If there are 100 tickets in an urn, and I'm only trying to see 10 of them while drawing with replacement, how do I find the result for that?
My intuition makes me want to say it's $n\sum_{i=1}^k\frac{1}{i}$ where $n$ is the total number of tickets, and $k$ is the number of tickets you're actually trying to collect.
You can use the usual technique. Let $N_{i}$ denote the number of tickets until you see the $i$-th new ticket among the $k$. Also, let $N$ denote the number of tickets untill all $k$ tickets are seen. Since $N=\sum_{i=1}^{k}{N_i}$, conclude that \begin{align*} E[N] &= E\left[\sum_{i=1}^{10}{N_i}\right] \\ &= \sum_{i=1}^{10}E[N_i] \end{align*} Note that $N_i$ follows a geometric distribution with parameter $\frac{k+1-i}{n}$. Therefore $E[N_i]=\frac{n}{k+1-i}$ and $E[N] = n\sum_{i=1}^{k}{\frac{1}{k+1-i}} = n\sum_{i=1}^{k}{\frac{1}{i}}$.