Came here from a link from expanding a float range. This one is more fun. Instead of how I got to the conclusion, it occurred to me that for a given random integer generating function f with "base" b (4 in this case,i'll tell why), it can be expanded like below:
(b^0 * f() + b^1 * f() + b^2 * f() .... b^p * f()) / (b^(p+1) - 1) * (b-1)
This will convert a random generator to a FLOAT generator. I will define 2 parameters here the b and the p. Although the "base" here is 4, b can in fact be anything, it can also be an irrational number etc. p, i call precision is a degree of how well grained you want your float generator to be. Think of this as the number of calls made to rand5 for each call of rand7.
But I realized if you set b to base+1 (which is 4+1 = 5 in this case), it's a sweet spot and you'll get a uniform distribution. First get rid of this 1-5 generator, it is in truth rand4() + 1:
function rand4(){
return Math.random() * 5 | 0;
}
To get there, you can substitute rand4 with rand5()-1
Next is to convert rand4 from an integer generator to a float generator
function toFloat(f,b,p){
b = b || 2;
p = p || 3;
return (Array.apply(null,Array(p))
.map(function(d,i){return f()})
.map(function(d,i){return Math.pow(b,i)*d})
.reduce(function(ac,d,i){return ac += d;}))
/
(
(Math.pow(b,p) - 1)
/(b-1)
)
}
This will apply the first function I wrote to a given rand function. Try it:
toFloat(rand4) //1.4285714285714286 base = 2, precision = 3
toFloat(rand4,3,4) //0.75 base = 3, precision = 4
toFloat(rand4,4,5) //3.7507331378299122 base = 4, precision = 5
toFloat(rand4,5,6) //0.2012288786482335 base = 5, precision =6
...
Now you can convert this float range (0-4 INCLUSIVE) to any other float range and then downgrade it to be an integer. Here our base is 4 because we are dealing with rand4, therefore a value b=5 will give you a uniform distribution. As the b grows past 4, you will start introducing periodic gaps in the distribution. I tested for b values ranging from 2 to 8 with 3000 points each and compared to native Math.random of javascript, looks to me even better than the native one itself:
http://jsfiddle.net/ibowankenobi/r57v432t/
For the above link, click on the "bin" button on the top side of the distributions to decrease the binning size. The last graph is native Math.random, the 4th one where d=5 is uniform.
After you get your float range either multiply with 7 and throw the decimal part or multiply with 7, subtract 0.5 and round:
((toFloat(rand4,5,6)/4 * 7) | 0) + 1 ---> occasionally you'll get 8 with 1/4^6 probability.
Math.round((toFloat(rand4,5,6)/4 * 7) - 0.5) + 1 --> between 1 and 7