We roll a die until each side appears at least once, then we stop.
What is the probability of rolling exactly $n$ dice?
I guess the answer is $$6-6\left(\dfrac{5}{6}\right)^n\;,$$
but this may be wrong.
In trying to solve this problem, I read this paper, but I couldn't solve the problem.
This is the coupon collector's problem. Let $N$ be the number of rolls of a $k$-sided die needed to see all $k$ different values. Then $$\mathbb{P}(N=n)= k^{-n}\, k! \,{n-1\brace k-1}.$$ Here ${n\brace k}$ refers to Stirling numbers of the second kind.
For example, the probability that we stop on the tenth roll of a six-sided die is $$\mathbb{P}(N=10)= 6^{-10}\, 6!\, {9\brace 5}={11585\over 139968}=.08277.$$
I'm going to assume that what you are asking is: if we roll a die until all sides (i.e. faces) appear, what is the probability that we will roll $n$ times?
In order for this to happen, we need exactly five faces to have appeared in the first $n-1$ rolls, and then the one missing face must appear on the $n$-th roll. Using inclusion-exclusion, we can express this, for $n>1$, as $$ p(n) = \frac{1}{6} \binom{6}{5} \left( \left(\frac{5}{6} \right)^{n-1} - \binom{5}{4} \left( \frac{4}{6} \right)^{n-1} + \binom{5}{3} \left( \frac{3}{6}\right)^{n-1} - \binom{5}{2} \left( \frac{2}{6} \right)^{n-1}+\binom{5}{1} \left( \frac{1}{6}\right) ^{n-1} \right) $$ So, for example, $p(6)=5/324$, $p(7)=25/648$, and $p$ hits a maximum of $875/10368$ at $n=11$.
As a check, you can (numerically) verify that $$ \sum_{i=6}^{\infty} ip(i) = 14.7,$$ the well-known expected number of throws needed to see all faces.
