Redoing this whole mess, once again I hate math XD.
This is dependent event, 32 faces to roll for specific face, things are rounded to 4, percentage rounded to 2.
Any 2 faces are guaranteed to get, so instead of needing to roll for 3 face I only need 1 face.
If 2 faces that are guaranteed to roll included in the rolling, instead of rolling 4 it is 2(will include the exclude).
3 face(only need 1 face)(first is excluded roll, second included roll);
(31/32)^4 = 0.8807, 1 - 0.8807 = 0.1193 × 100 = 11.93%
(31/32)^2 = 0.9385, 1 - 0.9385 = 0.615 × 100 = 6.15%
4F(2F) using binomial probability...; P(x) = (n/x)p^x q^n-x
n = 4 or 2 being rolled, x = 2 is same face, p = 1/32 successfull, q = 31/32 unsuccessfull
4/2 = 4C2 = 4!/(4-2)!4! = 6, 6(1/32)^2 (31/32)^4-2 = 6×961/1024^2=5766/1,048,576 = 0.55%
2/2 = 2C2 = 2!/(2-2)!2! = 1, 1(1/32)^2 (31/32)^2-2 = 0.098%
With how low the percentage is to achieve 10% you need to roll more successively(8/12/16/20)
8C2 = 28(1/32)^2 (31/32)^8-2 = 2.28%
12C2 = 66(1/32)^2 (31/32)^12-2 = 4.62%
16C2 = 120(1/32)^2 (31/32)^16-2 = 7.71%,
20C2 = 190((1/32)^2 (31/32)^20-2 = 11.13%, so you need to roll 20 times to achieve >10% to obtain 4 same faces, 18 rolls around 9.82%, 19 = somewhere 10%, probably this number....
