LogFAQs > #882757821

It's been ages since I've done any serious work with combinatorics, but what I'm thinking here is that we know the odds of a miss occurring at any given step (1/n+1), and we know at most one miss occurs if everyone dies. With that one miss occuring, the other ones all hit which has the same odds as the no-miss scenario.

So the answer would be (.2/5)+(.2/4)+(.2/3)+(.2/2)+.2, or roughly a 46% chance of everyone dying.
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JeffRaze, for all your random spellcasting needs