Agreed. I think the distribution would be like this (using the fact that the infinite sum of reciprocal powers of 2 is 1, and assuming steady-state population)

Agreed. I think the distribution would be like this (using the fact that the infinite sum of reciprocal powers of 2 is 1, and assuming steady-state population): In [1]: print('\n'.join([f'(age={100+i}, number={573000/2**(i+1):.3g})' for i in range(23)])) (age=100, number=2.86e+05) (age=108, number=1.12e+03) (age=116, number=4.37) (age=101, number=1.43e+05) (age=109, number=560) (age=117, number=2.19) (age=102, number=7.16e+04) (age=110, number=280) (age=118, number=1.09) (age=103, number=3.58e+04) (age=111, number=140) (age=119, number=0.546) (age=104, number=1.79e+04) (age=112, number=69.9) (age=120, number=0.273) (age=105, number=8.95e+03) (age=113, number=35) (age=121, number=0.137) (age=106, number=4.48e+03) (age=114, number=17.5) (age=122, number=0.0683) (age=107, number=2.24e+03) (age=115, number=8.74) I.e. the model predicts that on average if there are 573000 centenarians, ~1 is 118, and the probability of reaching 122 is .0683/573000 ≈ 1.19×10⁻⁷. The current real numbers [1] are 11 at 114, 1 at 115, 2 at 116, and 1 at 117 and 118, which are all a bit lower than the model, but not hugely. Surprisingly different from munchler's numbers though, so I may have something wrong. [1] https://en.wikipedia.org/wiki/List_of_the_oldest_living_people https://en.wikipedia.org/wiki/List_of_the_oldest_living_peop...