Thanks to Mikhail for posing this problem.
On average, how many rolls would it take to see every face on a particular die?
Consider a die with $n$ sides. To answer this question, we’ll want to break this problem down into $n$ smaller problems, each of which will be solved in the same manner.
Specifically, take a time in this process where we still need to see $k$ sides. Let the expected number of rolls it’ll take to see a unique side in this situation be $E_k$. With probability $\frac{k}{n}$ we will see a new side on the next roll, and with probability $\frac{n-k}{n}$ we will see a repeat face on the next roll. In the first situation, we just take 1 roll. In the second situation, we take 1 roll plus an additional $E_k$ rolls because rolling a die is independent and memoryless—no roll changes the outcome of future rolls.
So, we can write an equation for our expected value:
$$ E_k = \frac kn\cdot 1 + \frac{n-k}{n}\left(1 + E_k\right). $$
Solving for $E_k$ gives us an expected value of $\frac nk$. So in each stage we expect it to take $\frac nk$ rolls to see one of the $k$ faces we have not yet rolled.
Since we want this to happen for all values of $k$ from $1$ to $n$, we add this up to get our total expected number of rolls.
$$ \sum_{k=1}^n E_k = \sum_{k=1}^n \frac nk $$
In the case that Mikhail sent me, he was interested in $n = 20$. This gives us
$$ \sum_{k=1}^{20}\frac{20}{k}, $$
which is barely less than 72.
You can easily calculate by hand that for $n=6$ the expected number of rolls is 14.7.