Question

I want to generate a random number with a given probability but I’m not sure how to:

I need a number between 1 and 3

num = ceil(rand*3);

but I need different values to have different probabilities of generating eg.

0.5 chance of 1
0.1 chance of 2
0.4 chance of 3

I’m sure this is straightforward but I can’t think of how to do it.

Answer 1

The simple solution is to generate a number with a uniform distribution (using rand), and manipulate it a bit:

r = rand;
prob = [0.5, 0.1, 0.4];
x = sum(r >= cumsum([0, prob]));

or in a one-liner:

x = sum(rand >= cumsum([0, 0.5, 0.1, 0.4]));

Explanation

Here r is a uniformly distributed random number between 0 and 1. To generate an integer number between 1 and 3, the trick is to divide the [0, 1] range into 3 segments, where the length of each segment is proportional to its corresponding probability. In your case, you would have:

• Segment [0, 0.5), corresponding to number 1.
• Segment [0.5, 0.6), corresponding to number 2.
• Segment [0.6, 1], corresponding to number 3.

The probability of r falling within any of the segments is proportional to the probabilities you want for each number. sum(r >= cumsum([0, prob])) is just a fancy way of mapping an integer number to one of the segments.

Extension

If you’re interested in creating a vector/matrix of random numbers, you can use a loop or arrayfun:

r = rand(3); % # Any size you want
x = arrayfun(@(z)sum(z >= cumsum([0, prob])), r);

Of course, there’s also a vectorized solution, I’m just too lazy to write it.