2. In a $5 \times 5$ square grid, three chess pieces are placed arbitrarily at the centers of 25 smaller squares, with at most one piece per center. The probability that the centers of these three pieces form the vertices of a triangle is

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2. $\frac{537}{575}$.

Let $A$, $B$, and $C$ represent the centers of three chess pieces. The number of different ways to place $A$, $B$, and $C$ at the centers of 25 small squares is $25 \times 24 \times 23$.

The number of points forming a straight line in the centers of the small squares can be categorized as follows:
(1) There are 12 groups of five points in a straight line, including five rows, five columns, and two diagonals. In this case, the number of different ways to place $A$, $B$, and $C$ in a straight line is $12 \mathrm{~A}_{5}^{3}=720$;
(2) There are 4 groups of four points in a straight line, which are parallel to the diagonals. In this case, the number of different ways to place $A$, $B$, and $C$ in a straight line is $4 \mathrm{~A}_{4}^{3}=96$;
(3) There are 16 groups of three points in a straight line, including 4 groups parallel to the diagonals and 12 other groups. In this case, the number of different ways to place $A$, $B$, and $C$ in a straight line is $16 \mathrm{~A}_{3}^{3}=96$.

Therefore, the probability that the centers of the three chess pieces form the vertices of a triangle is
$$
P=1-\frac{720+96+96}{25 \times 24 \times 23}=\frac{537}{575} .
$$