Does binomial distribution have the smallest possible variance among all "reasonable" distributions that can model binary elections? Imagine an election where $n$ people make a binary choice: they vote for A or against it. The outcome is that $m$ people vote for A, and so A's result is $p=m/n$.
If I want to model these elections, I can assume that each person votes for A independently with probability $p$, leading to the binomial distribution of votes: $$\text{votes for A}\sim\mathsf{Binom}(n,p).$$ This distribution has mean $m=np$ and variance $np(1-p)$.
I can make other assumptions as well. For example, I can assume that probability $p$ is itself a random variable coming from some distribution (e.g. beta); this can lead to a beta-binomial distribution of votes for A. Or I can assume that people vote in groups of $k$, where each group of $k$ people makes the same choice and it is A with probability $p$. This will lead to a binomial distribution with larger variance. In all these cases, variance of the resulting distribution is larger than in the simplest binomial scheme.
Can I make a claim that binomial distribution has the smallest possible variance? In other words, can this claim be somehow made precise, e.g. by specifying some reasonable conditions on the possible distributions? What would these conditions be?
Or is there maybe some reasonable distribution that has lower variance?
I can imagine lower variance, e.g. when all $n$ people agree in advance on how they will vote, and so $\text{votes for A}$ is not really a random variable, but a fixed number $m$. Then the variance is zero. Or maybe almost all of them agreed but a few people did not, and then one can have tiny variance around $m$. But this feels like cheating. Can one have smaller-than-binomial variance without any prearrangements, i.e. when each person votes in some sense randomly?
 A: Double-no (it maximises the variance)
The answer from whuber is excellent.  To supplement that answer, it is also worth examining what happens if you assume that the votes are independent.  If we take the votes as mutually independent with probabilities $p_1,...,p_n$ then the mean and variance are:
$$\mathbb{E}(S_n) = \sum_{i=1}^n p_i
\quad \quad \quad \quad \quad 
\mathbb{V}(S_n) = \sum_{i=1}^n p_i - \sum_{i=1}^n p_i^2.$$
If we condition on a fixed expected value $\mu = \mathbb{E}(S_n)$ then it can be shown that the maximum variance is achieved when $p_1 = \cdots = p_n = \mu$.  (To demonstrate this you can set up the Lagrangian optimisation to attain this solution.)  So not only does the binomial distribution not minimise the variance, it maximises the variance out of all possible cases where we have independent votes.
A: No.
Suppose the voters consist of $n=2k$ married pairs.  The husbands get together and decide to vote against their wives, who themselves choose randomly.  The outcome is always $k$ votes for each of the candidates, with zero variance.
You might cry foul because the husbands are not voting randomly. Well, they are--they just happen to be tied closely with the random votes of their wives.  If that bothers you, change things a bit by having each husband flip ten fair coins.  If all ten are heads, he will vote with his wife; otherwise he votes against her.  You can check that the election outcome still has small (albeit nonzero) variance, even though every vote is unpredictable.
The crux of the matter lies in the negative covariance between two voting blocs, males and females.
