1. 5
  1. 3

    Normally when you’re partitioning a fixed number of slots into some number of bins, the smaller the bins (and therefore the more bins), the greater the variance of the count of balls/bin when you throw a fixed number of balls into random bins. I’d expect the same to be true here: the most-loaded word should have more of its bits set than the average, so an element not in the filter which mapped to that word should have a higher false positive probability. Not quite sure what this would mean in practice, though.