I am thinking about the math of two six sided dice, using a certain strategy, I would like to know the probability of a certain loosing event defined below.
Definitions Probability of A = rolling a 10 = 3/36 = 0.08333 Probability of B = rolling a 7 = 6/36 = 0.16666 Probability of C = rolling a 4,5,6,8,9 = 21/36 = 0.58533333
Probability of A before B = .333333
Probability of C before A = 0.875
To loose, a 10 must roll 5 times before a 7, but also if I roll a 4,5,6,8,9 seven times before a 10, OR a 4,5,6,8,9 14 times before two 10's roll, it wins (in addition to anytime a 7 is rolled as mentioned).
So what is the probability of loosing? How many times could you expect it to happen in 1000 rolls?
This is what I think I know. Probability of event A happening 5 times before event B = .333333$^5$ = .00412. Probability of event C happening 7 times before event A = .875$^7$ = .393
I don't know how to combine these two events, or figure in the third scenario (event C 14 times before event A twice or event B once).