, 2018

Solution:

For each $n, 1 \leq n \leq 20$, consider the first time that Fran chooses one of the multiples of $n$. It is in this move that $n$ is erased, and all the multiples of $n$ at most $20$ are equally likely to be chosen for this move. Hence this is the only move in which Fran could possibly choose $n$; since there are $\lfloor 20 / n\rfloor$ multiples of $n$ at most $20$, this means that the probability that $n$ is ever chosen is $1 / \lfloor 20 / n\rfloor$. Therefore the expected number of moves is
$$\begin{aligned} E & = \sum_{n=1}^{20} \frac{1}{\lfloor 20 / n\rfloor} \\ & = \frac{1}{20} + \frac{1}{10} + \frac{1}{6} + \frac{1}{5} + \frac{1}{4} + \frac{1}{3} + 4\left(\frac{1}{2}\right) + 10(1) = \frac{131}{10} . \end{aligned}$$
(This sum is easier to compute than it may seem, if one notes that $1 / 20 + 1 / 5 + 1 / 4 = 1 / 2$ and $1 / 6 + 1 / 3 = 1 / 2$.)