Probability Q on rolling two 8-sided dice Let's say I roll two 8-sided dice. I win if the sums '7' and '11' show up before we see the sum '9' TWICE. What is my probability of winning?
So this is my answer and please correct me if I am wrong:
Prob of getting either 7 or 9 —> 6/64 + 6/64 = 12/64 —> 18.75%
Prob of getting 9 twice —> 8/64 x 8/64 = 1/64 —> 1.6%
But now to find the probability of winning, would it be 18.75 - 1.6 = 17.15% ? Am I right or wrong?
Thanks!
EDIT
From the answers provided below, we have three different results:
84%
49%
60%
Which one would be the correct answer?
 A: This answer assumes the person wins if there is a sum of $\, 7\,$ or $\,11\,$ in any toss before a sum of $\,9\,$ appears twice.
Probability of winning in first toss $ = \frac{12}{64}$
a) If the person gets a sum of $9$ once, the chance of winning from there
$\displaystyle P(W1) = \frac{12}{64} + \frac{44}{64} \times (\frac{12}{64} + \frac{44}{64} \times ...)... = \frac{12}{64} (1 + \frac{44}{64} + (\frac{44}{64})^2 + ...)$
$\displaystyle = \frac{3}{5}$
[Using sum of infinite geometric series $a + ar + ar^2 + ... = \frac{a}{1-r}$]
b) If the first toss is any sum other than $7, 9, 11, \,$starting 2nd toss
$P(W2) = \frac{12}{64} + \frac{8}{64} \times P(W1) + \frac{44}{64} \times (\frac{12}{64} + \frac{8}{64} \times P(W1) + ...)... $
$$= (\frac{12}{64} + \frac{8}{64} \times P(W1))(1 + \frac{44}{64} + (\frac{44}{64})^2 + ...) = \frac{21}{25}$$
$\,$
So person's chance of winning $ \displaystyle = \frac{12}{64} + \frac{8}{64} \times P(W1) + \frac{44}{64} \times P(W2) = \frac{21}{25}$
A: Here is another approach, this time using an exponential generating function.  Readers interested in learning about generating functions can find many resources in the answers to this question: How can I learn about generating functions?
We want to find the probability that a sequence of rolls by two 8-sided dice contains at least one 7 and at least one 11 before it contains two 9s.  For now, let's assume that the final roll is a 7.  Since 11 has the same probability of 7, the probability of ending in 11 is the same as the probability of ending in 7.
Consider a successful sequence of rolls just before rolling the final 7.  Such a sequence must contain at least one 11, no 7, at most one 9, and any number of other rolls (not 7, 9, or 11).  The probability of rolling a 7 is $6/64$, the probability of rolling an 11 is $6/64$, the probability of rolling  9 is $8/64$, and the probability of rolling anything else is $44/64$. So the exponential generating function for the probability of such a sequence is
$$f(x) = (e^{(6/64) x} -1) \;(1+ (8/64)x)\; e^{(44/64)x}$$
Then the next roll must be a 7.  So the EGF for a successful sequence of $n+1$ rolls ending in 7 is
$$g(x) = (e^{(6/64) x} -1) \;(1+ (8/64)x)\; e^{(44/64)x}\;(6/64) \tag{1}$$
That is to say,
$$g(x) = \sum_{n=0}^{\infty} p_n \frac{x^n}{n!}$$
where $p_n$ is the probability of a successful series of $n+1$ rolls ending in 7. We want to know the sum of all the $p_n$'s. The trick here is that
$$\sum_{n=0}^{\infty} p_n = \int_0^{\infty} e^{-x}g(x) \;dx \tag{2}$$
which is based on the identity
$$\int_0^{\infty} e^{-x} x^n \;dx = n!$$
Substituting $(1)$ into equation $(2)$ and evaluating the integral, we find
$$\sum_{n=0}^{\infty} p_n = \frac{621}{2450}$$
This is the probability of a successful sequence of rolls ending in 7. The probability of a successful sequence ending in either 7 or 11 is twice that,
$$\frac{621}{1225}$$
A: Lemma: if you repeat an experiment with probability $P$ for event $A$ and probability $Q$ for event $B$ (constant in each repetition) then the probability that you will see an $A$ before a $B$ is $\tfrac{P}{P+Q}$.
Based on the above lemma, let $A$ be the event of getting 7 or 11, and $B$ the event of getting 9. The desired probability is
$$\tfrac{P}{P+Q}+\tfrac{Q}{P+Q}\tfrac{P}{P+Q}$$
The first term is getting 7 or 11 before the first occurrence of 9. The second term is that the first 9 came before the first 7 or 11, but then the second 9 came after. The idea is that you can after the first 9 restart the rolls, waiting for only one 9.
