Bob has five airplane tickets with prices $\$ 100, \$ 120, \$ 140, \$ 160$, and $\$ 180$. Bob gives an offer to Aerith: she can distribute his tickets among two bags, after which, without looking inside, Bob will randomly choose a bag and a ticket from it for Aerith to keep. What strategy should Aerith use to maximize the expected value of her ticket?
Let the first bag be the one with less tickets. Let $a<b$ and $S, T$ be the numbers and prices of tickets in the respective bags. We have $a+b=5$ and $S+T=\$ 700$. The expected values of random tickets for each bag is then $S / a$ and $T / b$, the average of which is

$$
\frac{S / a+T / b}{2}=\frac{b S+a T}{2 a b}=\frac{(b-a) S+a \cdot \$ 700}{2 a b} .
$$

For fixed $a<b$, this is maximized when $S$ is a large as possible. Thus if $(a, b)=(1,4)$, the first bag then should contain $\$ 180$, whereas for $(a, b)=(2,3)$, it should contain $\$ 180+\$ 160=\$ 340$.

Using the first strategy, she has an expected value of $\frac{\$ 180 / 1+\$(700-180) / 4}{2}=\$ 155$, and using the second, she has an expected value of $\frac{\$ 340 / 2+\$(700-340) / 3}{2}=\$ 145$, so the best strategy is the first strategy: put the $\$ 180$ ticket in one bag and the rest in another.