Xpectation Mania #2

Let X(n) be the number of rolls of the dice required to obtain all $n$ numbers. Then we can write $X(n) \; = \; X(n)_1 + X(n)_2 + \cdots + X(n)_n \hspace{2cm} \mathbb{E}[X(n)] \; = \; \sum_{j=1}^n \mathbb{E}[X(n)_j]$ where $X(n)_j$ is the number of additional rolls required to achieve the $j$ th different number after achieving $j-1$ different numbers. Then $X(n)_j$ has a geometric distribution with probability of success $\tfrac{n+1-j}{n}$ , and standard theory of the geometric distribution tells us that $\mathbb{E}[X(n)_j] = \tfrac{n}{n+1-j}$ , and hence $\mathbb{E}[X(n)] \; = \; \sum_{j=1}^n \frac{n}{n+1-j} \; = \; nH_n$ where $H_n$ is the $n$ th harmonic number. To $3$ significant figures, we have $\mathbb{E}[X(12)] = \boxed{37.239}$ .