4 ms·

Here's some intuition about why rand() * rand() is different than rand(). Let's say you are doing rand(2) which produces either 0 or 1, each equally often, the

Here's some intuition about why rand() * rand() is different than rand(). Let's say you are doing rand(2) which produces either 0 or 1, each equally often, then rand(2) * rand(2) will produce 0 much more often than 1. 0 * 0 = 0 0 * 1 = 0 1 * 0 = 0 1 * 1 = 1 Addition has a similar problem. 0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 2 If you use the right range and wrap the result. e.g. if you do mod(rand(2^n) + rand(2^n), 2^n) then addition will not affect the distribution of your numbers.

danger 16y ago
Or you could argue formally by pointing out that the maximum entropy distribution on the interval [0,1] is the uniform distribution, so modifying that distribution will only decrease its entropy, which means that it's "more predictable". http://en.wikipedia.org/wiki/Entropy_(information_theory) http://en.wikipedia.org/wiki/Entropy_(information_theory) http://en.wikipedia.org/wiki/Maximum_entropy_probability_distribution#Uniform_and_piecewise_uniform_distributions http://en.wikipedia.org/wiki/Maximum_entropy_probability_dis...