Biased Random Number Generator

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.

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.

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

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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.)