为什么来自stdlib的rand不遵循大数定律?

时间:2015-04-09 15:02:45

标签: c++ random std average

在下面的代码中,我期望一个角色数量达到平均结果正好为3.5的骰子,有时高于3.5的百分比是5%,其他时间(当然是不同的种子)是喜欢95.但即使你高达6040M左右,你也永远不会高于50%,低于3.5%?显然rand()中存在一点偏差......

我知道'真正的随机'不存在这一事实,但这真的很明显吗?

典型输出是:

  

平均值:3.50003计数器:3427000000以上百分比:83.2554 Perc abs高于计数器:50.0011
  平均值:3.49999计数器:1093000000以上百分比:92.6983 Perc abs以上计数器:50.0003

#include <stdio.h>      /* printf, scanf, puts, NULL */
#include <stdlib.h>     /* srand, rand */
#include <time.h>       /* time */
#include <unistd.h>
#include <iostream>
using namespace std;

int main ()
{
  long long int this_nr;
  long long int counter = 0;
  long long int above_counter = 0;
  long long int below_counter = 0;
  long long int above_counter_this = 0;
  long long int below_counter_this = 0;

  long long int interval_counter = 0;

  double avg = 0.0;
  srand (time(NULL));
  srand (time(NULL));
  srand (time(NULL));
  cout.precision(6);

  while(1) {
      this_nr = rand() % 6 + 1; // 0,1,2,3,4,5 or 6
      avg = ((double) this_nr + ((double)counter * (double) avg))
          / ((double) counter+1.0);
      if (this_nr <= 3) below_counter_this++;
      if (this_nr >= 4) above_counter_this++;
      if (avg < 3.5) below_counter++;
      if (avg > 3.5) above_counter++;
      if (interval_counter >= 1000000) {
        cout << "Average: " << avg << " counter: " << counter << " Percentage above: "
                 << (double) above_counter / (double) counter * 100.0
                 << " Perc abs above counter: " << 100.0 * above_counter_this / counter
                 << "                 \r";
        interval_counter = 0;
      }
      //usleep(1);
      counter++; 
      interval_counter++;
  }
}

1 个答案:

答案 0 :(得分:1)

众所周知,

rand()是一个糟糕的发生器,在低位时它特别糟糕。执行% 6仅取消低位。您也有机会体验一些modulo bias,但我希望这种效果相对较小。