This may not be quite what you had in mind, but: suppose you were trying to compute the absolute value of a Gauss sum $\sum_{a=0}^{p-1} \zeta^{a^2}$ where $\zeta = e^{ \frac{2 \pi i}{p} }, p$ an odd prime. Intuitively you might guess that quadratic residues are uniformly distributed and thus this sum should behave like a sum of $p$ random unit vectors. The expected value of the square root of the length of such a sum is just $p$, so one might expect that the absolute value of a Gauss sum should be about $\sqrt{p}$. In fact it is exactly $\sqrt{p}$!