Simulation of the Saier lottery to seek expectation

Nine cards , At first it was the opposite , Take three cards at random each time to change the face , The next extraction is based on this , Stop until all cards are timed . Find the expectation of extraction times . And calculate the probability of completing the extraction within 200 times .

There are three... Through binary bits 1 Iterative XOR of random numbers , Achieve the effect of continuous turnover . Because in binary 1 The number of digits is 3 There are only 84 Kind of , So you can use a three-tier loop to store it in s[84] in , You can sort and verify .

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

int s[84]={0};

int cmp(const void *a,const void *b){
    return *(int*)a-*(int*)b;}

int produce_random(void){
    int n=rand()&511;
    return n;
}

int randni(void){
    int n=rand()%84;
    return n;}

int expbi(const int n){
    int m=1<<n;
    return m;}

int main(int argc, char** argv){
    int i=0,j,k,l;
    int a,b,c;
    int v=1000000;
    for(j=0;j<7;j++){
       for(k=j+1;k<8;k++){
           for(l=k+1;l<9;l++){
               s[i++]=expbi(j)+expbi(k)+expbi(l);
               }
           }
       }
    float sum=0.0;
    //qsort(s,84,sizeof(int),cmp);
    srand((unsigned int)time(NULL));
    for(i=0;i<v;i++){
        a=0;
        for(k=0;k<100000;k++){
            b=randni();
            b=s[b];
            a=a^b;
            if(a==511)break;
            }
      // printf("%d\n",k);
        sum+=k;
       // if(k<200)
        //sum+=1;
        }
    sum/=v;
  printf("%f",sum);
   /* for(i=0;i<84;i++)
    printf("%d\n",s[i]);*/
    return 0;
  }

The result of calculating the expected mean value a million times

Probably 517 Times or so

Here is 200 Probability of success