Talk:Sybaris/@comment-68.35.32.214-20140518051040/@comment-173.227.104.102-20140518181738

as much as it shames me, My math did contain a single glaring mistake. There are indeed 16 outcomes for 4 bullets, but only 11 of them include 2 critical hits, not 12 as I originally stated. Therefore, the chance of scoring at least 2 critical hits is actually .625 x .625 x .625 x .625 x 11 = 1.6784 or 167.84%. My argument remains valid.