0,1,2,3,4 % 3
0 => 0
1 => 1
2 => 2
3 => 0
4 => 1
So, a uniform input will produce more 0s and 1s than it will 2s.
0 => 0
1 => 1
2 => 2
3 => 0
4 => 1
So, a uniform input will produce more 0s and 1s than it will 2s.
However, there are methods of converting uniform distributions that will preserve uniformity. I'll have to find a good reference.
Issue is that the root reply seemed to have taken it as a JS implementation issue (due to math.random) as it's quite obvious why you should not do the modulus (or when you can), if you know the underlying maths.
So did I, as a matter of fact.
You - on the other hand had given the answer to the wrong confusion.
I'm the root reply. jimktrains2's explanation exactly addressed my confusion.
To get a uniform distribution over an alphabet from random, I would do:
var bucket_size = (random.MAX_VALUE - random.MIN_VALUE) / count(alphabet)
return alphabet[random/bucket_size]max = 4, min = 0, count = 3
bucket_size = 4 / 3
0,1,2,3,4
0 / (4/3) = 0
1 / (4/3) = 3/4 = 0
2 / (4/3) = 6/3 = 2
3 / (4/3) = 9/4 = 2
4 / (4/3) = 12/4 = 3 = invalid index
but even if you adjust this to count - 1
bucket_size = 4 / 2 = 2
0 = 0/2 = 0
1 = 1/2 = 0
2 = 2/2 = 1
3 = 3/2 = 1
4 = 4/2 = 2
So, 0s and 1s are still more likely than 2 to appear.
One of the easiest, and slowest, ways is just to discard any trial in which your uniform RNG generates a number above the largest index and try again.
For a larger alphabet, you essentially just collect multiple trials, treat them as a single trial, and do similar tactics.
I have an off-by-one error… your random function has a range of 5, so it should be max - min + 1:
bucket_size = 5 / 3
0 / (5/3) = 0
1 / (5/3) = 3/5 = 0
2 / (5/3) = 6/5 = 1
3 / (5/3) = 9/5 = 1
4 / (5/3) = 12/5 = 2
Yeah, I planned for 65535 randoms and 52 alphabet. It would still mildly bias to the first range mod alphabet_size elements.