probability - A number of dice rolls to see every number at least once

We have a fair dice that can produce n different numbers. How many times should we roll the dice to see every number at least once with probability p?

Not a homework, just interesting. Tried to solve myself but with no luck.

I think it could be sort of coupon collector problem, but I can't get exact formula.

Answer

Let $S_k$ be all the outcomes in which a $k$ is not rolled. For each $k$, $|S_k|=(n-1)^r$ and there are $\binom{n}{1}$ choices for $k$. For each $j\ne k$, $|S_j\cap S_k|=(n-2)^r$ and there are $\binom{n}{2}$ choices for $j$ and $k$. Continuing in this manner and using Inclusion-Exclusion to count the number of outcomes missing at least $1$ number, we get
$$
\begin{align}
\left|\bigcup_{k=1}^nS_k\right|
&=\sum_{k=1}^n|S_k|-\sum_{j< k}|S_j\cap S_k|+\sum_{i&=\binom{n}{1}(n-1)^r-\binom{n}{2}(n-2)^r+\binom{n}{3}(n-3)^r-\dots
\end{align}
$$
Since there are a total of $n^r$ total outcomes, we get the number of outcomes in which all possible numbers are rolled is
$$n^r-\binom{n}{1}(n-1)^r+\binom{n}{2}(n-2)^r-\binom{n}{3}(n-3)^r+\dots$$
Thus, the probability of this occurring is
$$1-\binom{n}{1}\left(1-\frac1n\right)^r+\binom{n}{2}\left(1-\frac2n\right)^r-\binom{n}{3}\left(1-\frac3n\right)^r+\dots$$
So we need to find the smallest $r$ so that
$$p\le\sum_{k=0}^n(-1)^k\binom{n}{k}\left(1-\frac kn\right)^r$$

Expected Duration

The expected duration until all numbers are rolled can be computed using the formulas above; i.e.
$$\begin{align} \mathrm{E}[r] &=\color{#C00000}{\sum_{r=1}^\infty}\sum_{k=0}^n(-1)^k\binom{n}{k}\color{#C00000}{r\left[\left(1-\frac kn\right)^r-\left(1-\frac kn\right)^{r-1}\right]}\\ &=\sum_{k=1}^n(-1)^k\binom{n}{k}\color{#C00000}{\left(-\frac nk\right)}\\ &=n\sum_{k=1}^n(-1)^{k-1}\color{#00A000}{\binom{n}{k}}\frac1k\\ &=n\sum_{k=1}^n(-1)^{k-1}\color{#00A000}{\sum_{j=k}^n\binom{j-1}{k-1}}\frac1k\\ &=n\sum_{j=1}^n\sum_{k=1}^j(-1)^{k-1}\binom{j-1}{k-1}\frac1k\\ &=n\sum_{j=1}^n\color{#0000FF}{\sum_{k=1}^j(-1)^{k-1}\binom{j}{k}}\frac1j\\ &=n\sum_{j=1}^n\frac1j \end{align}$$
However, there is a simpler way. Since the expected duration for a binomial event with probability $p$ is $\frac1p$, we get that the expected duration for the $k^{\mathrm{th}}$ number is $\frac{n}{n-k+1}$. Therefore, the expected duration to get all numbers is
$$\sum_{k=1}^n\frac{n}{n{-}k{+}1}=n\sum_{k=1}^n\frac1{k}$$