In an election there are two competing candidates. Suppose that there are $G$ groups of voters, and that in each group there are $V$ voters (where $G\geq 2$ and $V\geq 3$). Provided that one candidate has very accurate information from polls and gerrymanders, what is the least number of favorable voters that such candidate needs to win the election? What is roughly the percentage of favorable voters that the candidate needs, if $G$ and $V$ are very large?
Let
$g$ be the smallest integer that is larger than $G/2$
$v$ be the smallest integer that is larger than $V/2$ .
We will prove that the number of needed favorable votes is $$gv$$ unless both $G$ and $V$ are even, in which case the number of needed favorable votes is $$gv-1$$
Then the requested percentage is greater than $25\%$: in the former case, this is because
$$gv>(G/2)(V/2)$$
while in the latter case this is because
$$gv-1=(G/2+1)(V/2+1)-1>(G/2)(V/2)$$
If $G$ and $V$ are very large, then $g$ is roughly $G/2$ and $v$ is roughly $V/2$ and $gv-1$ is roughly $gv$: we deduce that the requested percentage is close to $25\%$.
Suppose that $V$ is odd. No tie in a group is possible.
In this case, $g$ is the minimal number of groups needed to win the election, and $v$ is the minimal number of votes needed to win a group.
Then the minimum number of votes needed to win the election is $gv$.
Now suppose that $V$ is even. A tie in a group is possible: to win in a group, a candidate needs $V/2+1$ favorable votes, and to have a tie in a group, $V/2$ votes.
Suppose that candidate A wins in $w$ groups and gets a tie in $t$ groups. Then candidate B gets a tie in $t$ groups and wins in $G-w-t$ groups. Candidate A then wins if $w>G-w-t$, which means that the requested condition is $$2w+t>G\,.$$
Suppose that $G$ is odd. Then (up to decreasing $t$ by $1$) we may suppose that $t$ is even. If $t=0$, we can reason as in the previous case and find that a candidate precisely needs $gv$ votes to win the election. If $t\geq 2$, we can decrease $t$ by $2$ and increase $w$ by $1$ sparing on votes, so we can exclude this case.
Suppose that $G$ is even. Then $t=0$ or (up to decreasing $t$ by $1$) we may suppose that $t$ is odd. With $t=1$ we need one vote less than with $t=0$, namely $gv-1$ votes (because, in case $t=1$, decreasing $t$ by $1$ implies that the minimal $w$ must increase by $1$). If $t$ is odd and larger than $1$, then decreasing $t$ by $2$ and increasing $w$ by $1$ the candidate need less votes, so this case can be discarded.
In an election with $N$ voters (where $N\geq 3$) there are two candidates. Suppose that one candidate can gerrymander in an extreme way, meaning choosing the number of groups of voters and the number of voters that are in each group (beyond choosing which voters belong to which group). How many favorable votes does this candidate need to win the election with such extreme gerrymandering?
One favorable vote is not sufficient because with only one favorable voter the candidate can win in at most one group, and if there is only one group the candidate would not be able to have the majority in that group.
Two favorable votes are sufficient (independently of how large $N$ is). Indeed, the candidate can make three groups: two groups consist in one favorable voter each, and the third group contains all of the other voters. Then the candidate wins two out of the three groups and hence wins the election.