使用rand5()，生成rand7()(概率相同) [英] Use rand5(), to generate rand7() (with the same probability)

问题描述

可能的重复:
将随机范围从 1–5 扩展到 1–7

我在这里看到了这个问题:链接

I have seen the question in here: Link

作者提供的解决方案似乎没有产生相同的概率.

The solution the author provided didn't seem to generate the same probability.

例如，在对该函数的 10k 次调用中，数字 4 被返回了 1-2 次(而其他数字，如 2，每个都被返回了大约 2k 次).

For example, the number 4, out of 10k calls for the function, was returned 1-2 times (when the other numbers, like 2, were returned about 2k times each).

也许我理解错了，或者我写错了算法，但在这里:

Maybe I understood wrong, or I wrote the algorithm wrong, but here:

    static int rand5()
    {
        return new Random().Next(1, 6);
    }
    static int rand7()
    {
        while (true)
        {
            int num = 5 * (rand5() - 1) + rand5();
            if (num < 22) return ((num % 7) + 1);
        }
    }
    static void Main(string[] args)
    {
        int limit = 10000;
        int[] scores = new int[7];
        for (int i = 0; i < limit; i++)
        {
            scores[rand7() - 1]++;
        }
        foreach (int n in scores)
        {
            Console.Write(n + " ");
        }
        Console.WriteLine();
    }

提前致谢.

推荐答案

您没有在 Rand5 中生成随机数.

You are not generating random numbers in Rand5.

这样做:

static Random rand = new Random()
static int rand5()
{
    return rand.Next(1, 6);
}

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