# Straight route to coupon collection

Consider again the coupon collector's problem:

There are $$c$$ different types of coupon, and each coupon obtained is equally likely to be any one of the $$c$$ types. Find the probability that the first $$n$$ coupons which you collect do not form a complete set, and deduce an expression for the mean number of coupons you will need to collect before you have a complete set.

For the probability that the first $$n$$ coupons do not form a complete set, it will be the complementary probability that the first $$n$$ coupons do form a complete set, thus: $$\mathbb{P}(n\text{ coupons, not complete set}) = 1 - {n \choose c} \left(\frac{1}{c}\right)^c.$$ The idea for this expression is that the sequences of length $$n$$ that have at least the $$c$$ coupons are equal to the number of ways that $$c$$ elements can be disposed on a sequence of length $$n$$, thus $${n \choose c}$$. The sequence of $$c$$ distinct coupons has probability $$c^{-c}$$, and as I do not care about the remaining $$n-c$$ spots in the sequence, these can be anything.

For the mean number of coupons needed to collect before having a complete set, I would say: $$\sum_{m=c}^\infty m {m-1 \choose c-1} \left(\frac{1}{c}\right)^{c-1}\left(\frac{c-1}{c}\right)^{m-c-1}\frac{1}{c},$$ as in this case for every length $$m$$ we need to count the number of sequences which contain all the $$c$$ distinct coupons, but the last coupon of the collection must be in the $$m$$-th place, so we need to dispose $$c-1$$ coupons in $$m-1$$ places. In the remaining $$m-c$$ places we can allow any coupon except for the last one we need, so the probability is $$\frac{c-1}{c}$$.

What do you think? If I am correct, is there a way to compute the summation?