Four distinct points, $A$, $B$, $C$, and $D$, are to be selected from $1996$ points 
evenly spaced around a [circle](https://artofproblemsolving.com/wiki/index.php/Circle). All quadruples are equally likely to be chosen. 
What is the [probability](https://artofproblemsolving.com/wiki/index.php/Probability) that the [chord](https://artofproblemsolving.com/wiki/index.php/Chord) $AB$ intersects the chord $CD$? 
$\text{(A)}\ \frac{1}{4}\qquad\text{(B)}\ \frac{1}{3}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{2}{3}\qquad\text{(E)}\ \frac{3}{4}$
Let $WXYZ$ be a convex [cyclic quadrilateral](https://artofproblemsolving.com/wiki/index.php/Cyclic_quadrilateral) inscribed in a circle.  There are $\frac{\binom{4}{2}}{2} = 3$ 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 $\boxed{\text{B}}$.
Notice that $1996$ is irrelevant to the solution of the problem; in fact, you may pick points from the entire [circumference](https://artofproblemsolving.com/wiki/index.php/Circumference) of the circle.