Random number with Probabilities

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Yours is a pretty good way already and works well with any range.

Just thinking: another possibility is to get rid of the fractions by multiplying with a constant multiplier, and then build an array with the size of this multiplier. Multiplying by 10 you get

P(1) = 2
P(2) = 3
P(3) = 5

Then you create an array with the inverse values — ‘1’ goes into elements 1 and 2, ‘2’ into 3 to 6, and so on:

P = (1,1, 2,2,2, 3,3,3,3,3);

and then you can pick a random element from this array instead.

(Add.) Using the probabilities from the example in kiruwka’s comment:

int[] numsToGenerate           = new int[]    { 1,   2,    3,   4,    5   };
double[] discreteProbabilities = new double[] { 0.1, 0.25, 0.3, 0.25, 0.1 };

the smallest multiplier that leads to all-integers is 20, which gives you

2, 5, 6, 5, 2

and so the length of numsToGenerate would be 20, with the following values:

1 1
2 2 2 2 2
3 3 3 3 3 3
4 4 4 4 4
5 5

The distribution is exactly the same: the chance of ‘1’, for example, is now 2 out of 20 — still 0.1.

This is based on your original probabilities all adding up to 1. If they do not, multiply the total by this same factor (which is then going to be your array length as well).

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