我刚刚为冯·诺依曼无偏抛硬币法写了一段代码，理论上应该可以消除随机数生成过程中的任何偏差。更多信息请访问(http://en.wikipedia.org/wiki/Fair_coin)

int unbiased_random_bit() {    
    int x1, x2, prev;
    prev = 2;
    x1 = rand() % 2;
    x2 = rand() % 2;

    for (;; x1 = rand() % 2, x2 = rand() % 2)
    {
        if (x1 ^ x2)      // 01 -> 1, or 10 -> 0.
        {
            return x2;        
        }
        else if (x1 & x2)
        {
            if (!prev)    // 0011
                return 1;
            else
                prev = 1; // 1111 -> continue, bias unresolved
        }
        else
        {
            if (prev == 1)// 1100
                return 0;
            else          // 0000 -> continue, bias unresolved
                prev = 0;
        }
    }
}

我刚刚为冯·诺依曼无偏抛硬币法写了一段代码，理论上应该可以消除随机数生成过程中的任何偏差。更多信息请访问(http://en.wikipedia.org/wiki/Fair_coin)

int unbiased_random_bit() {    
    int x1, x2, prev;
    prev = 2;
    x1 = rand() % 2;
    x2 = rand() % 2;

    for (;; x1 = rand() % 2, x2 = rand() % 2)
    {
        if (x1 ^ x2)      // 01 -> 1, or 10 -> 0.
        {
            return x2;        
        }
        else if (x1 & x2)
        {
            if (!prev)    // 0011
                return 1;
            else
                prev = 1; // 1111 -> continue, bias unresolved
        }
        else
        {
            if (prev == 1)// 1100
                return 0;
            else          // 0000 -> continue, bias unresolved
                prev = 0;
        }
    }
}

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。

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。

RAND_MAX值为3(实际上它应该比这个值高得多，但偏差仍然存在)，从这些计算中可以看出存在偏差:

1% 2 = 1 2% 2 = 0 3% 2 = 1 Random_between(1,3) % 2 =更可能是1

在本例中，当您想要0到1之间的随机数时，不应该使用% 2。你可以通过% 3得到一个0到2之间的随机数，因为在这种情况下:RAND_MAX是3的倍数。

另一种方法

有更简单的方法，但要加上其他答案，这是我的解，得到一个0到n - 1之间的随机数，所以有n种不同的可能性，没有偏差。

编码可能性数量所需的比特数(不是字节数)就是您需要的随机数据的比特数 从随机位编码数字 如果这个数字是>= n，重新启动(不取模)。

真正随机的数据是不容易获得的，所以为什么使用比需要更多的比特。

下面是Smalltalk中的一个示例，使用伪随机数生成器的位缓存。我不是安全专家，所以请自担风险。

next: n

    | bitSize r from to |
    n < 0 ifTrue: [^0 - (self next: 0 - n)].
    n = 0 ifTrue: [^nil].
    n = 1 ifTrue: [^0].
    cache isNil ifTrue: [cache := OrderedCollection new].
    cache size < (self randmax highBit) ifTrue: [
        Security.DSSRandom default next asByteArray do: [ :byte |
            (1 to: 8) do: [ :i |    cache add: (byte bitAt: i)]
        ]
    ].
    r := 0.
    bitSize := n highBit.
    to := cache size.
    from := to - bitSize + 1.
    (from to: to) do: [ :i |
        r := r bitAt: i - from + 1 put: (cache at: i)
    ].
    cache removeFrom: from to: to.
    r >= n ifTrue: [^self next: n].
    ^r