Markov chain or another solution? Ok, here's the problem. I am rolling 2 fair dice repeatedly and I want to know what the chance is of rolling both a total of 5 and a total of 9 before rolling a total of 7. Let y equal the probability of a 5 or 9 or 7 being rolled. There are 4 ways each to roll a 5 and 9 and 6 ways to roll a seven for a total of 14/36 combinations. I calculate P(5y) = 1/9 / 7/18 or 1/9 * 18/7 = 18/63 = 2/7 P(9y) must equal P(5y) so also 2/7 P(7y) = 1  (P(5y) + P(9y)) = 3/7 So far so good? Now I'm not sure if I'm on the right track next or not but . . . If I imagine this problem as having 5 states: a) Having made 0 of the numbers yet. b) Having rolled a 5 but neither a 9 nor 7. c) Having rolled a 9 but neither a 5 nor 7. d) Having succeeded in rolling both a 5 and a 9 without the 7 (won). e) Having rolled a 7 and failed (lost). I then made a transition matrix from that data. a b c d e a 0 2/7 2/7 0 3/7 b 0 2/7 0 2/7 3/7 c 0 0 2/7 2/7 3/7 d 0 0 0 1 0 e 0 0 0 0 1 (EXCUSE THE FORMATTING) Since I always start in state 'a' I multiplied an initial distribution vector of [1 0 0 0 0] by the above matrix raised to the power of 50 to try and find an approximation of the steady state probabilities. This told me that P(5 and 9 before 7) = 0.2286 (ish) Anyone make any sense of that and tell me if I got the answer right? If yes, could I have done it in an easier way? If not, what should I have done? Thanks in advance for your help folks! 
This is how I did it. Consider sequences of rolls that lead to a win in $n$ rolls. Note $n \geq 2$ Roll $n$ will either be a $5$ or a $9$ Rolls $1 \text{ through }(n1)$ will not be roll $n$ and will contain no $7$ Now consider roll sequences of length $n1$ that satisfy this. There will be at least $1$, and up to $n1$ rolls that are either $5$ or $9$ whichever isn't roll $n$, and none of these will be $7$. Otherwise they can be any roll. so we end up with $P[\text{win}] = 2 \sum \limits_{n=2}^\infty \dfrac 1 9\left(\sum \limits_{k=1}^{n1}\dbinom{n1}{k}\left(\dfrac 1 9\right)^k\left(\dfrac{11}{18}\right)^{n1k}\right)$ Digesting this a bit. The $2$ in front is because roll $n$ can be either $5$ or $9$, and the probability is identical for both. The $\dfrac 1 9$ is the probability of roll $n$ being the $5$ (or the $9$) What's inside the parens is the sum of probabilities that there are $k$ occurrences of $5$ or $9$ whichever isn't roll $n$ This can be simplified to $P[\text{win}] = 2 \sum\limits_{n=2}^\infty \left(\left(\dfrac{13}{18}\right)^{n1}  \left(\dfrac{11}{18}\right)^{n1}\right)$ and then finally to $P[\text{win}] =\dfrac{8}{35}$ I leave it to you to digest all of this and reproduce the simplifications in more detail. ps. $p=\dfrac {8}{35}$ matches my sim of this very well so I'm pretty confident in this answer. 
THANKYOU!! Exactly what i was hoping to find. And also the answer i had was correct as well :) But a formula like yours is so much better. I will study it in detail tomorrow when i wake up. Thanks for your time and help. Really appreciate it! 
Trying to digest what you did I figured out that there is a lot I still don't know. I have had a look into geometric sequences and that is helping me get started. Are there any other topics you would recommend me learning more about to gain a better grasp of what's going on here? Haha Also, in playing with the problem I found another way to work it out I think. Since we only care about 5,9 and 7. Can I say that P(5 or 9  5,9,7) = 8/14 and then the probability of the next one is p(5  5,7) = 4/10 8/14 * 4/10 = 32/140 = 8/35 again. Think this only works is a case like this though where the P(n) is the same for both. How would your formula above change for a problem like rolling 4 and 5 before 7? 
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