piece has $k$ neighbours, the probability that it cannot be captured by the two white kings is $\frac{\binom{63-k}2}{\binom{63}2}=\frac{(63-k)(62-k)}{63\cdot62}$. Thus the probability that the black piece cannot be captured by either king is
$$ \frac{\frac9{16}(63-8)(62-8)+\frac1{16}(63-3)(62-3)+\frac38(63-5)(62-5)}{63\cdot62}=\frac{1193}{1488}\approx80.2\%\;. $$
For one black piece and two white rooks:
The number of squares from which a given square can be attacked by a rook is the same for all squares, namely $2\cdot7=14$. Thus the probability that the black piece cannot be captured by either rook is
$$ \frac{(63-14)(62-14)}{63\cdot62}=\frac{56}{93}\approx60.2\%\;. $$
For the cases with two black pieces, I’ll wait for your reply to my comment under the question.