We will show that “two of the three vertices selected are at a distance of $\sqrt{3}$” is a certain event. The regular hexagon with edge length $1$ is denoted by $A_1A_2A_3A_4A_5A_6$. If there exist two adjacent vertices taken out of the three vertices selected, it may be set as $A_1, A_2$. Note that $A_1A_3 = A_1A_5 = A_2A_4 = A_2A_6 = \sqrt{3}$, so the third vertex must be at a distance of $\sqrt{3}$ from one of the vertices $A_1, A_2$. If any two of the three vertices selected are not adjacent to each other, it can only be $A_1, A_3, A_5$ or $A_2, A_4, A_6$. At this point, there are clearly two vertices with a distance of $\sqrt{3}$.

Therefore, the desired probability is $1$. $\square$