We roll 7-sided die. We stop when all possible outcomes occur. What is the average sum of the outcomes?
I started with assigning $S_n=X_1+\ldots +X_n$, where $X_k$ is k-th result. Now put $$\tau=\min\left\{n \ : \ \left\{1,\ldots ,7\right\}\subset \left\{X_1,\ldots ,X_n\right\} \right\}.$$
I'll use Wald's equation $ES_{\tau}=E\tau EX_1$. I computed $EX_1=4$ and put $\tau=\tau_1+\ldots +\tau_7$ where $\tau_1=1$ and $\tau_i$ is a time of expecting $i$-th outcome, different from previous outcomes. So I would like to compute $E\tau =E(\tau_1+\ldots +\tau_7)=\ldots$?
My question is how to compute it?
Hint : for $\mathbb E[\tau_2]$. Suppose you have obtained $x$ on the first roll. Now you roll until you get a different result than $x$. You have a probability $6/7$ of getting $y\neq x$ and $1/7$ of getting $x$. In the first case, you stop and in the second you go on with the same probabilities. So you have $$ \mathbb E[\tau_2]=\frac67\times 1+\frac67\times\frac17\times 2+\frac67\left(\frac17\right)^2\times 3+\cdots.$$ You can write this as $$\mathbb E[\tau_2]=\sum_{k=0}^\infty \frac67\left(\frac17\right)^{k-1}\times k.$$ This series is well known. $\sum_k kx^k=x/(1-x)^2$ for $|x|<1$.
So we have the result $\mathbb E[\tau_2]=7/6$. As the $\tau_i$ are independent, you may find the final result easily from this.