模约化是一种常见的方法，可以使随机整数生成器避免永远运行的最坏情况。

When the range of possible integers is unknown, however, there is no way in general to "fix" this worst case of running forever without introducing bias. It's not just modulo reduction (rand() % n, discussed in the accepted answer) that will introduce bias this way, but also the "multiply-and-shift" reduction of Daniel Lemire, or if you stop rejecting an outcome after a set number of iterations. (To be clear, this doesn't mean there is no way to fix the bias issues present in pseudorandom generators. For example, even though modulo and other reductions are biased in general, they will have no issues with bias if the range of possible integers is a power of 2 and if the random generator produces unbiased random bits or blocks of them.)

这个答案的其余部分将显示随机生成器中运行时间和偏差之间的关系。从这里开始，我们将假设我们有一个“真正的”随机生成器，可以产生无偏和独立的随机比特

In 1976, D. E. Knuth and A. C. Yao showed that any algorithm that produces random integers with a given probability, using only random bits, can be represented as a binary tree, where random bits indicate which way to traverse the tree and each leaf (endpoint) corresponds to an outcome. In this case, we're dealing with algorithms that generate random integers in [0, n), where each integer is chosen with probability 1/n. The algorithm is unbiased if the same number of leaves appear in the tree for all outcomes. But if 1/n has a non-terminating binary expansion (which will be the case if n is not a power of 2), the algorithm will be unbiased only if—

二叉树具有“无限”深度，或者 二叉树的末端包含“排斥”叶，

无论哪种情况，算法都不会在常数时间内运行在最坏的情况下会一直运行下去。(另一方面，当n是2的幂时，最优二叉树的深度是有限的，并且没有拒绝节点。)

The binary tree concept also shows that any way to "fix" this worst-case time complexity will lead to bias in general. (Again, this doesn't mean there is no way to fix the bias issues present in pseudorandom generators.) For instance, modulo reductions are equivalent to a binary tree in which rejection leaves are replaced with labeled outcomes — but since there are more possible outcomes than rejection leaves, only some of the outcomes can take the place of the rejection leaves, introducing bias. The same kind of binary tree — and the same kind of bias — results if you stop rejecting after a set number of iterations. (However, this bias may be negligible depending on the application. There are also security aspects to random integer generation, which are too complicated to discuss in this answer.)

为了说明这一点，下面的JavaScript代码实现了J. Lumbroso(2013)提出的名为Fast Dice Roller的随机整数算法。请注意，它包括一个拒绝事件和一个循环，这是在一般情况下使算法无偏倚所必需的。

function randomInt(minInclusive, maxExclusive) {
 var maxInclusive = (maxExclusive - minInclusive) - 1
 var x = 1
 var y = 0
 while(true) {
    x = x * 2
    var randomBit = (Math.random() < 0.5 ? 0 : 1)
    y = y * 2 + randomBit
    if(x > maxInclusive) {
      if (y <= maxInclusive) { return y + minInclusive }
      // Rejection
      x = x - maxInclusive - 1
      y = y - maxInclusive - 1
    }
 }
}

Note

*这个答案不会涉及到C中的rand()函数，因为它有很多问题。这里最严重的问题可能是，C标准没有明确地为rand()返回的数字指定特定的分布，甚至没有统一的分布。

对于模的使用，有两种常见的抱怨。

one is valid for all generators. It is easier to see in a limit case. If your generator has a RAND_MAX which is 2 (that isn't compliant with the C standard) and you want only 0 or 1 as value, using modulo will generate 0 twice as often (when the generator generates 0 and 2) as it will generate 1 (when the generator generates 1). Note that this is true as soon as you don't drop values, whatever the mapping you are using from the generator values to the wanted one, one will occurs twice as often as the other. some kind of generator have their less significant bits less random than the other, at least for some of their parameters, but sadly those parameter have other interesting characteristic (such has being able to have RAND_MAX one less than a power of 2). The problem is well known and for a long time library implementation probably avoid the problem (for instance the sample rand() implementation in the C standard use this kind of generator, but drop the 16 less significant bits), but some like to complain about that and you may have bad luck

使用类似于

int alea(int n){ 
 assert (0 < n && n <= RAND_MAX); 
 int partSize = 
      n == RAND_MAX ? 1 : 1 + (RAND_MAX-n)/(n+1); 
 int maxUsefull = partSize * n + (partSize-1); 
 int draw; 
 do { 
   draw = rand(); 
 } while (draw > maxUsefull); 
 return draw/partSize; 
}

生成0到n之间的随机数将避免这两个问题(并且它避免RAND_MAX == INT_MAX溢出)

顺便说一句，c++ 11引入了标准方法来简化和rand()以外的其他生成器。

不断随机选取是去除偏差的好方法。

更新

如果我们在能被n整除的范围内搜索x，我们可以让代码更快。

// Assumptions
// rand() in [0, RAND_MAX]
// n in (0, RAND_MAX]

int x; 

// Keep searching for an x in a range divisible by n 
do {
    x = rand();
} while (x >= RAND_MAX - (RAND_MAX % n)) 

x %= n;

上面的循环应该非常快，平均1次迭代。

定义

模偏置是使用模算术将输出集缩减为输入集的子集时的固有偏置。一般来说，只要输入和输出集之间的映射不是均匀分布的，就会存在偏置，例如当输出集的大小不是输入集大小的除数时使用模算术。

这种偏差在计算中尤其难以避免，在计算中，数字被表示为比特串:0和1。找到真正随机的随机性来源也非常困难，但这超出了本文讨论的范围。对于这个答案的其余部分，假设存在无限的真正随机比特的来源。

问题的例子

让我们考虑使用这些随机比特来模拟掷骰子(0到5)。有6种可能，所以我们需要足够的位来表示数字6，也就是3位。不幸的是，3个随机比特会产生8种可能的结果:

000 = 0, 001 = 1, 010 = 2, 011 = 3
100 = 4, 101 = 5, 110 = 6, 111 = 7

我们可以通过取模6的值来将结果集的大小减小到恰好6，但是这就出现了模偏问题:110产生0,111产生1。这个骰子上膛了。

可能的解决方案

方法0:

从理论上讲，人们可以雇佣一支小部队整天掷骰子，并将结果记录在数据库中，然后每个结果只使用一次，而不是依赖随机比特。这听起来很实际，而且很可能不会产生真正随机的结果。

方法1:

不使用模量，一个简单但在数学上正确的解决方案是丢弃产生110和111的结果，并简单地重新尝试3个新比特。不幸的是，这意味着每一次掷骰子都有25%的几率需要重新掷一次，包括每一次重新掷骰子本身。这显然是不切实际的，除了最微不足道的用途。

方法2:

使用更多的位:使用4位而不是3位。这产生了16种可能的结果。当然，每次结果大于5时重新滚动会使情况变得更糟(10/16 = 62.5%)，因此仅靠这一点是没有帮助的。

请注意，2 * 6 = 12 < 16，所以我们可以安全地取任何小于12的结果，并将其取模6以均匀分布结果。其他4个结果必须被丢弃，然后像前面的方法一样重新滚动。

一开始听起来不错，但让我们来计算一下:

4 discarded results / 16 possibilities = 25%

在这种情况下，1个额外的比特根本没有帮助!

这个结果很不幸，但让我们再次尝试5位:

32 % 6 = 2 discarded results; and
2 discarded results / 32 possibilities = 6.25%

确实有了改进，但在许多实际情况下还不够好。好消息是，添加更多比特永远不会增加需要丢弃和重新滚动的几率。这不仅适用于骰子，而且适用于所有情况。

然而，如前所述，增加1个额外的位可能不会改变任何东西。事实上，如果我们将点数增加到6位，概率仍然是6.25%。

这就引出了另外两个问题:

如果我们添加足够多的比特，是否能保证丢弃的概率会降低? 一般情况下多少位才够呢?

通解

幸运的是，第一个问题的答案是肯定的。6的问题在于，2^x mod 6在2和4之间翻转而2和4恰好是2的倍数，所以对于偶数x > 1，

[2^x mod 6] / 2^x == [2^(x+1) mod 6] / 2^(x+1)

因此，6是一个例外，而不是规则。有可能找到更大的模，以同样的方式产生连续的2次幂，但最终这必须环绕，弃牌的概率将会降低。

在不提供进一步证明的情况下，一般使用两倍的数字 将提供一个较小的，通常不重要的， 弃牌的机会。

概念证明

下面是一个示例程序，它使用OpenSSL的libcrypo提供随机字节。在编译时，请确保使用-lcrypto链接到大多数人都可以使用的库。

#include <iostream>
#include <assert.h>
#include <limits>
#include <openssl/rand.h>

volatile uint32_t dummy;
uint64_t discardCount;

uint32_t uniformRandomUint32(uint32_t upperBound)
{
    assert(RAND_status() == 1);
    uint64_t discard = (std::numeric_limits<uint64_t>::max() - upperBound) % upperBound;
    RAND_bytes((uint8_t*)(&randomPool), sizeof(randomPool));

    while(randomPool > (std::numeric_limits<uint64_t>::max() - discard)) {
        RAND_bytes((uint8_t*)(&randomPool), sizeof(randomPool));
        ++discardCount;
    }

    return randomPool % upperBound;
}

int main() {
    discardCount = 0;

    const uint32_t MODULUS = (1ul << 31)-1;
    const uint32_t ROLLS = 10000000;

    for(uint32_t i = 0; i < ROLLS; ++i) {
        dummy = uniformRandomUint32(MODULUS);
    }
    std::cout << "Discard count = " << discardCount << std::endl;
}

我鼓励使用MODULUS和ROLLS值来查看在大多数情况下实际发生了多少次重新滚动。持怀疑态度的人也可能希望将计算值保存到文件中，并验证分布是否正常。

正如公认的答案所示，“模偏置”的根源在于RAND_MAX的低值。他使用一个非常小的RAND_MAX(10)值来表明，如果RAND_MAX为10，那么您尝试使用%生成一个0到2之间的数字，将导致以下结果:

rand() % 3   // if RAND_MAX were only 10, gives
output of rand()   |   rand()%3
0                  |   0
1                  |   1
2                  |   2
3                  |   0
4                  |   1
5                  |   2
6                  |   0
7                  |   1
8                  |   2
9                  |   0

所以有4个0的输出(4/10的概率)，只有3个1和2的输出(各3/10的概率)。

所以这是有偏见的。数字越小，出来的几率越大。

但这只在RAND_MAX很小的时候才会很明显。或者更具体地说，当你modding的数字比RAND_MAX大的时候。

一个比循环更好的解决方案(循环效率非常低，甚至不应该被建议使用)是使用输出范围大得多的PRNG。梅森Twister算法的最大输出为4,294,967,295。这样做MersenneTwister::genrand_int32() % 10，将是均匀分布的，模偏效应将几乎消失。

模约化是一种常见的方法，可以使随机整数生成器避免永远运行的最坏情况。

When the range of possible integers is unknown, however, there is no way in general to "fix" this worst case of running forever without introducing bias. It's not just modulo reduction (rand() % n, discussed in the accepted answer) that will introduce bias this way, but also the "multiply-and-shift" reduction of Daniel Lemire, or if you stop rejecting an outcome after a set number of iterations. (To be clear, this doesn't mean there is no way to fix the bias issues present in pseudorandom generators. For example, even though modulo and other reductions are biased in general, they will have no issues with bias if the range of possible integers is a power of 2 and if the random generator produces unbiased random bits or blocks of them.)

这个答案的其余部分将显示随机生成器中运行时间和偏差之间的关系。从这里开始，我们将假设我们有一个“真正的”随机生成器，可以产生无偏和独立的随机比特

In 1976, D. E. Knuth and A. C. Yao showed that any algorithm that produces random integers with a given probability, using only random bits, can be represented as a binary tree, where random bits indicate which way to traverse the tree and each leaf (endpoint) corresponds to an outcome. In this case, we're dealing with algorithms that generate random integers in [0, n), where each integer is chosen with probability 1/n. The algorithm is unbiased if the same number of leaves appear in the tree for all outcomes. But if 1/n has a non-terminating binary expansion (which will be the case if n is not a power of 2), the algorithm will be unbiased only if—

二叉树具有“无限”深度，或者 二叉树的末端包含“排斥”叶，

无论哪种情况，算法都不会在常数时间内运行在最坏的情况下会一直运行下去。(另一方面，当n是2的幂时，最优二叉树的深度是有限的，并且没有拒绝节点。)

The binary tree concept also shows that any way to "fix" this worst-case time complexity will lead to bias in general. (Again, this doesn't mean there is no way to fix the bias issues present in pseudorandom generators.) For instance, modulo reductions are equivalent to a binary tree in which rejection leaves are replaced with labeled outcomes — but since there are more possible outcomes than rejection leaves, only some of the outcomes can take the place of the rejection leaves, introducing bias. The same kind of binary tree — and the same kind of bias — results if you stop rejecting after a set number of iterations. (However, this bias may be negligible depending on the application. There are also security aspects to random integer generation, which are too complicated to discuss in this answer.)

为了说明这一点，下面的JavaScript代码实现了J. Lumbroso(2013)提出的名为Fast Dice Roller的随机整数算法。请注意，它包括一个拒绝事件和一个循环，这是在一般情况下使算法无偏倚所必需的。

function randomInt(minInclusive, maxExclusive) {
 var maxInclusive = (maxExclusive - minInclusive) - 1
 var x = 1
 var y = 0
 while(true) {
    x = x * 2
    var randomBit = (Math.random() < 0.5 ? 0 : 1)
    y = y * 2 + randomBit
    if(x > maxInclusive) {
      if (y <= maxInclusive) { return y + minInclusive }
      // Rejection
      x = x - maxInclusive - 1
      y = y - maxInclusive - 1
    }
 }
}

Note

*这个答案不会涉及到C中的rand()函数，因为它有很多问题。这里最严重的问题可能是，C标准没有明确地为rand()返回的数字指定特定的分布，甚至没有统一的分布。