Suppose you have a series of bytes randomly selected with a uniform distribution over [0, 256). Take seven of these bytes, for example 0 , a 1 , ... a 6 . Calculate ((((((((< 6 % 32) / 32 + a 5 ) / 256 + a 4 ) / 256 + a 3 ) / 256 + a 2 ) / 256 + a 1 ) / 256 + a 0 ) / 256.
Explanation: a 6 % 32 denotes a remainder of 6 modulo 32. This takes up five bits of 6 . Then dividing by 32 “shifts” these bits to the right of the notation, eight new bits are added, dividing by 256 “shifts” the right eight bits, and so on, until we have 53 bits.
This gives 2 53 possible results in [0, 1). More can be done because the resolution of the floating point is smaller, because the values are close to zero, but then there are problems with uniformity and other problems.