o = rolling a 1 or 3 (1/3 probability)
e = rolling a 2, 4, or 6 (1/2 probability)
f = rolling a 5 (1/6 probability)
So we have 3 cases. Rolling a bunch of "o"s before rolling an f. Rolling a bunch of "o"s and 1 "e" before rolling an f. Rolling an f.
Rolling a bunch of "o"s before rolling an f.
((1/3) * 1/6) + ((1/3)^2 * 1/6) + ((1/3)^3 * 1/6).... + ((1/3)^infinity * 1/6)
This is a geometric sequence.
a = 1/18, r = 1/3, a/(1-r)
(1/18)/(2/3) = 1/12
Rolling a bunch of "o"s and 1 "e" before rolling an f.
2((1/3) * 1/6) + 3((1/3)^2 * 1/6) + 4((1/3)^3 * 1/6)... + (infinity - 1)((1/3)^infinity * 1/6)
I'm not actually sure how to calculate this, but hopefully someone else can help with that. #It probably is around 1/5 to 1/4.
Rolling an f.
1/6
1/6 + 1/12 + (2((1/3) * 1/6) + 3((1/3)^2 * 1/6) + 4((1/3)^3 * 1/6)... + (infinity - 1)((1/3)^infinity * 1/6))
1/4 + (2((1/3) * 1/6) + 3((1/3)^2 * 1/6) + 4((1/3)^3 * 1/6)... + (infinity - 1)((1/3)^infinity * 1/6))
Sorry I wasn't able to help more. :((
=^._.^=
