Aerith rolls a fair die until she gets a roll that is greater than or equal to her previous roll. Find the expected number of times she will roll the die before stopping.
The number of rolls is always at least 2 and at most 7 . For there to be at least $k$ rolls, the first $k-1$ rolls have to be distinct and in decreasing order. The number of ways this can happen is $\left(\begin{array}{c}6 \\ k-1\end{array}\right)$, and the total number of ways to have $k-1$ rolls is $6^{k-1}$. Thus, the expected value is

$$
\begin{aligned}
E(\# \text { of rolls }) & =P(\geq 1 \text { roll })+P(\geq 2 \text { rolls })+\cdots+P(\geq 7 \text { rolls }) \\
& =1+\frac{\left(\begin{array}{c}
6 \\
1
\end{array}\right)}{6}+\frac{\left(\begin{array}{c}
6 \\
2
\end{array}\right)}{6^{2}}+\frac{\left(\begin{array}{c}
6 \\
3
\end{array}\right)}{6^{3}}+\frac{\left(\begin{array}{c}
6 \\
4
\end{array}\right)}{6^{4}}+\frac{\left(\begin{array}{c}
6 \\
5
\end{array}\right)}{6^{5}}+\frac{\left(\begin{array}{c}
6 \\
6
\end{array}\right)}{6^{6}} .
\end{aligned}
$$

By the binomial theorem, this is $\left(1+\frac{1}{6}\right)^{6} \approx 2.52$.