What is the smallest number of tosses for all outcomes? What is the smallest number of tosses that need to be done to get all the possible outcomes $\{1, 2, 3, 4, 5,6\}$ of a right dice with a reliability of $0.99$?
My attempt is to use the Bernoulli scheme and the Moivre–Laplace theorem:
$$P\left(\left|\frac{\mu}{n}-p\right|<\delta\right)=2\Phi\left(\delta \sqrt{\frac{n}{p \cdot q}}\right)=0.99$$ but I don't know how to define $\delta$ here.
 A: you want to calculate the probability that 6 events all happen. This can be done with inclusion exclusion.
We get $\sum\limits_{i=0}^6 \binom{6}{i} (-1)^i (\frac{i}{6})^N$ which we can approximate from above as $1-6(5/6)^N$ so we are going to require $6(\frac{5}{6})^N < 1/100$. we can solve for $N$ and we get $N=\frac{-log(600)}{log(5)-log(6)}$ which is approximately $35.086$, so if I had to guess I would say $35$ is the first value, although apparently I got it wrong and it's $36$, but to be fair with $35$ you get approximately $0.9898$. Unless there is something wrong with my code (which I include next).
#include<bits/stdc++.h>
using namespace std;

const int MAX = 15;
int bino[MAX][MAX];

long double pot(long double b,int e){
        long double res = 1;
        while(e){
                if( e%2) res = res*b;
                b = b*b;
                e/=2;
        }
        return res;
}

int main(){
        int n;
        cin >> n;
        for(int i=0;i<MAX;i++){
                bino[i][0] = 1;
        }
        for(int i=1;i<MAX;i++){
                for(int j=1;j<MAX;j++){
                        bino[i][j] = bino[i-1][j] + bino[i-1][j-1];
                }
        }
        long double res = 0;
        for(int i = 0;i<=6;i++){
                long double summand = 1;
                summand = summand*bino[6][i];
                summand = summand*pot(-1,i);
                summand = summand*pot((1.0/6)*i,n);
                res += summand;
        }
        cout << res << endl;


}

A: The probability that at least one number does not appear after n tosses is $Q=6(\frac{5}{6})^n-15(\frac{4}{6})^n+20(\frac{3}{6})^n-15(\frac{2}{6})^n+6(\frac{1}{6})^n$, so the desire probability is $P=1-Q$.  Solve for $n$ with $Q=0.01$
