What is the expected number of rolls to get a subset of all possibilities? I need to know how many times I should expect to need to roll an $n$-sided fair die to get each of $m$ (where $m < n$) outcomes at least once.
In my specific example I have a $7$ possible outcomes, of which I want to achieve $3$, with order being irrelevant.
Please excuse if this has been answered before, I searched but all I could find were answers for how many rolls you need to get one specific outcome, or every possible one, nothing for a subset of the possible outcomes and I couldn't figure out how to generalize those answers to apply to my problem.
 A: Define the event that any occurance of the $m$ desired sides to be success. The probability of success would be $\frac{m}{n}$. This is a geometric distribution.
After the occurance of one the desired side, remove that side from our success set. Now define our event of success to be any occurance of the remaining $m-1$ sides. Again this is a geometric distribution, but with probability of success being $\frac{m-1}{n}$.
We can repeat this process for all the $m$ sides.
Hence expected number of trials is
$$\frac{n}{m}+\frac{n}{m-1}+\ldots \frac{n}{m-(m-1)}=\sum_{l=0}^{m-1}\frac{n}{m-l}=\sum_{l=1}^m\frac{n}{l}$$
For your specific example, 
the answer should be $$\frac{7}{1}+\frac{7}{2}+\frac{7}{3}$$
A: Method I (Markov):  Let $E_i$ be the expected number assuming you have already seen $i$ of your list (so the answer you want is $E=E_0$.  Then We have the recursion $$E_i=\frac {n-i}m\times \left(E_{i+1}+1\right)+\frac {m-(n-i)}m\times \left(E_{i}+1\right)$$  Thus $$ E_i=\frac m{n-i}+E_{i+1}$$
In your case $(m,n)=(7,3)$, that gives $E=12.8\overline 3$
Method II: (Coupon Collector Problem) If you only had the $n$ desired coupons the answer would be $nH_n$, where $H_n$ denotes the $n^{th}$ Harmonic Number.  Here, only $\frac nm$ of the tosses count, so the answer would be $$\frac mn \times nH_n=mH_n$$  In your case that gives $$E=7\times \frac {11}{6}=12.8\overline 3$$
