What is the expected number of rolls to get 5 different numbers from a 10-face die?

Question

Suppose we have a 10-face fair die such that the expected number of rolls to get a particular number from $\{1, 2, \ldots, 10\}$ is $10.$ Then, I am wondering how would I go about finding the expected number of rolls needed to get 5 different numbers. Or more generally, the expected number of rolls needed to get n different numbers where $n < 10.$

Answer 1

This is the coupon collector’s problem. The first number takes $1$ roll. Then you have probability $\frac9{10}$ per roll to get a new number, so the next new number takes $\frac{10}9$ rolls, and so on. The expected number of rolls needed to get $n$ different numbers is

$$ \sum_{k=0}^{n-1}\frac{10}{10-k}=\sum_{j=11-n}^{10}\frac{10}j=10\left(H_{10}-H_{10-n}\right)\;, $$

where $H_n$ is the $n$-th harmonic number.

Answer 2

Let $X_i$ denote the number of rolls needed to get the $i$th new number. Then $X_1=1$, and $X_i-X_{i-1} \sim Geo((10-(i-1))/10)$. Hence $$E[X_i] = E[X_1 + \sum_{j=2}^i X_{j}-X_{j-1} =\sum_{j=1}^i \frac{10}{10-(i-1)} ]$$