4 ms·

You may have missed the fact that not all these 573K living centenarians are 100 year olds, only half of them are, the rest are already 101 or more, and of thos

• mkl 5y ago
  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...