, 2018

Solution:

Consider the general case of $n$ frogs. Let $E_{i}$ be the expected time for all frogs to enter the pool when $i$ frogs are on the shore and $n-i$ frogs are in the pool. We have $E_{0}=0$, $E_{n}=1+E_{n-1}$, and
$$E_{i} = \frac{i}{n} E_{i-1} + \frac{n-i}{n} E_{i+1} + 1$$
for $0 < i < n$. Define $f_{i}$ so that
$$E_{i} = \frac{f_{i}}{(n-1)(n-2) \cdots (i)} + E_{i-1}$$
Then by plugging this equation into the first equation, we can show that
$$f_{i} = n(n-1) \cdots (i+1) + (n-i) f_{i+1}$$
Furthermore, we know that $f_{n} = 1$. Therefore
$$\begin{aligned} f_{1} & = \sum_{i=1}^{n} \frac{n!}{i!} \frac{(n-1)!}{(n-i)!} \\ & = (n-1)! \sum_{i=1}^{n} \binom{n}{i} \\ & = (n-1)!\left(2^{n}-1\right) . \end{aligned}$$
Therefore
$$E_{1} = \frac{(n-1)!\left(2^{n}-1\right)}{(n-1)!} + E_{0} = 2^{n}-1$$
Plugging in $n=2018$ yields $E_{1} = 2^{2018}-1$.