# Examining bias in non-majoritarian random pair

One of the nice features of using random ballot for electing representatives is the strong proportionality property: The expected fraction of parliament made up by some group is the fraction of people who vote for someone from that group.

I’d like to consider how this looks in the following voting system:

1. Pick two random members of the populace
2. Everyone votes for one of them without the possibility of abstention.
3. A candidate wins with probability equal to the fraction of people who voted for them

This isn’t the majoritarian random pair in which the candidate of the two with the majority vote wins. Why? Well partly because it’s just easier to reason about this one, but partly also because hard cutoffs like that create a vastly more biased system.

The proportionality property here is a bit more complicated. There are two factors that determine what the probability of electing someone from a specific group are. We’ll call them $$g$$ and $$e$$. $$g$$ is the probability of a random person belonging to the group (i.e. the fraction of the population who do). $$e$$ is the probability that given a randomly chosen member of that group and a randomly chosen member not of that group a randomly chosen voter, they will vote for the in-group member. Call the probability that a member of the group is elected $$r$$.

$$e$$ might need a bit of elaboration. It basically measures how much the populace likes that group – if $$e = 1$$ then they’ll always vote for that group, if $$e = 0$$ they will never vote for that group, if $$e = \frac{1}{2}$$ they don’t really care whether you’re a member of that group or not.

What is the probability of a member of the group being elected? Well, consider first whether they were picked. If both of the chosen candidates are group members, one must be elected. If neither are then one is not elected, and if it’s mixed with one in-group and one out-group then a group member is elected with probability $$e$$

This means that $$r= g^2 + 2g(1-g)e$$.

Lets start with the $$e = \frac{1}{2}$$ case. This means that the probability is $$g^2 + g – g^2 = g$$. So for any group where there is no popular opinion pro or against the group, you get representation proportional to its presence in the populace.

By considering $$e = 0$$ and $$e = 1$$ we can get some general bounds: $$g^2 \leq r \leq 1 – (1 – g)^2 = 2g – g^2$$. This means that no matter how much everyone else hates them, any group with a reasonable representation in the populace will get representation. For comparison, if we’re trying to elect a house of $$650$$ out of a populace of $$63$$ million as in the UK, any group with about 250 thousand members expects to get a representative regardless of how much people hate them.

An example where this is relevant is race. Currently just shy of 80% of the UK population are white. Unfortunately just shy of 96% of our MPs are white. The lower bound this would put on the percentage of non-white MPs (in probability anyway) is 4% (this is suspiciously close to the actual percentage, but that’s just a coincidence). So even in the circumstance where our country was so profoundly racist that every single one of us only ever voted for white people (I’d like to think we’re a bit better than this) we would still not be doing worse than the status quo. This is however an example of this system doing worse than sortition – a sortition will elect in proportion to the population, regardless of the prevailing biases.

This lower bound is probably not realistic. In general it seems unlikely that any group will be so hated that no-one will vote for them, because when this happens that group will tend to close ranks and vote for in-group members. So the more common lower bound is likely to be that $$e = g$$. This gives us $$r = g^2 + 2g^2(1-g) = 3g^2 – 2g^3$$. At the $$g = 0.2$$ mark this gives us about $$10\%$$ of the house. For much smaller $$g$$, which will correspond to fringe groups (e.g. racist groups like the EDL), $$g^3$$ is small enough as to be negligible this doubles the base probability. It’s also worth noting that this sort of effect means that although the lower bound makes this better than random ballot in principle, in practice under random ballot many people would likely be voting in-group and thus get much the same benefit.

Gender is another interesting example to consider. If we consider the $$g^2$$ lower bound, that should give us at least a quarter of our representatives being women. Sadly, that’s actually slightly better than our current representation of women in parliament – it gives us an expected number of $$162.5$$ compared to our current $$146$$.

And this is a worst-case scenario: It’s only that bad if every single person will always vote for a man over a woman (which is obviously untrue given that we’ve managed to elect 146 female candidates under a first past the post system). With $$g = \frac{1}{2}$$ we have $$r = \frac{1}{4} + \frac{1}{2}e$$. it’s almost impossible to know what an accurate value for $$e$$ is, but if we take $$e = \frac{1}{4}$$ as a pessimistic lower bound this gives us $$r = \frac{3}{8}$$, which is surprisingly not bad.

Another case worth considering is what you do when a group is really popular outside of all proportion to its membership. For example, if you live in the top 1% of wealth distribution and want electing you can get yourself a pretty decent chance of winning the election against someone who isn’t just by throwing money at advertising and PR (this isn’t strictly true, but it’s about the worst case scenario). The result of this is that they get… about 2% of the seats. That’s not so bad. I had trouble finding salaries for MPs prior to their elections (post election their salary puts them in the top 5%), but I bet a lot more than 13 of them broke £150k. I would also expect this bias to be much stronger in a random ballot system where the set of candidates was unrestricted.

So where am I going with this? I’m not really sure. I just wanted to see how heavily this was distorted compared to a sortition. The conclusion seems to be that it’s bad but not too bad – in biasing the sortition towards what society thinks makes a good candidate we’re naturally biasing towards society’s prejudices, but the randomization of the candidates puts bounds on how badly it can do that. It seems like this might actually be a nice middle ground between sortition and random ballot.

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