Introduction to probability Dice We roll a fair die repeatedly until we see the number four appear and then we stop.
What is the probability that we needed an even number of die rolls? 
For this problem I said that it would be the Sum of (5/36)*(25/36)^j and I got an answer of 5/9? Does that seem correct. 
Thanks 
 A: Hint: The probability that it takes exactly $k$ rolls is
$$
\frac16\left(\frac56\right)^{k-1}
$$
Add those up for even $k$. Your setup looks okay, but
$$
\begin{align}
\sum_{k=1}^\infty\frac16\left(\frac56\right)^{2k-1}
&=\overbrace{\frac{5}{36}}^{\text{first term}}\quad\overbrace{\frac1{1-\frac{25}{36}}}^{\Large\frac1{1-\text{common ratio}}}\\
&=\frac5{11}
\end{align}
$$

Another Approach: Given that we end either on roll $2n-1$ or on roll $2n$, there are $6$ ways to end on roll $2n-1$: $4$-$1$, $4$-$2$, $4$-$3$, $4$-$4$, $4$-$5$, $4$-$6$; and $5$ ways to end on roll $2n$: $1$-$4$, $2$-$4$, $3$-$4$, $5$-$4$, $6$-$4$. Therefore, no matter what $n$ is, the probability of ending on roll $2n$ would be
$$
\frac5{6+5}=\frac5{11}
$$
A: We want the probability of having an even number of rolls. That means we want an odd number of non-fours, and then a four. So, we could have 1, 3, 5, 7, etc. non-fours, followed by a four. Given that the die is fair, we will assume that $P(4) = \frac{1}{6}$ and that the rolls are independent, so we can multiply when appropriate. If I am not mistaken, this becomes:
$$
\begin{align}
&\frac{1}{6}\times\sum_{i=0}^\infty \left(\frac{5}{6}\right)^{2i + 1}\\
&=\frac{1}{6}\times\sum_{i=0}^\infty \frac{5}{6}\left(\frac{5}{6}\right)^{2i}\\
&=\frac{1}{6}\times\frac{5}{6}\times\sum_{i=0}^\infty \left(\frac{5}{6}\right)^{2i}\\
&=\frac{1}{6}\times\frac{5}{6}\times\sum_{i=0}^\infty \left(\left(\frac{5}{6}\right)^2\right)^i\\
&=\frac{1}{6}\times\frac{5}{6}\times\sum_{i=0}^\infty \left(\frac{25}{36}\right)^i\\
&=\frac{5}{36}\times\frac{1}{1 - \frac{25}{36}}\\
&=\frac{5}{36}\times \frac{36}{11} = \frac{5}{11}
\end{align}
$$
