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How would one calculate the probability of rolling a six (at least once) on seven dice? Would I be correct in proposing $1 - P(\text{no sixes})$ which is $1 - (5/6) ^ 7$?

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marked as duplicate by mookid, M Turgeon, Sami Ben Romdhane, Magdiragdag, user127096 Mar 26 at 19:08

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up vote 2 down vote accepted

Yes, that is correct, since "no sixes" is the complementary event to "at least one 6".

$$P(\text{at least one 6}) = 1 - P(\text{no 6 at all}) = 1 - (5/6)^7$$

If the question was "exactly one six" it would be different, but that does not seem to be the question.

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