Indeed!

And the number of choices that leaves you with is rather tiny.

Worse (or better), from throwing cases at the code, I bet it is possible to prove that all such orderings are optimal so that the one that sort gives you is optimal.

I think it might not be hard to prove and $a <= $b <= $c implies $a <= $c (assuming non-negative cost and selectivity between 0 and 1). And that should be enough.