random number near 1

Question

I know I have done this before, but now I'm struggling with it again. I am trying to set a random number on either side of 1: .98, 1.02, .94, 1.1, etc. If I get a random number between 0 and 100, how can I use that to get within the range I want?

The programming language doesn't really matter here. I'm using Pure Data. Just looking for the math involved.

Answer 1

Divide by 100 and add 1. (I assume you are looking for a range from 0 to 2?)

Answer 2

return 0.9 + rand(100) / 500.0

or am I missing something?

Answer 3

low + (random() / 100) * range

So for example:

0.90 + (random() / 100) * 0.2

Answer 4

How near? You could use a Guassian (a.k.a. Normal) distribution with a mean of 1 and a small standard deviation.

A Gaussian is suitable if you want numbers close to 1 to be more frequent than numbers a bit further away from 1.

Some languages (such as Java) will have support for Gaussians in the standard library.

Answer 5

If rand() returns you a random number between 0 and 100, all you need to do is:

(rand() / 100) * 2

to get a random number between 0 and 2.

If on the other hand you want the range from 0.9 to 1.1, use the following:

0.9 + ((rand() / 100) * 0.2)

Answer 6

Uniform

If you want a (psuedo-)uniform distribution (evenly spaced) between 0.9 and 1.1 then the following will work:

  range = 0.2
  return 1-range/2+rand(100)*range/100

Adjust the range accordingly.

Pseudo-normal

If you wanted a normal distribution (bell curve) you would need special code, which would be language/library specific. You can get a close approximation with this code:

sd = 0.1
mean = 1
count = 10
sum = 0
for(int i=1; i<count; i++) 
  sum=sum+(rand(100)-50)
}
normal = sum / count
normal = normal*sd + mean

Answer 7

You want a range from -1 to 1 as output from your rand() expression.

( rand(2) - 1 )

Then scale that -1 to 1 range as needed. Say, for a .1 variation on either side:

(( rand(2) - 1 ) / 10 )

Then just add one.

(( rand(2) - 1 ) / 10 ) + 1