User blog comment:Naruyoko/PEGG detailed log/M/@comment-32994025-20190704032006/@comment-39541634-20191216041330

There's actually a way to tell, with relatively good precision, when we'll hit every new letter:

1. The average value of n^2, in the long run, is (12+22+32+42+52+62+72+82+92+102)/10 = 38.5.

2. Each letter (except E and J) have X-values from 2 to 10. That's a required increase of 8. For clarity, let's call this number k. For every letter besides E and J, k=8. For E, k=10 (because it runs from E0 to E10). For J, k=6 (because it runs from J4 to J10).

3. If we give every letter a number (E=1, F=2, G=3, ...) then the Z-value for a given letter is given by: 0.01*0.7m-1 where m is the number assigned to that letter. So the average increase in Z per day is 38.5*0.01*0.7m-1 = 0.385*0.7m-1.

End result:

Te average number of days we would expect to spend on a given letter is: k/(0.385*0.7m-1)

So here is the expected time for each letter (rounded to the nearest integer):