Before I start, I should say that I intend to vote in favour of changing our electoral system to AV, when the alternative is to stick with the First Past the Post system. I have my reasons, and I may well try to explain those reasons at a later date, but suffice it to say for now that, although AV is far from the perfect system, I would view its adoption as progress. That said, the purpose of this post is to illuminate what I see as a problem in the vote-counting process, and to suggest an improvement, purely for the sake of argument. If it’s already been suggested elsewhere, that wouldn’t surprise me at all, but I think it’s worthwhile regardless.
Being reductive, a victory under FPTP signifies that the victor has the support of the largest group; but of course, this can be achieved with the support of the largest minority, even if the majority consider him or her to be abhorrent. Under AV, the significance of a victory is quite different: that more voters have expressed a clear preference for the winner over any remaining candidates than vice versa. I’m pretty comfortable with that aspect of AV.
What I’m less comfortable with is the process of elimination that reduces the field down to the final set of candidates in the first place. For those who don’t know the system, a very quick summary:
- Voters choose candidates in order of preference, numbering them “1”, “2”, “3” and so on. Once they are no longer interested in expressing a preference between the remaining candidates, they stop – so, for instance, you could mark a single candidate as your first preference and leave it at that, as in FPTP, or you could rank every single candidate, or anything inbetween.
- If one candidate has over 50% of all first preferences, they are declared the winner.
- If no candidate has over 50% of all first preferences, the candidate with the fewest first preference votes is eliminated, and their votes are transferred to the relevant voters’ second preferences.
- If a candidate now has 50% of the vote, they are declared the winner. If not, the candidate who now has the fewest votes is eliminated, and their votes transferred.
- If the candidates are reduced to a field of two without either one gathering over 50% of the vote, the winner is the candidate who is left with the largest number of votes. Bear in mind that it is possible for no candidate to achieve a majority even in a final field of two, since some voters will have had all of their preferences eliminated before the final round of counting. Their votes will therefore not count for any of the remaining candidates.
The problem is that the elimination of a candidate after each inconclusive round of counting serves only to resolve the impasse; the possibility of the eliminated candidate winning would otherwise remain. Because I have a feeling that this is not very clear, I’ll condescend (with apologies) to illustrate it with an example involving three candidates whom – in the spirit of wild creativity – I’ll call Mr Ay, Ms Bee, and Lady Cee of Epping Forest. They are standing for election under AV in a constituency with population 100. It’s an island group, obvs.
In the first round of counting, Mr Ay picks up 35 first-preference votes. Ms Bee also picks up 35 first-preference votes. Lady Cee of Epping Forest picks up just 30 first-preference votes, and is eliminated, and her votes transferred. Of her 30 voters, 10 had ranked Mr Ay as their second preference, and 20 had ranked Ms Bee as their second preference. Therefore, in the second round of counting, Ms Bee has 35 + 20 = 55 votes (to Mr Ay’s 35 + 10 = 45), and is elected as the new MP for Unnamed Island Group Inexplicably Located Somewhere in Epping Forest! Hurrah! Although democracy is, of course, the real winner.
BUT remember the premise: Ms Bee was elected because more electors preferred her to Mr Ay (55) than preferred Mr Ay to Ms Bee (45). The question then is: did more voters prefer Ms Bee to Lady Cee than preferred Lady Cee to Ms Bee? Well the thing is that, in the event, every single person who ranked Mr Ay as their first preference ranked Lady Cee as their second choice. So, in fact, not only did those 30 electors who ranked Lady Cee first prefer her to Ms Bee, but so did the 35 voters who gave their first preferences to Mr Ay – 65 electors in all. Only 35 voters preferred Ms Bee to Lady Cee, meaning that more voters preferred Lady Cee to Ms Bee than preferred Ms Bee to Lady Cee. And actually, every single person who ranked Ms Bee as their first preference had ranked Lady Cee as their second choice too. So, again, not only did the 30 electors who ranked Lady Cee first prefer her to Mr Ay, but so did the 35 voters who gave their first-choice votes to Ms Bee; more voters preferred Lady Cee to Mr Ay (65) than preferred Mr Ay to Lady Cee (35). Should Lady Cee not therefore have won?
The issue is the process of elimination. The only way to discover the above result would have been to eliminate one or other of Mr Ay and Ms Bee after the first round of voting – but this would have seemed grossly inappropriate when, at that stage, they lay in joint first place. Indeed, if you did so and then just continued the process to completion as above, it would be completely inappropriate. But what I am suggesting is that there should be not one counting process, but many – one for each possible final pairing of candidates, with everyone else’s votes transferred in each case. Only that way can we discover which candidate truly has the broadest support by comparison with each of the rest. In an election with three candidates, that means three separate counts – not too unwieldy. In an election with six candidates though, there are 15 possible pairings, and it’s starting to get a bit more arduous (if you’re bothered, the formula to calculate the number of pairings for n candidates is n × (n-1) ÷ 2). But how arduous, really? Given the technology that we have access to, this shouldn’t be any kind of issue.
Unfortunately, this is where my mathematical abilities fail me, because although it seems intuitive that there would always be one candidate who wins each of their counts, I can’t demonstrate that conclusively. If anyone who reads this can either confirm or refute this assumption, I’d be very grateful. To be honest, if anyone reads this at all, I’d be very grateful, and even if you can’t read, I’m grateful to you for casting your tearful gaze on my blog rather than someone else’s. But if you can read and can do that little bit of maths, I will be especially grateful.
That’s it, really. Haven’t we had fun? Oh come on, you don’t mean that.