16. (1996 American High School Mathematics Examination) On a circle, there are 1996 points evenly distributed. Four different points $A, B, C, D$ are chosen equally from these points. The probability that chords $AB$ and $CD$ intersect is ( ).
A. $\frac{1}{4}$
B. $\frac{1}{3}$
C. $\frac{1}{2}$
D. $\frac{2}{3}$
E. $\frac{3}{4}$
16. B. Reason: Consider the order of points $A, B, C, D$.

For any convex quadrilateral, if $AB$ and $CD$ are exactly the diagonals, they will intersect. When $A, B$ are adjacent vertices, there are 8 possible orders; $C, D$ have 2 possible orders. When $A, B$ are opposite vertices, there are 4 possible orders; $C, D$ have 2 possible orders. Thus, the required probability is $\frac{4 \times 2}{4 \times 2 + 8 \times 2} = \frac{1}{3}$.