What would the odds be of (1) of those 24% chances panning out with (5) attempts ? I never learned how to calculate that in school, which is some BS.
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i'm thinking it's 0.24 times the probability of 4 failures, which is 0.3336, so 0.08
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I guess you *would* get 1.2 of them on average, at least
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yeah, what i really want is the approximately 20% chance of getting none of them. phrasing it that from that direction doesn't seem to be helping it either
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