I like randomized voting. I like majority judgement. Naturally I’ve wondered for a while whether there is a natural randomized variant of majority judgement.

One option is to do this just by resampling majority judgement, but this doesn’t seem like a good option. You have to decide how many people to sample, and if it’s all of them then the answer is just going to asymptotically approach deterministic majority judgement.

I came up with a solution recently that I think holds a certain amount of appeal as a design, but after some analysis I think it doesn’t hold much appeal as a voting system. These are some notes on it for posterity.

The idea is as follows: People vote as in majority judgement, assigning a grades to each candidate.

We then  run the following process:

1. If there is only one remaining candidate, they win
2. If there are multiple remaining candidates but they all have exactly the same grades, pick one at random
3. From each candidate, independently pick a voter and assign that candidate the grade that voter gave them (so if 90% of people ranked a candidate excellent we give them the grade excellent with 90% probability). Of the grades assigned this way, pick the highest. Any candidate which got less than that grade drops out.
4. Return to 1 with the new set of candidates.

When I came up with this I thought it was obvious and natural and clearly the randomized way to do MJ. In reality it doesn’t track the behaviour of MJ well at all.

In particular it’s very easy to show that the relation “A beats B in a two candidate race” is non-transitive. How? You just construct grades which constitute non-transitive dice.

Further, as a result of this, it lacks the property of majority judgement that adding a new candidate to the list doesn’t affect whether A beats B (because you can add a candidate which knocks A out of the race earlier than B).

It also doesn’t track random-ballot very well I think. Consider the case with three candidates, A, B and C and three grades. Suppose candidate A has 50% good and B and C have 25% each. In random ballot that would also be the probability of them winning. In this system:

The possible outcomes for the second round are {A}, {B}, {C}, {A, B}, {B, C}, {A, C}, {A, B, C}. Let WA be the event that A wins and let \(p = P(WA)\). Then:

\[\begin{align*}
P(WA|{A}) &= 1 \\
P(WA|{B}) &= P(WA|{C}) = P(WA|{B, C})= 0\\
P(WA|{A, B}) &= \frac{3}{4}\\
P(WA|{A, C}) &= \frac{3}{4}\\
P(WA|{A, B, C}) &= p \\
\end{align*}\]

So
\[\begin{align*}
p &= \frac{1}{2}*\frac{3}{4}*\frac{3}{4} \\
& + \frac{3}{4}(2 * \frac{1}{2} * \frac{1}{4} * \frac{3}{4}) \\
& + p ( \frac{1}{2} * \frac{1}{4} * \frac{1}{4} + \frac{1}{2} * \frac{3}{4} * \frac{3}{4})\\
p &= \frac{9}{16} + \fraction{5}{16}p\\
p &= \frac{9}{11} \\
\end{align*}\]

I made multiple arithmetic mistakes in doing that, so although I’ve double checked it it could still be wrong, but it’s at least in the plausible direction: In most likely scenarios, A gets several chances to win it can either win immediately by getting a high grade and the other two getting a low grade (which happens with about a 28% chance), or it can be put into a head to head which it is likely to win (which happens with about 19% chance) or it can restart and get those chances again. So this system significantly distorts random ballot in favour of the majority.

Due to, ah, reasons which will soon become apparent, I’ve been interested in opening up the voting data from elections which use more appropriate voting systems than FPTP. Unfortunately it turns out that this is a bad idea.

You should read the paper. It’s short, well written and very interesting. But the summary of it is that when you’ve got something like Majority Judgment or a ranked voting system like AV there are a lot of redundant bits of information in your vote – so much so that this means that pretty much any individual vote will be unique. Malicious parties can then force you to encode messages in your vote – e.g. by saying “Vote for this person first, then vote for the remainder in this distinctive order”. If they then have access to the anonymised set of votes cast they can look for that distinctive pattern in the results and thus confirm that you have voted as you were told to vote. Its absence proves that you didn’t, and its presence proves that with very high probability you did.

This attack is pretty universal – it holds for almost any voting system that allows you to express a complicated enough set of preferences to produce a fair result. I do however feel the need to point out that there is at least one fair voting system that is immune to this…

P.S. This post represents two new things I’m planning to try with this blog in the near future: More promotion of “this is a thing that I found interesting” that is a little bit better curated than you just reading my pinboard feed and shorter blog posts which just point something out rather than constituting a fully formed essay. We’ll see how this plan goes.

A while ago I designed a voting system which is basically how you apply random ballot to multi-member elections. To recap, here’s how it works:

You are voting to elect N candidates. Each voter selects N of the candidates who are running and lists them in order of preference. You then repeatedly select (with replacement!) a random voter and elect the candidate they most prefer who has not already been elected. Once you have done this N times, you have elected N candidates and your job is done.

I was thinking about this again recently, and I’ve realised that it’s a much better idea than I initially realised. In particular:

1. It is indeed strategy proof – you have no incentive to vote outside your true order (note: See edit below. This is wrong)
2. It has a very nice strong proportionality property
3. It can be used to patch the single biggest objection people have to my towards a more perfect democracy proposal.

That it’s strategy proof is essentially obvious: At each point you are performing a random ballot, which is strategy proof, for the set of people who have not already be elected. So you you should always be voting for your most preferred person who is not already elected. This means that if you prefer A to B then you should rank A higher than B – if A has already been elected, you will vote for B, if not you will vote for A.

Edit: Wrong! You need to take into account the effect of other peoples’ votes. Suppose there are three candidates A, B, C and we’re electing two of them. Suppose everyone other than me is voting in order A, C. Now, suppose that I really really hate C and think A is a little better than B. This means that I strongly want the set of elected candidates to be A and B. Therefore I should vote B, A. Conditioning on my being picked on the first round (if I’m not it doesn’t matter what I vote – my second round selection will always be B, because A was elected in the first round), if I put A first then we will end up with {A, C} with very high probability (whenever I’m not picked for the second round too) . If I put B first then we will end up with {A, B} with probability 1. Therefore I should put B first even though I would prefer A.

This example can be extended further so that I may not even vote for my top N candidates.

The proportionality property it has is kinda interesting. It has the following property:

Suppose some group fields at least N candidates. If x% of the population prefer every member of that group to everyone not in that group, that group will have an expected number of seats at least x% of N.

So, for example, if x% of the population fill out their ballot entirely with members of a single party, that party will expect to have at least x% of the elected seats.

Why at least x%? Because the remaining voters can  mix it up by e.g. preferring some members of that group to people not in it, but some people not in it to some people in it. If no-one outside those core x% of voters vote for the group then the expected proportion is exactly x%.

Finally, the patch to the perfect democracy proposal: One of the biggest things people object to about it is that people with a strong mandate can still fail to be elected – even if you have 90% of the vote, that’s still a one in ten chance of not being elected. This significantly depletes the chances of having an experienced person who people love staying in power, which is a bit of a shame.

The nice thing about this system instead of that it is that it handles this case much better by giving each candidate multiple chances to get in. As a result, the probability of a candidate with a fraction \(p\) of the first votes getting elected is at least \(1 – (1 – p)^N\) (it may be more depending on the distribution of second and higher votes). So if they have 90% of the first votes, with N = 2 they have at least a 99% chance of getting in, with N=3 they have 99.9%, etc. With a more modest 50% of the vote, they have a 99% chance of getting it at \(N \geq 7\) seats.

So this is the fix: Rather than having a large number of single member constituencies, you instead have a smaller number of multi-member constituencies. 10 is probably a good number – it’s on the cusp of what is feasible to get people to vote on, but it means that anyone with at least 40% of the first votes is extremely likely to get a seat, and anyone with 50% of them is virtually certain to get a seat (about every 10-20 general elections you will expect one candidate in the entire house with 50% of the vote in their constituency to fail to get a seat).

It definitely complicates the initial proposal, but I don’t think it’s that difficult to get people to reasonably rank 10 candidates they like. A more difficult thing might be persuading people to go for multi-member constituencies.

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.