Jamie rolls a die multiple times. find the probability that she rolls her first 5 before she rolls her second even number Jamie rolls her fair 6-sided die multiple times. 
Find the probability that she rolls her first 5 before she rolls her second (not necessarily distinct) even number?
This is what I have so far...
the probability that she rolls a 5 is $\frac{1}{6}$
the probability that she rolls an even number if $\frac{1}{2}$
Now I'm stuck...can anyone help me get to the final product? please be clear and concise.
 A: First, we compute the probability that Jamie rolls a 5 before she rolls an even number. Let this probability equal $q$.
We can write down an equation involving $q$ by considering the different things that could happen on Jamie's first roll. If she rolls a $5$ on her first roll, the game ends and she "wins"; this happens with probability $1/6$. If she rolls an even number on her first roll, the game ends and she "loses"; this happens with probability $1/2$. Otherwise, the game restarts. Thus, $$q = \frac{1}{6} \cdot 1 + \frac{1}{2} \cdot 0 + \frac{1}{3} q \implies q = \frac{1}{4}$$
Now, let $p$ be the probability that Jamie rolls a 5 before she rolls her second even number. Again, we look at cases. If Jamie rolls a 1 or 3, the game resets; this happens with probability $1/3$. If she rolls an even number, then we're reduced to the game that Jamie wins with probability $q$; this happens $1/2$ of the time. And last, if she rolls a 5, then she just wins, and the game ends. So, $$p = \frac{1}{3} p + \frac{1}{2} q + \frac{1}{6} \implies p = \frac{7}{16}$$
A: Hints:
It's probably easier to find the probability of the complement: two even numbers before a $5$.
Consider rolling until she gets an even number or a $5$ (so ignore any $1$s or $3$s).  Since there are three possible even numbers and only one $5$, what is the probability that she gets an even before a $5$?
Then what's the probability that she gets two evens before a $5$?
A: You roll the dice until you roll an even number or a five. Then you write down the letter E or a 5. You continue until you roll another even number or a five. Again, you write down the letter E or a 5. 
Everytime you write something down, the probability is 3/4 that you write an E, and 1/4 that you write a 5. The chance that you wrote down two Es is 9/16. The probability that you didn't write two Es is 7/16, and that's what we were asked about. 
A: Consider the following two events:


*

*$A:=\left\{\mbox{sequence of rolls containing one or no even number, and ending in a 5}\right\}$

*$B:=\left\{\mbox{sequence of rolls containing some even number}\right\}$


The probability of interest, in a sequence of independent dice rolls, is
$$
P\left(A \mbox{ followed by } B\right){}={}P\left(B\, |\, A\right)P\left(A\right)\,.
$$
Note that, because the rolls are independent and the event $B$ is an almost-sure event (that is, the probability of it not occurring is zero), we have
$$
P\left( B \, | \, A\right){}={}P\left( B \right){}={} 1\,.
$$
So, 
$$
P\left(A \mbox{ followed by } B\right){}={}P\left(A\right)\,.
$$
We proceed, therefore, to compute the event $A$ as follows (making liberal use of independence):
$$
\begin{eqnarray*}
P\left(A\right)&{}={}&\sum\limits_{k=1}^{\infty} \left(P\left(k\mbox{ rolls, no even, ending in }5\right) {}+{} P\left(k\mbox{ rolls, one even, ending in }5\right)\right) \newline
&{}={}&\sum\limits_{k=1}^{\infty} \left(\frac{1}{6}\left(\frac{1}{3}\right)^{k-1}{}+{}\frac{1}{6}{k-1\choose 1}\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)^{k-2}\right) \newline
&{}={}&\frac{1}{6}\sum\limits_{k=1}^{\infty} \left(\left(\frac{1}{3}\right)^{k-1}{}+{}\frac{1}{2}\left(k-1\right)\left(\frac{1}{3}\right)^{k-2}\right) \newline
&{}={}&\frac{1}{6}\left(\frac{1}{\left(1-\frac{1}{3}\right)}{}+{}\frac{1}{2\left(1-\frac{1}{3}\right)^2}\right)\newline
&{}={}&\frac{7}{16}\,.
\end{eqnarray*}
$$   
A: I just noticed that this follows directly from paw88789's excellent hint. However, just to get attention to that hint, perhaps we need to see how nicely it works:
Given that the only rolls that count are $2,4,5,6$, what is the probability that two evens are rolled first? $\left(\frac34\right)^2=\frac9{16}$. Thus, the probability that the $5$ comes before the second even face is
$$
1-\frac9{16}=\frac7{16}
$$
