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Comment by ColinWright | Hacker News Reader

The second is what I say about being evenly distributed.

The first is easy to solve. Get a result that's sufficiently fine-grained and in the range [0,1). Then scale up by X and round up.

You can add randomness by pushing your number through a cryptographic hash function and treating the result as a binary number in the range [0,1).

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Excuse my ignorance, but what is the meaning of the differing brackets in the notation you're using: "[0,1)"?

Standard mathematical notation to indicate whether the endpoint is or is not included.

  { x : a < x && x < b }  <=>  (a,b)
  { x : a <=x && x < b }  <=>  [a,b)
  { x : a < x && x <=b }  <=>  (a,b]
  { x : a <=x && x <=b }  <=>  [a,b]

In the example above the questioner is asking about getting a number in the range 1 to X inclusive. That would be [1,X]. Random number generators can return numbers in the range 0 to 1, including 0, not including 1. You can write that as [0,1).

If you want an integer n in [1,X] and you get reals r [0,1), you can create a function:

    f : [0,1) -> [1,X], f(r) = 1+floor(X*r)

Always check the endpoints, and the expected frequencies.

Thank you, that's helpful!

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