@cfbolz I've been a little bothered (mostly theoretically) by the fact that allocations are sampled at most once, regardless of their size. Turns out the correction factor isn't that hard to compute.
An allocation of n bytes is sampled pn times in expectation. The probability of being sampled 0 times is (1 - p)^n, so the expected sampling count given the allocation is sampled at least once is pn / [1 - (1 - p)^n] (sanity check: reduces to 1 if n == 1)