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Message: <blockquote data-quote="Achan hiArusa" data-source="post: 4875077" data-attributes="member: 2597">What you were calculating is the percent chance of getting a number of failures equal to your die pool.  Since you only need a single success, probabilities are additive (really they use the binomial theorem).  So you chances of failure are (not including 10 agains):luck die--90% (+10% of really bad thing happening)1 die--70%2 dice--40%3 dice--10%4 or more dice--automaticHowever that is not neccessarily true, so we must use the binomial theorem to get the correct answer for the percentage chance of failure:luck die--90%1 die--70%2 dice--42%3 dice--19%4 dice--7.6%5 dice--2.8%6 dice--1.0%7 dice--0.36%8 dice--0.12%9 dice--0.041%10 dice--0.014%Now this does not include the fact that 8, 9, or 10 agains can bump the die probability to the next level.  There is a 10% this will happen on a 10 again, 20% on 9 agains, and 30% on 8 agains which would require feeding these numbers back into the binomial theorem and calculating chance of rerolls (and would require the sum of a Riemann-Zeta function since its for every 8, 9, or 10).But that doesn't count the fact that I know people whom the dice are out to get.</blockquote>