Cheers. I am trying to get an eually distributed random double from the interval (-1,1). Here's what I have tried:


This only gives me the interval [-1,1).


This makes x a double from the interval (-1,1), but I think the numbers that can be drawn are not equally distributed. I'm just a chemist, so someone better at math than me fill me in, please.


It's (-1,1), but the 0 should be twice a probable as any other number.

rand.nextDouble() returns a value from [0, 1). So you'll never get the value 1, which means that you'll never get (2*rand.nextDouble()) == 2. therefore, you'll never get 1 which is why your interval is from [-1, 1)



When you say that it's not equally distributed, what do you mean? rand.nextDouble() itself returns a pseudorandom, uniformly distributed double value between 0.0 and 1.0. This means that the numbers will be uniformly distributed but not actually random. Pseudorandom means that the numbers generated are random but are the same random numbers each run (I think, maybe that's true in c++ and not java but i don't really remember)



I'm pretty sure you're right about this one.

can you explain what you're trying to accomplish? keep in mind that doubles only have 64 bit precision, so obviously infinitely many numbers are skipped. and ofc ^ already mentioned these aren't really random

2*rand-1 should be fine for practical purposes.

iirc from my probability course in high school, the PDF does not change whether the range is inclusive or exclusive. so (-1, 1) has the same pdf and same cdf as [-1, 1]

This post was edited by carteblanche on Apr 8 2013 06:09pm

I know, thanks.
I also know, as you have mentioned, that any randomly generated number is only pseudo-random (but equally distributed), but let's leave that aside.


I mean that not every number in (-1,1) has the same probability to be drawn with the second approach.
I'll explain with two dice:

rand.nextInt(6)+1 + rand.nextInt(6)+1 will give integers from [2,12], where some numbers, such as 7, aber more probable than others like 2 or 12. There are several outcomes that roll a 7, (3+4; 4+3; 5+2 etc.) but only one that rolls a 12 (6+6).

rand.nextInt(11)+2 will give Integers from [2,12], but they are equally distributed, meaning each number from 2 to 12 is drawn with the same probability.

So, rand.nextDouble()-rand.nextDouble() will give me a number from (-1,1), but I am not sure whether they are equally distributed.


Which leaves the problem of 0 being twice as probable as any other number from (-1,1) with this approach.


I want to generate random numbers from the interval (-1,1), with the same probability for each number that can be drawn to be drawn. It's a programming exercise.

Another try:


Idea: if it is 0, give it a 50% of not being zero.

This post was edited by tt_toby on Apr 9 2013 03:53am