Keep selecting a random is a good way to remove the bias.

Update

We could make the code fast if we search for an x in range divisible by n.

// 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; 

The above loop should be very fast, say 1 iteration on average.

Definition

Modulo Bias is the inherent bias in using modulo arithmetic to reduce an output set to a subset of the input set. In general, a bias exists whenever the mapping between the input and output set is not equally distributed, as in the case of using modulo arithmetic when the size of the output set is not a divisor of the size of the input set.

This bias is particularly hard to avoid in computing, where numbers are represented as strings of bits: 0s and 1s. Finding truly random sources of randomness is also extremely difficult, but is beyond the scope of this discussion. For the remainder of this answer, assume that there exists an unlimited source of truly random bits.

Problem Example

Let's consider simulating a die roll (0 to 5) using these random bits. There are 6 possibilities, so we need enough bits to represent the number 6, which is 3 bits. Unfortunately, 3 random bits yields 8 possible outcomes:

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

We can reduce the size of the outcome set to exactly 6 by taking the value modulo 6, however this presents the modulo bias problem: 110 yields a 0, and 111 yields a 1. This die is loaded.

Potential Solutions

Approach 0:

Rather than rely on random bits, in theory one could hire a small army to roll dice all day and record the results in a database, and then use each result only once. This is about as practical as it sounds, and more than likely would not yield truly random results anyway (pun intended).

Approach 1:

Instead of using the modulus, a naive but mathematically correct solution is to discard results that yield 110 and 111 and simply try again with 3 new bits. Unfortunately, this means that there is a 25% chance on each roll that a re-roll will be required, including each of the re-rolls themselves. This is clearly impractical for all but the most trivial of uses.

Approach 2:

Use more bits: instead of 3 bits, use 4. This yield 16 possible outcomes. Of course, re-rolling anytime the result is greater than 5 makes things worse (10/16 = 62.5%) so that alone won't help.

Notice that 2 * 6 = 12 < 16, so we can safely take any outcome less than 12 and reduce that modulo 6 to evenly distribute the outcomes. The other 4 outcomes must be discarded, and then re-rolled as in the previous approach.

Sounds good at first, but let's check the math:

4 discarded results / 16 possibilities = 25% 

In this case, 1 extra bit didn't help at all!

That result is unfortunate, but let's try again with 5 bits:

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

A definite improvement, but not good enough in many practical cases. The good news is, adding more bits will never increase the chances of needing to discard and re-roll. This holds not just for dice, but in all cases.

As demonstrated however, adding an 1 extra bit may not change anything. In fact if we increase our roll to 6 bits, the probability remains 6.25%.

This begs 2 additional questions:

1. If we add enough bits, is there a guarantee that the probability of a discard will diminish?
2. How many bits are enough in the general case?

General Solution

Thankfully the answer to the first question is yes. The problem with 6 is that 2^x mod 6 flips between 2 and 4 which coincidentally are a multiple of 2 from each other, so that for an even x > 1,

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

Thus 6 is an exception rather than the rule. It is possible to find larger moduli that yield consecutive powers of 2 in the same way, but eventually this must wrap around, and the probability of a discard will be reduced.

Without offering further proof, in general using double the number of bits required will provide a smaller, usually insignificant, chance of a discard.

Proof of Concept

Here is an example program that uses OpenSSL's libcrypo to supply random bytes. When compiling, be sure to link to the library with -lcrypto which most everyone should have available.

#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; } 

I encourage playing with the MODULUS and ROLLS values to see how many re-rolls actually happen under most conditions. A sceptical person may also wish to save the computed values to file and verify the distribution appears normal.

@user1413793 is correct about the problem. I'm not going to discuss that further, except to make one point: yes, for small values of n and large values of RAND_MAX, the modulo bias can be very small. But using a bias-inducing pattern means that you must consider the bias every time you calculate a random number and choose different patterns for different cases. And if you make the wrong choice, the bugs it introduces are subtle and almost impossible to unit test. Compared to just using the proper tool (such as arc4random_uniform), that's extra work, not less work. Doing more work and getting a worse solution is terrible engineering, especially when doing it right every time is easy on most platforms.

Unfortunately, the implementations of the solution are all incorrect or less efficient than they should be. (Each solution has various comments explaining the problems, but none of the solutions have been fixed to address them.) This is likely to confuse the casual answer-seeker, so I'm providing a known-good implementation here.

Again, the best solution is just to use arc4random_uniform on platforms that provide it, or a similar ranged solution for your platform (such as Random.nextInt on Java). It will do the right thing at no code cost to you. This is almost always the correct call to make.

If you don't have arc4random_uniform, then you can use the power of opensource to see exactly how it is implemented on top of a wider-range RNG (ar4random in this case, but a similar approach could also work on top of other RNGs).

Here is the OpenBSD implementation:

/* * Calculate a uniformly distributed random number less than upper_bound * avoiding "modulo bias". * * Uniformity is achieved by generating new random numbers until the one * returned is outside the range [0, 2**32 % upper_bound). This * guarantees the selected random number will be inside * [2**32 % upper_bound, 2**32) which maps back to [0, upper_bound) * after reduction modulo upper_bound. */ u_int32_t arc4random_uniform(u_int32_t upper_bound) { u_int32_t r, min; if (upper_bound < 2) return 0; /* 2**32 % x == (2**32 - x) % x */ min = -upper_bound % upper_bound; /* * This could theoretically loop forever but each retry has * p > 0.5 (worst case, usually far better) of selecting a * number inside the range we need, so it should rarely need * to re-roll. */ for (;;) { r = arc4random(); if (r >= min) break; } return r % upper_bound; } 

It is worth noting the latest commit comment on this code for those who need to implement similar things:

Change arc4random_uniform() to calculate 2**32 % upper_bound as -upper_bound % upper_bound. Simplifies the code and makes it the same on both ILP32 and LP64 architectures, and also slightly faster on LP64 architectures by using a 32-bit remainder instead of a 64-bit remainder.

Pointed out by Jorden Verwer on tech@ ok deraadt; no objections from djm or otto

The Java implementation is also easily findable (see previous link):

public int nextInt(int n) { if (n <= 0) throw new IllegalArgumentException("n must be positive"); if ((n & -n) == n) // i.e., n is a power of 2 return (int)((n * (long)next(31)) >> 31); int bits, val; do { bits = next(31); val = bits % n; } while (bits - val + (n-1) < 0); return val; } 

Mark's Solution (The accepted solution) is Nearly Perfect.

int x; do { x = rand(); } while (x >= (RAND_MAX - RAND_MAX % n)); x %= n; 

edited Mar 25 '16 at 23:16

Mark Amery 39k21170211

However, it has a caveat which discards 1 valid set of outcomes in any scenario where RAND_MAX (RM) is 1 less than a multiple of N (Where N = the Number of possible valid outcomes).

ie, When the 'count of values discarded' (D) is equal to N, then they are actually a valid set (V), not an invalid set (I).

What causes this is at some point Mark loses sight of the difference between N and Rand_Max.

N is a set who's valid members are comprised only of Positive Integers, as it contains a count of responses that would be valid. (eg: Set N = {1, 2, 3, ... n } )

Rand_max However is a set which ( as defined for our purposes ) includes any number of non-negative integers.

In it's most generic form, what is defined here as Rand Max is the Set of all valid outcomes, which could theoretically include negative numbers or non-numeric values.

Therefore Rand_Max is better defined as the set of "Possible Responses".

However N operates against the count of the values within the set of valid responses, so even as defined in our specific case, Rand_Max will be a value one less than the total number it contains.

Using Mark's Solution, Values are Discarded when: X => RM - RM % N

EG: Ran Max Value (RM) = 255 Valid Outcome (N) = 4 When X => 252, Discarded values for X are: 252, 253, 254, 255 So, if Random Value Selected (X) = {252, 253, 254, 255} Number of discarded Values (I) = RM % N + 1 == N IE: I = RM % N + 1 I = 255 % 4 + 1 I = 3 + 1 I = 4 X => ( RM - RM % N ) 255 => (255 - 255 % 4) 255 => (255 - 3) 255 => (252) Discard Returns $True 

As you can see in the example above, when the value of X (the random number we get from the initial function) is 252, 253, 254, or 255 we would discard it even though these four values comprise a valid set of returned values.

IE: When the count of the values Discarded (I) = N (The number of valid outcomes) then a Valid set of return values will be discarded by the original function.

If we describe the difference between the values N and RM as D, ie:

D = (RM - N) 

Then as the value of D becomes smaller, the Percentage of unneeded re-rolls due to this method increases at each natural multiplicative. (When RAND_MAX is NOT equal to a Prime Number this is of valid concern)

EG:

RM=255 , N=2 Then: D = 253, Lost percentage = 0.78125% RM=255 , N=4 Then: D = 251, Lost percentage = 1.5625% RM=255 , N=8 Then: D = 247, Lost percentage = 3.125% RM=255 , N=16 Then: D = 239, Lost percentage = 6.25% RM=255 , N=32 Then: D = 223, Lost percentage = 12.5% RM=255 , N=64 Then: D = 191, Lost percentage = 25% RM=255 , N= 128 Then D = 127, Lost percentage = 50% 

Since the percentage of Rerolls needed increases the closer N comes to RM, this can be of valid concern at many different values depending on the constraints of the system running he code and the values being looked for.

To negate this we can make a simple amendment As shown here:

 int x; do { x = rand(); } while (x > (RAND_MAX - ( ( ( RAND_MAX % n ) + 1 ) % n) ); x %= n; 

This provides a more general version of the formula which accounts for the additional peculiarities of using modulus to define your max values.

Examples of using a small value for RAND_MAX which is a multiplicative of N.

Mark'original Version:

RAND_MAX = 3, n = 2, Values in RAND_MAX = 0,1,2,3, Valid Sets = 0,1 and 2,3. When X >= (RAND_MAX - ( RAND_MAX % n ) ) When X >= 2 the value will be discarded, even though the set is valid. 

Generalized Version 1:

RAND_MAX = 3, n = 2, Values in RAND_MAX = 0,1,2,3, Valid Sets = 0,1 and 2,3. When X > (RAND_MAX - ( ( RAND_MAX % n ) + 1 ) % n ) When X > 3 the value would be discarded, but this is not a vlue in the set RAND_MAX so there will be no discard. 

Additionally, in the case where N should be the number of values in RAND_MAX; in this case, you could set N = RAND_MAX +1, unless RAND_MAX = INT_MAX.

Loop-wise you could just use N = 1, and any value of X will be accepted, however, and put an IF statement in for your final multiplier. But perhaps you have code that may have a valid reason to return a 1 when the function is called with n = 1...

So it may be better to use 0, which would normally provide a Div 0 Error, when you wish to have n = RAND_MAX+1

Generalized Version 2:

int x; if n != 0 { do { x = rand(); } while (x > (RAND_MAX - ( ( ( RAND_MAX % n ) + 1 ) % n) ); x %= n; } else { x = rand(); } 

Both of these solutions resolve the issue with needlessly discarded valid results which will occur when RM+1 is a product of n.

The second version also covers the edge case scenario when you need n to equal the total possible set of values contained in RAND_MAX.

The modified approach in both is the same and allows for a more general solution to the need of providing valid random numbers and minimizing discarded values.

To reiterate:

The Basic General Solution which extends mark's example:

// Assumes: // RAND_MAX is a globally defined constant, returned from the environment. // int n; // User input, or externally defined, number of valid choices. int x; do { x = rand(); } while (x > (RAND_MAX - ( ( ( RAND_MAX % n ) + 1 ) % n) ) ); x %= n; 

The Extended General Solution which Allows one additional scenario of RAND_MAX+1 = n:

// Assumes: // RAND_MAX is a globally defined constant, returned from the environment. // int n; // User input, or externally defined, number of valid choices. int x; if n != 0 { do { x = rand(); } while (x > (RAND_MAX - ( ( ( RAND_MAX % n ) + 1 ) % n) ) ); x %= n; } else { x = rand(); } 

In some languages ( particularly interpreted languages ) doing the calculations of the compare-operation outside of the while condition may lead to faster results as this is a one-time calculation no matter how many re-tries are required. YMMV!

// Assumes: // RAND_MAX is a globally defined constant, returned from the environment. // int n; // User input, or externally defined, number of valid choices. int x; // Resulting random number int y; // One-time calculation of the compare value for x y = RAND_MAX - ( ( ( RAND_MAX % n ) + 1 ) % n) if n != 0 { do { x = rand(); } while (x > y); x %= n; } else { x = rand(); } 

There are two usual complaints with the use of modulo.

• 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

Using something like

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; } 

to generate a random number between 0 and n will avoid both problems (and it avoids overflow with RAND_MAX == INT_MAX)

BTW, C++11 introduced standard ways to the the reduction and other generator than rand().

With a RAND_MAX value of 3 (in reality it should be much higher than that but the bias would still exist) it makes sense from these calculations that there is a bias:

1 % 2 = 1 2 % 2 = 0 3 % 2 = 1 random_between(1, 3) % 2 = more likely a 1

In this case, the % 2 is what you shouldn't do when you want a random number between 0 and 1. You could get a random number between 0 and 2 by doing % 3 though, because in this case: RAND_MAX is a multiple of 3.

Another method

There is much simpler but to add to other answers, here is my solution to get a random number between 0 and n - 1, so n different possibilities, without bias.

• the number of bits (not bytes) needed to encode the number of possibilities is the number of bits of random data you'll need
• encode the number from random bits
• if this number is >= n, restart (no modulo).

Really random data is not easy to obtain, so why use more bits than needed.

Below is an example in Smalltalk, using a cache of bits from a pseudo-random number generator. I'm no security expert so use at your own risk.

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 

Modulo reduction is a commonly seen way to make a random integer generator avoid the worst case of running forever.

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.)

The following answer of mine discusses the relationship between running time and bias in random generators, assuming we have a "true" random generator that can produce unbiased and independent random bits. The answer doesn't even involve the rand() function in C because it has many issues. Perhaps the most serious here is the fact that the C standard does not explicitly specify a particular distribution for the numbers returned by rand(), not even a uniform distribution.

• How to generate a random integer in the range [0,n] from a stream of random bits without wasting bits?

As the accepted answer indicates, "modulo bias" only really shows up very much because of the very low value of RAND_MAX from stdlib.h.

That answer shows that for an extremely small value of RAND_MAX (10), then you tried to generate a number between 0 and 2 using %, the following outcomes would result:

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 10 | 1 

So there are 4 outputs of 0's (4/10 chance), 4 outputs of 1 (4/10 chance) but only 3 outputs of 2 (3/10 chance).

So it's biased because the lower numbers have a better chance of coming out.

Practically, that only shows very much obviously when the number you are modding by is close to RAND_MAX.

For RAND_MAX=32767 and N=3, the actual chances are:

0: 10923 33.3353% 1: 10923 33.3353% 2: 10922 33.3323% 

So it won't really show up if your N is small.

I should point out that <random> has been in since C++11, and there's really no excuse to be using rand() from stdlib.h anymore.

#include <random> #include <time.h> #include <stdio.h> int main() { std::default_random_engine random; //mt19937 // usual practice to seed RNG with start time random.seed( time( 0 ) ); printf( "Rand Max: %llu\n", random._Max ); // 4,294,967,295 for( int i = 0; i < 25; i++ ) { printf( "%u, ", random() % 3 ); } } 

With a RAND_MAX of 4,294,967,295 and N=3, your numbers are:

0 : 1431655766 33.33333335% 1 : 1431655765 33.33333333% 2 : 1431655765 33.33333333% 

So the modulo bias effect all but disappears on it's own then

mt19937_64

Using a RAND_MAX of 18,446,744,073,709,551,615

If that's not good enough for you yet, there's even mt19937_64, which has a RAND_MAX of 18,446,744,073,709,551,615

Hello large RAND_MAX, goodbye modulo bias!

#include <map> #include <time.h> #include <random> #include <stdio.h> int main() { std::mt19937_64 random; //RAND_MAX of 18,446,744,073,709,551,615 printf( "RAND_MAX: %llu\n", random.max() ); random.seed( time( 0 ) ); uint64_t Iters = 16'000'000; map<uint64_t, uint64_t> vals; for( uint64_t i = 0; i < Iters; i++ ) { vals[ random() % 3 ]++; } printf( "%llu iters\n", Iters ); for( std::pair<const uint64_t, uint64_t>& p : vals ) { printf( "%llu : %llu (%f%%)\n", p.first, p.second, (double)p.second/Iters ); } } 

Typical run:

RAND_MAX: 18446744073709551615 16000000 iters 0 : 5333947 (0.333372%) 1 : 5332659 (0.333291%) 2 : 5333394 (0.333337%) 

The jitter in the values just shows you that the modulo bias effect all but disappears with a large enough RAND_MAX