[naniwaduni]

> then average case O is worst case O because the worst case term dominates

No, you're overgeneralizing. It's true that the worst case term dominates if e.g. a constant fraction of inputs exhibit the worst-case behavior, so that average complexity is lower bounded by a constant fraction of the worst-case behavior.

This isn't true for the kind of algorithms where it's customary to talk about average case behavior. We only normally talk about average cases when as input size increases, the fraction of inputs with the higher complexity bound shrinks faster. For example, we can imagine an algorithm which, on inputs of size n, requires n^2 operations on 1/n of possible inputs, and n operations on the other (n-1)/n inputs. Then the average runtime is, formally, (n^2)*(1/n)+n*(n-1)/n=2n-1, which is still O(n), even though the worst-case runtime is still O(n^2).

Where this goes wrong is, e.g. perhaps the O(n^2) runtime occurs on "trivial" inputs which are disproportionately likely to appear. For example, if we the above hypothetical function is idempotent and runs in O(n^2) on all of its possible outputs, but O(n) on all other inputs, it'd be easy to accidentally quadratic by calling it on its result repeatedly under the assumption that it's safe to do so.*