三门问题的简单c语言代码

int i, n, stayWins = 0, switchWins = 0;
srand(time(NULL));
printf("请输入尝试次数：");
scanf("%d", &n);
for(i = 0; i < n; i++)
{
    prizeDoor = rand() % 3 + 1;
    guess = rand() % 3 + 1;
    do
    {
        open = rand() % 3 + 1;
    } while(open == prizeDoor || open == guess);
    remainingDoor = 6 - guess - open;
    // 不换门
    if(guess == prizeDoor)
        stayWins++;
    // 换门
    if(remainingDoor == prizeDoor)
        switchWins++;
}

printf("不换门胜率：%f\n", (double)stayWins / n);
printf("换门胜率：%f\n", (double)switchWins / n);

return 0;

上计算生物学提到的三门问题，第一反应有点反常规，仔细思考过后通过c语言用大量重复次数验证了正确答案（换门胜率2/3，不换门胜率1/3）。