There was a thread on this problem but It was closed and I am still having problems even having read it. 1. The problem statement, all variables and given/known data In successive rolls of a pair of fair dice, what is the probability of getting 2 sevens before 6 even numbers? 2. Relevant equations Let A, B and C be the events that you get a seven, and even number, or neither on a given roll, respectively. Thus P(A) = 6/36 P(B) = 18/36 P(C) = 1 − P(A) − P(B) = 1/3 3. The attempt at a solution This had a solution http://www.math.duke.edu/~wka/math135/until.pdf and there was one at this forum https://www.physicsforums.com/showthread.php?t=656455 but I could not fully understand the answers (doing an independent study to I do not have a teacher I can just ask) UC davis had an answer that made a bit more sense https://www.math.ucdavis.edu/~gravner/MAT135A/hw4.pdf but still fell short of a coherent explanation. I'll address my problem with that first. I get why we subtract the first part from one. probability 6 evens 0 sevens plus probability 1 seven exactly multiplied by 6 evens ALL subtracted from one. But shouldn't that be .5^6 for both of them? The duke one was even more troublesome for this part = 1 P(A) + 0 P(B) + P(E0)P(C) C should be 1/3* probability I roll j number of evens on the first roll given l neither evens nor 7 on the first roll? Why do I treat it as though C is unknown and as though (E0lC)(C) is the same as E0 I was at a complete loss for the second = 0 P(A) + P(Ej−1)P(B) + P(Ej )P(C) why is P(Ej |even on the first roll)P(even on the first roll) equal to P(Ej−1)P(B)...so I factor out the first even? what are the steps that take (1 − P(C))P(Ej ) = P(Ej−1)P(B) to the next part? Why is it all to the j power? I think i might be missing some rules of Geometric series. Or something.