Puzzling Dice Roll Probability Problem 
Mark and Jacob are taking turns rolling a fair die. Mark rolls first. What is the probability that Mark will roll an odd number before Jacob rolls a $4$?

This just has me stumped... I'm not too familiar with summations, so simple answers are appreciated
 A: We can think of this as a game where Mark wins (and the game stops) if he rolls an odd number, and Jacob wins if he rolls a $4$. 
A good way to approach these sort of problems is as follows. Let $p$ be the probability that Mark wins. What happens after one turn for each player? Either Mark wins immediately (with probability $\frac12$) or Mark doesn't win and then Jacob does (probability $\frac12\times\frac16=\frac1{12}$), or neither of them win (probability $\frac5{12}$). Now if neither of them win on their first turn, we are back at Mark's turn to throw, and this is the same situation we started in, so the probability of Mark winning from here is still $p$. Thus
$$p=\frac12+\frac5{12}p,$$
which can be easily solved for $p$.
A: Hint: what is the probability of Mark rolling only even numbers until Jacob rolls a $4$, assuming he rolls the $4$ for the first time in the $n^{th}$ try? And what is the probability of Jacob rolling a $4$ for the first time in the $n^{th}$ try?
With this you should get a convergent series whose value is $1-P$, where $P$ is the probability that you are looking for.
More explicitly:


*

*Probability of Mark rolling only even numbers until Jacob rolls a 4 is $(\frac{1}{2})^{n}$, assuming Jacob does so for the first time in the nth try.

*Probabiligy of Jacob doing so for the first time only in the nth try is $(\frac{5}{6})^{n-1}\frac{1}{6}$.

*Probability of Mark rolling only even numbers until Jacob rolls a 4 for the first time is therefore the following sum (because the possibilities are disjoint, as Jacob cannot roll a 4 for the first time both in the third and in the fourth try):
$$ \sum_{i=1}^{\infty}(\frac{1}{2})^{n}(\frac{5}{6})^{n-1}\frac{1}{6}= \frac{1}{6}\sum_{i=1}^{\infty}(\frac{1}{2})^{n}(\frac{5}{6})^{n}\frac{6}{5}=\frac{1}{5}\sum_{i=1}^{\infty}(\frac{5}{12})^{n}=\frac{1}{5}\frac{\frac{5}{12}}{1-\frac{5}{12}}=\frac{1}{7}$$

*Probability of Mark rolling at least one odd number before Jacob rolls a 4 is $1-\frac{1}{7}=\frac{6}{7}$.
