java random percentages

Make an array. Randomly drop 100 %'s into each of the parts of that array. Example shows n=7.

import java.util.Random;

public class random100 {
    public static void main (String [] args) {
        Random rnd = new Random();
            int percents[] = new int[7];
            for (int i = 0; i < 100; i++) {
                int bucket = rnd.nextInt(7);
                percents[bucket] = percents[bucket] + 1;
            }
        for (int i = 0; i < 7; i++) {
            System.out.println("bucket " + i + ": " + percents[i]);
        }

    }

}

To be precise it depends on exactly how you want the samples to be unbiased. Here is a rough way which will roughly give you a good result.

1. Generate n-1 integers from 0,..100, say a[i] for i = 0, to n-2.
2. Let total be the sum of these numbers
3. Compute b[i] = floor(100*a[i]/total) for i = 0, to n-2
4. Set b[n-1] = 100 - (b[0] + ... b[n-2]).

Then b is your resulting array of percentages.

The last one will be biased, but the rest should be uniform.

Of course if you want to do this in a more accurate way you'll have to use Gibbs sampling or Metropolis hastings.

First, obvious solution.

do
    int[] a = new int[n];
    for (int i = 0; i < n; ++i) {
        a[i] = random number between 0 and 100;
    }
until sum(a) == 100;

It's not perfect in terms of complexity (number of iterations to reach sum 100 can be quite large), but distribution is surely 'unbiased'.

edit
Similar problem: how to generate random point in a circle with radius 1 and center in (0, 0)? Solution: continue generating random points in range (square) [-1..1,-1..1] until one of them fits the circle :)

I had a similar issue and ended up doing as you said, generating random integers up to the difference of the sum of the existing integers and the limit. I then randomized the order of the integers. It worked pretty well. That was for a genetic algorithm.