6. In the Cartesian coordinate system $x O y$, the set of points
$$
K=\{(x, y) \mid x, y \in\{-1,0,1\}\} \text {. }
$$

If three points are randomly selected from the set $K$, the probability that there exist two points among these three with a distance of $\sqrt{5}$ is $\qquad$
6. $\frac{4}{7}$.

It is known that there are 9 points in the point set $K$, so the number of ways to randomly select three points from the point set $K$ is $\mathrm{C}_{9}^{3}=84$.

Label the points in the point set $K$ as $A_{1}$, $A_{2}$, ..., $A_{8}$, $O$ according to Figure 2, where there are 8 pairs of points with a distance of $\sqrt{5}$.

By symmetry, consider the case of taking points $A_{1}$ and $A_{4}$. Then, there are 7 ways to choose the remaining point, resulting in $7 \times 8=56$ sets of three points (the order of the three points in each set is not considered).

For each $A_{i}(i=1,2, \cdots, 8)$, there are exactly $A_{i+3}$ and $A_{i+5}$ in the point set $K$ that are at a distance of $\sqrt{5}$ from it (the subscripts are understood modulo 8), thus, there are exactly $\left\{A_{i}, A_{i+3}, A_{i+5}\right\}(i=1,2, \cdots, 8)$ these 8 sets of three points that are counted twice.
Therefore, the number of sets of three points that meet the condition is $56-8=48$.
Thus, the required probability is $\frac{48}{84}=\frac{4}{7}$.