Fair die rolled 20 times What is the probability that each number of 1 to 6 shows up at least once in those rolls?
My guess: 20*(6/6^20) 
20 rolls times 6 outcomes in 1 roll over total different sequences.
the 1-6 is throwing me off
 A: Hint: We find the probability of the complement, the probability that $1$ or more numbers are missing. We use Inclusion/Exclusion.
The probability $1$ is missing is $(5/6)^{20}$. Same for $2$ missing, and so on. But adding these up, that is, finding $\binom{6}{1}(5/6)^{20}$, counts twice the situations where $1$ and $2$ are missing.  
The probability $1$ and $2$ are missing is $(4/6)^{20}$. It is the same for all the other pairs, so our second estimate for the probability at least one is missing is $\binom{6}{1}(5/6)^{20}-\binom{6}{2}(4/6)^{20}$.
But we have subtracted too much, for we have subtracted one too many time the probability that for example $1,2,3$ are all missing. So we must add back $\binom{6}{3}(3/6)^{20}$.
Thus our third estimate is $\binom{6}{1}(5/6)^{20}-\binom{6}{2}(4/6)^{20}+\binom{6}{3}(3/6)^{20}$.
Continue. Two more terms, and we will have the exact answer. But we already have a very good approximation.
A: In how many ways can this happen, this is the same as the number of surjections from $\{1,2,3,4,\dots,19,20\}$ to $\{1,2,3,4,5,6\}$. How many of these are there?
there are $20\brace 6$ ways to partition the domain and $6!$ ways to select which part of the domain goes to which element of the domain. Since there are $6^{20}$ possible outcomes of throwing $20$ die the final answer is  
$$\dfrac{6! {20\brace 6}}{6^{20}}=\frac{2691299309615}{3173748645888}\approx0.847988$$
