jmc

There are $\left(\genfrac{}{}{0ex}{}{7}{2}\right)=21$ pairs of points in the heptagon, and all but 7 (the sides of the heptagon) are diagonals, which means there are 14 diagonals. So there are $\left(\genfrac{}{}{0ex}{}{14}{2}\right)=91$ pairs of diagonals. Any four points on the heptagon uniquely determine a pair of intersecting diagonals. (If vertices $A,B,C,D$ are chosen, where $ABCD$ is a convex quadrilateral, the intersecting pair of diagonals are $AC$ and $BD$.) So the number of sets of intersecting diagonals is the number of combinations of 4 points, which is $\left(\genfrac{}{}{0ex}{}{7}{4}\right)=35$. So the probability that a randomly chosen pair of diagonals intersect is $\frac{35}{91}=\boxed{\frac{5}{13}}$.