We set up a light and a dark bishop on the chessboard randomly (e.g., based on row shuffling). What is the probability that, in 3 attempts, at least once neither bishop attacks the other.
The first rook can be placed in 64 different ways. The second rook, since it cannot stand on the field occupied by the first, can only be placed on 63 fields. Therefore, the total number of possible placements for 2 rooks is \( l = 64 \cdot 63 \). (Alternatively, \( l = V_{64}^{2} = 64 \cdot 63 \)).

A rook on a chessboard can attack 21, 23, 25, or 27 fields, depending on whether it is 0, 1, 2, or 3 fields away from the edge of the board. The number of such fields is 28, 20, 12, and 4, respectively.

Therefore, the number of favorable placements in terms of attackability is:

\[
\begin{gathered}
k = 28 \cdot 21 + 20 \cdot 23 + 12 \cdot 25 + 4 \cdot 27 = 4(147 + 115 + 75 + 27) = \\
= 4 \cdot 364 = 16 \cdot 7 \cdot 13
\end{gathered}
\]

Thus, the probability that two randomly placed rooks can attack each other is:

\[
v_{1} = \frac{k}{l} = \frac{16 \cdot 7 \cdot 13}{64 \cdot 63} = \frac{13}{4 \cdot 9} = \frac{13}{36}
\]

The probability that in 3 attempts, all three times the rooks can attack each other is:

\[
v_{3} = \left(\frac{13}{36}\right)^{3}
\]

Therefore, the probability that in 3 attempts, at least once neither rook attacks the other is:

\[
v = 1 - \left(\frac{13}{36}\right)^{3} = 0.953
\]

Zsombok Zoltán (Bp., IV., Konyves Kalman g. II. o. t.)