Probability to see all 6 numbers on a die after n throws I am trying to work out the probability of seeing all 6 numbers on a fair die at least once after n throws, where n > 6.
So I found a related question:
Probability of rolling a dice 8 times before all numbers are shown.
and a part of the provided answer seems to work for n = 7, but the problem I have is I don't know how to generalize this to work with all n > 6.
Would be nice if someone could explain the general approach to such a problem.
 A: First, use inclusion/exclusion principle in order to count the number of combinations:


*

*Include the number of combinations with at most $\color\red{6}$ values showing: $\binom{6}{\color\red{6}}\cdot\color\red{6}^n$

*Exclude the number of combinations with at most $\color\red{5}$ values showing: $\binom{6}{\color\red{5}}\cdot\color\red{5}^n$

*Include the number of combinations with at most $\color\red{4}$ values showing: $\binom{6}{\color\red{4}}\cdot\color\red{4}^n$

*Exclude the number of combinations with at most $\color\red{3}$ values showing: $\binom{6}{\color\red{3}}\cdot\color\red{3}^n$

*Include the number of combinations with at most $\color\red{2}$ values showing: $\binom{6}{\color\red{2}}\cdot\color\red{2}^n$

*Exclude the number of combinations with at most $\color\red{1}$ values showing: $\binom{6}{\color\red{1}}\cdot\color\red{1}^n$


Then, divide by the total number of combinations in order to compute the probability:
$$\frac{\sum\limits_{k=0}^{6-1}(-1)^k\cdot\binom{6}{6-k}\cdot(6-k)^n}{6^n}$$
A: Roll the die $n$ times and for $i\in\{1,\dots,6\}$ let $E_i$ denote the event that number $i$ does not show up. 
Then you are looking for: $$1-\Pr(E_1\cup E_2\cup E_3\cup E_4\cup E_5\cup E_6)$$
Now attack this using inclusion/exclusion and symmetry.
