Write a truly inclusive random method for javascript

Question

Javascript\'s MATH object has a random method that returns from the set [0,1) 0 inclusive, 1 exclusive. Is there a way to return a truly random method that includes 1.

Answer 1

Since this question has been asked again, and I didn't read this approach here I'll add another answer.

IMO the best you can do, without too much hassle would be:

exclusive:

//simply ignore the 0
for(var rand=0; !rand;rand = Math.random());

//or simpler:
var rand = Math.random() || Math.random();
//because the probability for two consecutive `0` is pretty much non existant.

this doesn't even introduce an error, since we just excluded the possibility of returning 0, every other value between 0 and 1 has the same probability

inclusive:

var rand = Math.random() * 2;
if(rand > 1) rand-=1;
//so the 0 shares it's probability with the 1.

just to be clear about how tiny the "error" is:

• the probability for a 0 or a 1 is
  1 / Math.pow(2, 54) or about 5.55e-17
• the probability for any other value between 0 and 1 is
  1 / Math.pow(2, 53) or about 11.1e-17

and the whole random-function would be:

function random(min, max, inclusive){
    var r = Math.random();
    if(inclusive)
        r = r>0.5? 2*r-1: 2*r;
    else 
        while(!r) r = Math.random();

    return r * (max - min) + min;
}

Edit: I'm not sure wether I make a mistake, but shouldn't the probability be fixed on the inclusive approach, if I add another bit to the zeroes and ones, and therefore duplicate their probability:

var rand = Math.random() * 4;
rand = (rand % 1) || (rand & 1);