Probability number comes up before another In craps, say I roll a 5 for the place bet. What is the probability that I roll another 5 before rolling a 7?
Is this correct?: $P(5 \text{ before 7}) = P(5) + P(\neg 7 \neg 5, 5) + P(\neg 7 \neg 5, \neg 7 \neg 5, 5) + ....$. This becomes $\frac{4}{36}+\frac{26}{36}\frac{4}{36} + ...\left( \frac{26}{36} \right)^n\frac{4}{36} = \frac{1}{9} \sum \limits_{i=0}^n \left( \frac{26}{36} \right)^i = \frac{1}{9} \frac{36}{10} = \frac{2}{5}$ ?
 A: Your notation $\neg 7 \neg 5$ confused me a bit and so I composed a long
answer, but now that I understand what you wrote a little better,
yes, what you have written is correct.
When you roll a 5, that becomes your point and then you repeatedly roll the dice until either your point shows up and you win, or you roll a 7 and you lose. Thus, having established a point of 5, your (conditional) win probability is 
$$P(5)+P(N,5) + P(N,N,5) + \cdots = 
\frac{1}{9} + \frac{13}{18}\times \frac{1}{9}
+ \left(\frac{13}{18}\right)^2\times \frac{1}{9} + \cdots
= \frac{1}{9}\times \frac{1}{1-\frac{13}{18}} = \frac{2}{5}$$
where $N$ is the event that the roll is neither 5 nor 7 (what you
have written as $\neg 7 \neg 5$).
More generally, if $A$ and $B$ are mutually exclusive events,
then on a sequence of independent trials, the probability that
$A$ occurs before $B$ is
$\displaystyle \frac{P(A)}{P(A)+P(B)}.$
See, for example, this answer.
A: Here is an alternative way of using conditional probability, beside the answer via conditional probability by @Dilip Sarwate in another question. (This answer is adapted from the book "An Introduction to Stochastic Modeling" [Section 2.2 The Dice Game Craps] by Pinsky and Karlin)
Consider repeated rolls of the pair of dice and let $Z_n$ ($n \in \mathbb{N}$) be the sum observed on the $n$-th roll.
Denote the event of "$\textrm{5 before 7}$" by $B$ and its probability $\alpha$. 
According to the law of total probability, we obtain
$$\alpha = P \{ B\} = \sum_{k=2}^{12} P \{B \mid Z_1 = k \} P \{ Z_1 = k\}$$
Now we have
$$P \{ B \mid Z_1 = 5 \} = 1, \quad P \{ B \mid Z_1 = 7 \} = 0.$$
If the first roll results in anything other than 5 or 7, the problem is repeated in a statistically identical setting. That is, 
$$P \{ B \mid Z_1 = k \} = \alpha, \forall k \neq 5, 7.$$
Therefore, we have
$$\alpha = P \{ Z_1 = 5 \} \times 1 + P \{ Z_1 = 7 \} \times 0 + \sum_{k \neq 5,7} P \{ Z_1 = k \} \times \alpha \\ = P \{ Z_1 = 5 \} + [1 - P \{ Z_1 = 5 \} - P \{ Z_1 = 7 \}] \alpha$$
Solving this equation, we get
$$\alpha = \frac{P \{ Z_1 = 5 \}}{P \{ Z_1 = 5 \} + P \{ Z_1 = 7 \}} = \frac{\frac{4}{36}}{\frac{4}{36} + \frac{6}{36}} = \frac{2}{5}.$$
