Dice probability where sum of seven doesn't precede sum of eight. Given the question: 
A pair of fair dice is rolled until the first sum of 8 appears. What is the probability that a sum of seven doesn't precede sum of 8?
My solution is:
P(sum of 7 doesn't precede 1st sum of 8) = 1 - P(sum of 7 precedes 1st sum of 8) $= 1 - \left(\frac{6}{36}\right)\left(\frac{5}{36}\right) = \frac{211}{216}$ 
This solution seems right to me but I thought why can't it be solved using conditional probability: 1 - P(first sum of 8 comes given sum of seven is rolled) = 1 - P(8|7) = 1 - P(8 and 7)/P(7) =  1 - (5/36)(6/36)/6(/36) = 1 - (5/36) = 31/36 or is that just for 8 after 7 on any roll following 7, that is not necessarily 8 right after 7?
 A: Sums other than $7$ or $8$ are irrelevant: if we do not record anything that is not a $7$ or $8$, the game ends in one "step." The ordinary probability of a $7$ is $\frac{6}{38}$, and the probability of an $8$ is $\frac{5}{36}$. Thus if we record only a $7$ or $8$, the probability of $8$ is $\frac{5}{11}$.
More formally, the probability that a toss is an $8$ given that it is a $7$ or an $8$ is, by the usual formula for conditional probabilities, equal to $\frac{\frac{5}{36}}{\frac{6}{36}+\frac{5}{36}}$, that is, $\frac{5}{11}$. 
A: You'll need to account for all possible numbers of rolls.
The probability that you roll anything but $7$ or $8$, $n$ times in a row, is $(\frac{25}{36})^n.$
The probability that you roll an $8$ is $\frac{5}{36}$.
So the total probability is
$$\sum_{k=0}^{\infty}\left(\frac{25}{36}\right)^{k}\left(\frac{5}{36}\right)=\frac{5}{36}\sum_{k=0}^{\infty}\left(\frac{25}{36}\right)^k = \frac{5}{36}\frac{36}{11} = \frac{5}{11}.$$
A: Let the chance getting a 7 be q, the chance getting an 8 be p
The probability is $p+(1-p-q)p+(1-p-q)^2p+...=\frac{p}{p+q}$
What are $p$ and $q$? Do you see why this is the answer?
Final answer: $5/11$
