I would have thought this approach would have been highly susceptible to the Birthday Paradox[0]; it seems to me to be nothing more than modern day numerology.
they're using 200 bits. look at the generalized birthday problem section on your cited link--the number of samples for there to be greater than a 50% probability of a collision is > (3-2ln2)/6 + sqrt(2(2^200)ln2) = 1.4910^30. For a probability of 0.01 of a collision you expect at least sqrt((2^200)ln(1/(1-0.01))=1.27*10^29 samples.