这是我在看过别人的答案后得出的最简单的答案:

def r5tor7():
    while True:
        cand = (5 * r5()) + r5()
        if cand < 27:
            return cand

Cand在[6,27]范围内，如果r5()的可能结果是均匀分布的，则可能结果是均匀分布的。你可以用下面的代码来测试我的答案:

from collections import defaultdict

def r5_outcome(n):
    if not n:
        yield []
    else:
        for i in range(1, 6):
            for j in r5_outcome(n-1):
                yield [i] + j

def test_r7():
    d = defaultdict(int)
    for x in r5_outcome(2):
        s = sum([x[i] * 5**i for i in range(len(x))])
        if s < 27:
            d[s] += 1
    print len(d), d

R5_outcome(2)生成r5()结果的所有可能组合。我使用与解决方案代码中相同的筛选器进行测试。你可以看到所有的结果都是相等的，因为它们有相同的值。

亚当·罗森菲尔德正确答案的前提是:

X = 5^n(在他的例子中，n=2) 操作n个rand5次调用以获得范围[1,x]内的数字y Z = ((int)(x / 7)) * 7 如果y > z，再试一次。否则返回y % 7 + 1

当n = 2时，有4种可能:y ={22,23,24,25}。如果你使用n = 6，你只有1个扔掉的东西:y ={15625}。

5^6 is 15625 7 times 2232 is 15624

你又给rand5个电话。但是，您获得一个丢弃值(或无限循环)的机会要低得多。如果有办法让y没有可能的一次性值，我还没有找到它。

你需要的函数是rand1_7()，我写了rand1_5()，这样你就可以测试它并绘制它。

import numpy
def rand1_5():
    return numpy.random.randint(5)+1

def rand1_7():
    q = 0
    for i in xrange(7):  q+= rand1_5()
    return q%7 + 1

下面是一个利用c++ 11特性的答案

#include <functional>
#include <iostream>
#include <ostream>
#include <random>

int main()
{
    std::random_device rd;
    unsigned long seed = rd();
    std::cout << "seed = " << seed << std::endl;

    std::mt19937 engine(seed);

    std::uniform_int_distribution<> dist(1, 5);
    auto rand5 = std::bind(dist, engine);

    const int n = 20;
    for (int i = 0; i != n; ++i)
    {
        std::cout << rand5() << " ";
    }
    std::cout << std::endl;

    // Use a lambda expression to define rand7
    auto rand7 = [&rand5]()->int
    {
        for (int result = 0; ; result = 0)
        {
            // Take advantage of the fact that
            // 5**6 = 15625 = 15624 + 1 = 7 * (2232) + 1.
            // So we only have to discard one out of every 15625 numbers generated.

            // Generate a 6-digit number in base 5
            for (int i = 0; i != 6; ++i)
            {
                result = 5 * result + (rand5() - 1);
            }

            // result is in the range [0, 15625)
            if (result == 15625 - 1)
            {
                // Discard this number
                continue;
            }

            // We now know that result is in the range [0, 15624), a range that can
            // be divided evenly into 7 buckets guaranteeing uniformity
            result /= 2232;
            return 1 + result;
        }
    };

    for (int i = 0; i != n; ++i)
    {
        std::cout << rand7() << " ";
    }
    std::cout << std::endl;

    return 0;
}

首先，我在1点上移动ramdom5() 6次，得到7个随机数。 其次，将7个数相加得到公和。 第三，除法的余数是7。 最后加1，得到从1到7的结果。 这个方法给出了在1到7的范围内获得数字的相等概率，除了1。1的概率略高。

public int random7(){
    Random random = new Random();
    //function (1 + random.nextInt(5)) is given
    int random1_5 = 1 + random.nextInt(5); // 1,2,3,4,5
    int random2_6 = 2 + random.nextInt(5); // 2,3,4,5,6
    int random3_7 = 3 + random.nextInt(5); // 3,4,5,6,7
    int random4_8 = 4 + random.nextInt(5); // 4,5,6,7,8
    int random5_9 = 5 + random.nextInt(5); // 5,6,7,8,9
    int random6_10 = 6 + random.nextInt(5); //6,7,8,9,10
    int random7_11 = 7 + random.nextInt(5); //7,8,9,10,11

    //sumOfRandoms is between 28 and 56
    int sumOfRandoms = random1_5 + random2_6 + random3_7 + 
                       random4_8 + random5_9 + random6_10 + random7_11;
    //result is number between 0 and 6, and
    //equals 0 if sumOfRandoms = 28 or 35 or 42 or 49 or 56 , 5 options
    //equals 1 if sumOfRandoms = 29 or 36 or 43 or 50, 4 options
    //equals 2 if sumOfRandoms = 30 or 37 or 44 or 51, 4 options
    //equals 3 if sumOfRandoms = 31 or 38 or 45 or 52, 4 options
    //equals 4 if sumOfRandoms = 32 or 39 or 46 or 53, 4 options
    //equals 5 if sumOfRandoms = 33 or 40 or 47 or 54, 4 options
    //equals 6 if sumOfRandoms = 34 or 41 or 48 or 55, 4 options
    //It means that the probabilities of getting numbers between 0 and 6 are almost equal.
    int result = sumOfRandoms % 7;
    //we should add 1 to move the interval [0,6] to the interval [1,7]
    return 1 + result;
}

这个表达式足以得到1 - 7之间的随机整数

int j = ( rand5()*2 + 4 ) % 7 + 1;