, 2023

Solution:

To get within an order of magnitude, one approach is to let $X_{n}$ be a random variable which takes the value $\pm \sqrt[3]{F_{n}}$, with the sign chosen uniformly at random. We want the probability that $S=\sum_{i=2}^{31} X_{i}$ is in $[-1,1]$. We can attempt to approximate the distribution of $S$ as normal (this is loosely justified because it is the sum of many independent random variables). Using the approximation $F_{n} \approx \frac{1}{\sqrt{5}} \varphi^{n}$ for $\varphi=\frac{1+\sqrt{5}}{2}$, the variance of $S$ is:
$$\begin{aligned} \operatorname{Var}(S) & =\sum_{i=2}^{31} \operatorname{Var}\left(X_{i}\right) \\ & =\sum_{i=2}^{31} F_{i}^{2 / 3} \\ & \approx \sum_{i=2}^{31} 5^{-1 / 3} \varphi^{2 i / 3} \\ & \approx 5^{-1 / 3} \cdot\left(\frac{\varphi^{62 / 3}}{1-\varphi^{-2 / 3}}\right) \end{aligned}$$
Now, we use the fact that if $\frac{1}{\sqrt{\operatorname{Var}(S)}} S$ is standard normal, then the probability that $S \in[-1,1]$ is approximately
$$\frac{1}{\sqrt{2 \pi}} \cdot \frac{2}{\sqrt{\operatorname{Var}(S)}} \approx \sqrt{\frac{2 \cdot 5^{1 / 3}}{\pi}} \cdot \frac{\sqrt{1-\varphi^{-2 / 3}}}{\varphi^{31 / 3}}$$
When we multiply this by $2^{30}$, we get an approximation of $E \approx 4064598$, which achieves $A / E \approx 0.96$ and would score 17 out of 25 points.