<s>[INST] 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 
 [/INST] 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:
$$
\begin{align}
&\frac{1}{6}\times\sum_{i=0}^\infty \left(\frac{5}{6}\right)^{2i + 1}\\
&=\frac{1}{6}\times\sum_{i=0}^\infty \frac{5}{6}\left(\frac{5}{6}\right)^{2i}\\
&=\frac{1}{6}\times\frac{5}{6}\times\sum_{i=0}^\infty \left(\frac{5}{6}\right)^{2i}\\
&=\frac{1}{6}\times\frac{5}{6}\times\sum_{i=0}^\infty \left(\left(\frac{5}{6}\right)^2\right)^i\\
&=\frac{1}{6}\times\frac{5}{6}\times\sum_{i=0}^\infty \left(\frac{25}{36}\right)^i\\
&=\frac{5}{36}\times\frac{1}{1 - \frac{25}{36}}\\
&=\frac{5}{36}\times \frac{36}{11} = \frac{5}{11}
\end{align}
$$
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