Probability Dice Game Paul, Dave and Sarah are rolling a fair six sided die. Paul will go first, always followed by Dave, who is always followed by Sarah, who is always followed by Paul, and so on...
What is the probability that Sarah will be the first one to roll a six?
Now I know the probability of Sarah rolling a six first on the first sequence would be
(5/6)(5/6)(1/6)
But what would be the probability of her rolling a six, if she failed to on the first sequence, the second sequence.
This lead to me finding the equation
(((5/6)(5/6)(5/6))^s)((5/6)(5/6)(1/6))
The first part being the number of sequences, s, she did not roll a six and the second part being when she finally rolls a six.
So I am doing this right?
Could someone help me figure out what to do next, or if I am completely wrong point me in the right direction?
 A: Hint.  There are many ways in which Sarah could be the first to roll a six:


*

*Paul doesn't, Dave doesn't, Sarah does;


OR


*

*Paul doesn't, Dave doesn't, Sarah doesn't, Paul doesn't, Dave doesn't, Sarah does;


OR


*

*Paul doesn't, Dave doesn't, Sarah doesn't, Paul doesn't, Dave doesn't, Sarah doesn't, Paul doesn't, Dave doesn't, Sarah does;


OR etc.  I think you can work out the probability of each case - you have pretty much done that already.  Now how do you combine all the different cases to get your final answer?
A: Let $p$ be the probability that Sarah (ultimately) wins. This can happen in two ways:
(i) Paul and Dave strike out on their first tosses, and Sarah gets a $6$. This has probability $\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{1}{6}$.
(ii) Paul, Dave, and Sarah strike out on their first tosses, but Sarah ultimately wins. Since after the three failures, the game in essence starts again, this has probability $\left(\frac{5}{6}\right)^3p$.
Thus
$$p=\frac{5^2}{6^3}+\frac{5^3}{6^3}p.$$
we have a linear equation for $p$. Solve. After some simplification we get $\frac{5^2}{6^3-5^3}$. 
A: The probability that Sarah is the first to roll a $6$ is $\frac{25}{91}$. Think it this way: Suppose $n$ people roll a die, one by one, in a fixed order, until someone gets a $6$. Let $P_{ij}$ be the probability that player $i$ gets that first $6$ at round $j$, we have:
· Round $1$:
$P_{11} = \frac{1}{6}, \, P_{21} = \frac{1}{6}\left(\frac{5}{6}\right), \, P_{31} = \frac{1}{6}\left(\frac{5}{6}\right)^{2}, \, \ldots \,, \, P_{n1} = \frac{1}{6}{\left(\frac{5}{6}\right)}^{n-1}$
· Round $2$:
$P_{12} = {\left(\frac{5}{6}\right)}^{n}\frac{1}{6}, \, P_{22} = {\left(\frac{5}{6}\right)}^{n+1}\frac{1}{6}, \, P_{32} = {\left(\frac{5}{6}\right)}^{n+2}\frac{1}{6}, \, \ldots \,, \, P_{n2} = {\left(\frac{5}{6}\right)}^{2n-1}\frac{1}{6}$
· Round $3$:
$P_{13} = {\left(\frac{5}{6}\right)}^{2n}\frac{1}{6}, \, P_{23} = {\left(\frac{5}{6}\right)}^{2n+1}\frac{1}{6}, \, P_{33} = {\left(\frac{5}{6}\right)}^{2n+2}\frac{1}{6}, \, \ldots \,, \, P_{n3} = {\left(\frac{5}{6}\right)}^{3n-1}\frac{1}{6}$
· Round $k$:
$P_{1k} = {\left(\frac{5}{6}\right)}^{(k-1)n}\frac{1}{6}, \, P_{2k} = {\left(\frac{5}{6}\right)}^{(k-1)n+1}\frac{1}{6}, \, P_{3k} = {\left(\frac{5}{6}\right)}^{(k-1)n+2}\frac{1}{6}, \, \ldots \,, \, P_{nk} = {\left(\frac{5}{6}\right)}^{kn-1}\frac{1}{6}$
So $P_{ij} = \frac{1}{6}{\left(\frac{5}{6}\right)}^{(j-1)n + (i-1)}$. The probability that player $i$ gets the first $6$ is the sum of the probabilities over each of his turns: $P_i = \displaystyle \sum_{j = 1}^{\infty} \frac{1}{6}{\left(\frac{5}{6}\right)}^{(j-1)n + (i-1)} = \frac{{\left(\frac{5}{6}\right)}^{i-1}}{6\left(1 - {\left(\frac{5}{6}\right)}^{n}\right)}$
In your particular case, $n=3$ and Sarah is the third player so $i=3$.
