[lifthrasiir]

The proposed generator has, assuming the true uniform integer generator, the mean expected value of 0.5 + 2^-152 [1]. If you can ever distinguish this difference you are more likely to see a biased RNG output much earlier.

[1] Start with the first `rand_between_zero_and_one` snippet. `x = ((x + 1) >> 1) + (e << 52)` can be rewritten as `d = (1.0 + ((x + 1) >> 1) * 2^-52) * 2^(e - 1023)` (since it always generates a normal number). `E[(x + 1) >> 1]` exactly equals to 2^-51, and `E[2^e] = 2^1022 * 2^-1 + 2^(1022-1) * 2^-2 + ... + 2^(1022-74) * 2^-75 + 2^(1022-75) * 2^-75 = 2^1023 (1/4^1 + 1/4^2 + ... + 1/4^75 + 0.5/4^75) = 2^1023 (1/3 + 1/(6*4^75))`. So `E[d] = (1 + 2^-51 * 2^52) * (1/3 + 1/(6*4^75)) = 1/2 + 1/4^76`.