, 2019

Solution:

Let $P_{n}(m)$ be the probability that after $n$ voters have voted, Alice gets $m$ votes. We show by induction that for $n \geq 3$, the ratio $P_{n}(2): P_{n}(3): \cdots: P_{n}(n-1)$ is equal to $1: 2: \cdots:(n-2)$. We take a base case of $n=3$, for which the claim is obvious. Then suppose the claim holds for $n=k$. Then $P_{k}(m)=\frac{2 m-2}{(k-1)(k-2)}$. Then
$$P_{k+1}(i)=\frac{k-i}{k} P_{k}(i)+\frac{i-1}{k} P_{k}(i-1)=\frac{(k-i)(2 i-2)+(i-1)(2 i-4)}{k(k-1)(k-2)}=\frac{2 i-2}{k(k-1)} .$$
Also, we can check $P_{k+1}(2)=\frac{2}{k(k-1)}$ and $P_{k+1}(k)=\frac{2}{k}$, so indeed the claim holds for $n=k+1$, and thus by induction our claim holds for all $n \geq 3$. The probability that Ceila wins the election is then
$$\frac{\sum_{m=2}^{1009} P_{2019}(m)}{\sum_{m=2}^{2018} P_{2019}(m)}=\frac{1008 \cdot(1+1008) / 2}{2017 \cdot(1+2017) / 2}=\frac{504}{2017}$$
and thus the probability that Alice wins is $\frac{1513}{2017}$.