One of the issues that often gets cited as an objection to random ballot is that it’s too random. What happens if a lunatic fringe gets voted in to power? People might get a majority simply by random chance.

This is just a brief note on how unlikely that is.

Suppose there are \(N\) seats, and we have a party with a fraction \(q\) of the vote. Let \(W_i\) be a random variable which is 1 if the party wins seat \(i\) and 0 if it loses. Let \(W = \sum W_i\) be the number of votes won and \(q_i = P(W_i = 1)\).

\(W\) is a sum of independent random variables. I’m going to assume that \(N\) is large enough that we can apply the law of large numbers and approximate it as a normal random variable.

So from now on we’ll pretend \[W \sim \mathrm{Normal}(qN, \sigma^2)\]

Then the probability of getting a majority is
\[
\begin{align*}
P(W > \frac{1}{2}N) & = P(Z > \frac{(\frac{1}{2} – q)N}{\sigma})\\
&= 1 – \mathrm{erf}\left((\frac{1}{2} – q)\sqrt{\frac{N}{q(1-q)}} \right) \\
\end{align*}\]

erf is monotonic increasing, so this value is maximized by making the argument as small as possible. Everything in there other than \(\sigma\) is fixed, so this is done by maximizing \(\sigma\), or equivalently \(\sigma^2\)

When does this happen?

Well \(W\) is a sum of independent random variables, so its variance is the sum of their variances. So \[\sigma^2 = \sum q_i(1 – q_i)\]

We want to maximize this subject to the constraint that \(\sum q_i = qN\). We can turn this constrained optimisation problem into an unconstrained one by adding a lagrange multiplier t. So we’re optimising the function \[f(q_i, t) = \sum q_i(1 – q_i) + t (\sum q_i – qN)\]

Taking partial derivatives we have

\[\frac{\partial f}{\partial q_i} = 1 – 2 q_i + t\]

Setting this to 0 we get

\[ q_i = \frac{1 – t}{2} \]

The significant part of this being that it’s a constant that doesn’t depend on i. So applying the constraint we have \(q = q_i\), and \(\sigma^2 = N q (1 – q)\).

(I think it’s probably possible to reach this conclusion without the normality assumption, but I haven’t thought through the details).

Thus
\[
\begin{align*}
P(\mathrm{majority}) & = 1 – \mathrm{erf}\left(\frac{(\frac{1}{2} – q)N}{\sigma}\right)\\
& = 1 – \mathrm{erf}\left((\frac{1}{2} – q)\sqrt{\frac{N}{q(1-q)}}\right)\\
\end{align*}
\]

At this point we’ll just plug some numbers in. Here’s some python code:

import math
from scipy.special import erf
 
def probability_of_majority(n,p):
    sd = math.sqrt(n * p * (1 - p))
    deviations_to_majority = n * (0.5 - p) / sd
    return 1 - erf(deviations_to_majority)

At this point let’s just plug some numbers in.

Consider the UK house of commons with its 650 seats. Suppose we have 40% of the vote. What are the chances of us getting a majority?

Plugging them into our python code we get that the answer is 1.84e-13. i.e. if we ran a general election 5 times a year between now and the point when homo sapiens first started to appear, it would still be extremely surprising if any parties with 40% of the vote had ever achieved a majority.

At 45% of the vote we have a probability of 0.00029. If we ran an annual election every year since the birth of Christ, it’d still be more likely than not this would never have happened but it wouldn’t be very surprising.

At 48% of the vote we’re up to 0.15. At this point we expect this to happen every few general elections.

So the lunatic fringe? They’re not going anywhere. The lunatic majority? Well, that’s what you get in a Democracy. But unless someone actually has a majority they’re not getting in.

Compare this to the percentage of the popular vote with which parties typically reach a majority in the UK under a deterministic system.

From this I can only conclude that deterministic voting (in a constituency based system) is simply too random to be practical.