Biased Random Number Generator

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: http://stackoverflow.com/questions/1761626/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

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 4

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 5

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 6

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 .

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 7

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.