Quoted By:
I did some math:
first, there are 64 possible characters for ids (A-Z, a-z, 0-9, +, /)
ok, consider if ids were only 3 characters. Then the first letter would have to be K or k, the second letter could be o, O, or 0, and the third letter could be T or t.
Then the probability of k0t id would be
(2/64 * 3/64 * 2/64) = 0.00004578 (1 in every 21845 ids)
but really we have 8 character ids. so there are 6 sets of 3 consecutive letters (k0t????? through ?????k0t)
so you might think "just multiply by 6!" that's close, but they aren't independent sets (so you mistakenly double-count ids with multiple k0ts)
we want the reciprocal of the probability that *none* of the 6 are k0t
so:
1-((1-(2/64 * 3/64 * 2/64))**6) = 0.0002746 (1 in every 3641 ids)
tldr; k0t id is about 2.25 times MORE LIKELY than encountering a shiny pokemon (1/8192)
