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Probability that numbers 1...6 show up at least once when rolling 8 dice

How can this be solved using the inclusion-exclusion principle.

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1 Answer 1

  1. There are $6^8$ outcomes in total.

  2. We subtract the $5^8$ outcomes that avoid "1", the $5^8$ outcomes tha tavoid "2", ..., the $5^8$ outcomes that avoid "6".

  3. Then we have to add back the $4^8$ outcomes that avoid both "1" and "2" as they have been subtracted twice; similarly for all $6\choose 2$ sombinations of two avoided numbers.

How did we treat outcomes that avoid three numbers so far? Each such outcome was counted once in (1), subtracted three times in (2), added back ${3\choose2}=3$ times in (3), so th enet count is one, which is one to many. Therefore we need to subtract $3^8$ for each of $6\choose 3$ choices of three numbers.

Thus the count of successful outcomes is $6^8-6\cdot 5^8+{6\choose 2}\cdot 4^8-{6\choose 3}3^8\pm\ldots$ (where you still have to fill in how to treat the ases that four or more numbers are avoided). Divide this by the total number $6^8$ to obtain the probability.

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I know that I am not suppose to comment on other peoples answers, but just curious, When you say "(where you still have to fill in how to treat the cases that four or more numbers are avoided)" What do you mean by this? Does this mean that you would have to add (6 choose 4)2^8 - (6 choose 5)1^8? –  Jason10 Apr 7 '13 at 19:37
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