Nonsense. It requires roughly 1+log(10) bits in expectation and has a geometric tail.
Thus, I recant "around 20, 40, 60 or more" and replace it by "around 9, 18, 27 or more" numbers. Probability that you have to reject any resulting hex digit is obviously 3/8.
With b bits you are subdividing the (0,1) interval into 2^b regions, and only need more bits if one of the 9 multiples of 0.1 land in the interval you've picked. As b increases this probability drops exponentially.
It's a fair point that the number of "numbers" depends on how low the entropy of the source is, but in the link the probability of collision wasn't massive.
The OP is silly though; if you know the distribution there are way better techniques. They don't seem to worry about how they measure it.