09-13-2017, 11:00 PM
with those bitmask probability checks, chance of success = 1 / (2 ^ # of bits set) .
so 1Fh is worse than 0Fh, and (assuming no RNG manipulation or such) 2Fh is the same as 1Fh.
so 1Fh is worse than 0Fh, and (assuming no RNG manipulation or such) 2Fh is the same as 1Fh.

