Biased Random Number Generator - c++

Question

I am looking for a random number generator that can be biased. For instance, say I want a random number between 1-5, with the probability being:
1: Comes up 20% of the time
2: Comes up 10% of the time
3: Comes up 40% of the time
4: Comes up 25% of the time
5: Comes up 5% of the time
Is there anything in the standard library, or other libraries out there that would do this? Alternatively, is there an efficient way to do this myself?

Answer 1

For your problem, just pick a random element from this list uniformly:
[1, 1, 1, 1, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5]
In general, check this answer: Weighted random numbers
In TR1 and C++0x, there is <random> header which contains the discrete_distribution class to generate such numbers, among others.
You may also want to check out GSL which contains much more random distributions (and random number generators) than the standard <random> library. (But note that GSL uses GPLv3.)

Answer 2

Best way's probably to just take the normal unbiased random generator then return based on the interval its value falls into.
Just an if statement that gives 1 for 0:0.2, 2 for 0.2:0.3, 3 for 0.3:0.7, 4 for 0.7:0.95 and 5 for 0.95:1. Best to make either the lower or upper limit of the interval inclusive and the other exclusive.
int biasedRandom(){
double i = randomNumber();
if(i<= 0.2){return 1;}
else if(i <= 0.3){return 2;}
else if(i <= 0.7){return 3;}
else if(i <= 0.95){return 4;}
else{return 5;}
}
Something like that.

Answer 3

The Boost random number library provides the ability to specify different shaped distributions for your generator. It's a great library - see http://www.boost.org/doc/libs/1_42_0/libs/random/index.html.

Answer 4

Coming late to the party on this one. Here is the C++0x answer:
#include <iostream>
#include <random>
#include <iterator>
int main()
{
// Set up distribution
double interval[] = {1, 2, 3, 4, 5, 6};
double weights[] = { .2, .1, .4, .25, .05};
std::piecewise_constant_distribution<> dist(std::begin(interval),
std::end(interval),
std::begin(weights));
// Choose generator
std::mt19937 gen; // seed as wanted
// Demonstrate by pouring into avg[rand-1]
const unsigned N = 1000000;
double avg[sizeof(weights) / sizeof(weights[0])] = {0};
for (unsigned i = 0; i < N; ++i)
avg[static_cast<unsigned>(dist(gen)) - 1]++;
// Comute averages
for (double* i = std::begin(avg); i < std::end(avg); ++i)
*i /= N;
// Display
for (unsigned i = 1; i <= sizeof(avg)/sizeof(avg[0]); ++i)
std::cout << "avg[" << i << "] = " << avg[i-1] << '\n';
}
Which for me outputs:
avg[1] = 0.199779
avg[2] = 0.100002
avg[3] = 0.400111
avg[4] = 0.250257
avg[5] = 0.049851

Answer 5

What you are describing is the implementation of a random number generator that draws from a particular probability distribution. For example, drawing numbers from a Gaussian distribution should draw random numbers such that the probability of a particular draw, x is proportional to
(source: wikimedia.org)
.
In general, the approach is to draw from a uniform random distribution and then pick the value of the desired distribution's cumulative distribution function (CDF) at that drawn location. In the case of a Normal Gaussian, draw a random number, x from a uniform distribution (this is what standard random number generators should give) and then choose as the random, Gaussian distributed value. For your case, the CDF you describe is a piece-wise continuous stair-step function which could be implemented using any of the many (correct) answers you have already received.
Of course, this is all trivia. What you should be doing is using a library that already handles this for you. Statistics and random number generation are not trivial and there's no need to re-invent the wheel. See Neil's answer (and check out the Boost random number library).

Answer 6

Why don't you just use a regular random number generator that return number between 0.0 and 1.0, and wrap it with another function that returns a number according to your requirements?
like
double biased (double seed) {
if (seed >= 0.0 && seed <0.2) return 1;
else if ...
}

Answer 7

Throw a random real number x in [0,1], if 0< x<0.2 return 1, if 0.2<x <0.3 return 2, etc.
See here for the general problem.

Answer 8

Kenny gave an appropriate answer tailored to your particular frequency distribution.
The more general answer works with a CDF - Cumulative Distribution Function - for the data, and uses a uniform random number to pick a value within the distribution.

Answer 9

I am doing to do the same thing and I found this:
http://eli.thegreenplace.net/2010/01/22/weighted-random-generation-in-python/
Seems good enough for the purpose you stated.

Answer 10

#include <boost/random/discrete_distribution.hpp>
#include <boost/random/mersenne_twister.hpp>
#include <boost/random/variate_generator.hpp>
#include <iostream>
int main()
{
unsigned int seed = 42;
boost::mt19937 generator(seed);
// return 0 with probability 10%
// 1 40%
// 2 50%
boost::random::discrete_distribution<int> custom_dist{1,4,5};
boost::variate_generator<boost::mt19937&,
boost::random::discrete_distribution<int> > rndn(generator, custom_dist);
for (unsigned int i = 0; i<10000; i++) {
std::cout << rndn() << std::endl;
}
return 0;
}
And here is a plot of the result:

Answer 11

I was looking for something like this for TypeScript, but only found this question for C.
So here is a biased random number generator in TypeScript I came up with, in case anybody needs something like this in TypeScript. I am sure you can translate it to C somehow.
export async function weightedRandomItem<T>(list: { weight: number; item: T }[]): Promise<T> {
const weightSum = sumBy(list, (item) => item.weight)
const randomIndex = await randomIntegerBetween(1, weightSum)
let currentIndex = 1
for (const listItem of list) {
if (randomIndex >= currentIndex && randomIndex < currentIndex + listItem.weight) {
return listItem.item
}
currentIndex += listItem.weight
}
throw new Error("No item selected. Impossible.")
}
where randomIntegerBetween(minInclusive: number, maxInclusive: number) returns a random integer from the specified range (min and max inclusive) from the RNG of your choice.
sumBy() is the lodash function in this case, and it should be self-explanatory.
As input you could pass in something like:
[{
weight: 10,
item: 1,
},
{
weight: 50,
item: 2,
},
{
weight: 30,
item: 3,
},
{
weight: 10,
item: 4,
}]
Then, the result would most probably be 2.