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 $\frac{4}{36}+\frac{26}{36}\frac{4}{36} + ...\left( \frac{26}{36} \right)^n\frac{4}{36} = \frac{1}{9} \sum \limits_{i=0}^n \left( \frac{26}{36} \right)^i = \frac{1}{9} \frac{36}{10} = \frac{2}{5}$ ?
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.