How often are two permutations comparable?

Two permutations of $\left[ n \right]$ are comparable in the Bruhat order if one is closer, in a natural way, to the identity permutation, $1 \, 2 \, \cdots \, n$, than the other. We show that the number of comparable pairs is of order $\left(n!\right)^2/n^2$ at most, and $\left(n!\right)^2\left(0.708\right)^n$ at least. For the related weak order, the corresponding bounds are $\left(n!\right)^2\left(0.362\right)^n$ and $\left(n!\right)^2\prod_{i=1}^n \left(H\left(i\right)/i\right)$, where $H\left(i\right):=\sum_{j=1}^i 1/j$. In light of numerical experiments, we conjecture that for each order the upper bound is qualitatively close to the actual number of comparable pairs.