A Fair Die Is Rolled N Times. What Is The Probability That At Least 1 Of The 6 Values Never Appears?
Answer : I think via inclusion/exclusion the probability that at least one of the six values never appears after n rolls of the die would be: p ( n ) = ( 6 1 ) ( 5 6 ) n − ( 6 2 ) ( 4 6 ) n + ( 6 3 ) ( 3 6 ) n − ( 6 4 ) ( 2 6 ) n + ( 6 5 ) ( 1 6 ) n p(n) = {6 \choose 1}({5 \over 6})^n - {6 \choose 2}({4 \over 6})^n + {6 \choose 3}({3 \over 6})^n - {6 \choose 4}({2 \over 6})^n + {6 \choose 5}({1 \over 6})^n p ( n ) = ( 1 6 ) ( 6 5 ) n − ( 2 6 ) ( 6 4 ) n + ( 3 6 ) ( 6 3 ) n − ( 4 6 ) ( 6 2 ) n + ( 5 6 ) ( 6 1 ) n To understand, first just consider the probability of a 1 never showing up: ( 5 6 ) n ({5 \over 6})^n ( 6 5 ) n Easy enough. Now what are the chances of either a 1 never showing up OR a 2 never showing up. To first order it's just twice the above, but by simply doubling the above, you've double-counted the events where neither a 1 nor a 2 show up, so you have to subtract that off to correct the double counting: 2 ( 5 6 ) n − ( 4 6 ) ...