smc

Let $WXYZ$ be a convex cyclic quadrilateral inscribed in a circle. There are $\frac{\left(\genfrac{}{}{0ex}{}{4}{2}\right)}{2}=3$ ways to divide the points into two groups of two. If you pick $WX$ and $YZ$, you have two sides of the quadrilateral, which do not intersect. If you pick $XY$ and $ZW$, you have the other two sides of the quadrilateral, which do not intersect. If you pick $WY$ and $XZ$, you have the diagonals of the quadrilateral, which do intersect. Any four points on the original circle of $1996$ can be connected to form such a convex quadrilateral $WXYZ$, and only placing $A$ and $C$ as one of the diagonals of the figure will form intersecting chords. Thus, the answer is $\frac{1}{3}$, which is option $\boxed{\frac{1}{3}}$. Notice that $1996$ is irrelevant to the solution of the problem; in fact, you may pick points from the entire circumference of the circle.