In this regard SHA-1 despite it's weaknesses can be viewed as a pseudo random function. This meamsmeans we are are.frowing n throwing $n$ balls into n$n$ bins. An output bin remains empty if all the balls miss it. Which happens with probability (1-1/n)^n$(1-1/n)^n$ which is 1/e$1/e$ and that is the portion of output bins which are empty. We also can estimate that the most populated bin has aproxinately log(n)approximately $log(n)$ balls.