HOW MANY PACKETS DO I HAVE TO BUY? Hi all, I'm posing this question because I'm interested in the answer from a purely theoretical point of view - I'm not going to go out and test the answer afterwards, and it's not for my homework (I last did homework more than 10 years ago). Anyway: A certain breakfast cereal manufacturer is giving away a free toy inside each pack. There are ten toys in the series, and I'd like to collect them all. However, I can't tell which toy I'm going to get when I buy the pack. Assuming that each toy is distributed equally, how many packs would I have to buy to be 50%, 70% or 90% sure of having all ten toys? Work I've done already: I can see that in order to calculate the number of possible combinations of toys I would have after buying n packs of cereal, I would need the 'combination with repitition' formula. However, I can't find an expression for the 'successful' combinations, which is where I'm struggling. I suspect that the number of successful combinations after buying 10 packs is equal to the number of permutations, as this will list all the possible combinations with one of each toy. I'm at a complete loss for buying 11 packs, or 12 or any n(packs) > n(toys). I've done some tests with just two toys (start small!): After buying two packs, the probability of success is 0.5. The two successful combinations are AB and BA, and the two unsuccessful are AA and BB (i.e. I get the same toy twice). After buying three packs, the probability of success rises to 0.75. There are eight total combinations, but only two are unsuccessful (AAA and BBB), leaving six successful. After buying four packs, the probability of success rises to 7/8, as the number of unsuccessful combinations remains at two, but the number of successes rises to 14. Generally, for the two-toy problem, the probability of getting a successful combination after n turns is equal to (2^n - 2) / 2^n or, in words, the total number of combinations minus unsuccessful, divided by the total number of combinations. But I'm not sure how to proceed, and I can't see how I'd expand to significantly larger numbers (such as the 10 I posed at the start).