Sunday, 24 April 2011

I'm Backing AV 110%

(Sorry to go on about it, but I'm just fascinated by mathematics.)

One of the headline reasons being put forward in favour of AV is that it means that no one can become an MP without winning more than 50% of the vote (assuming every voter uses all of their votes, which is a detail I'm not too fussed about). I'm not sure this is such a great thing, since some of that support might be, to say the least, grudging, coming from voters who could actually think of six or seven candidates that they would rather have had than the winner for whom their vote was finally counted.

However 50% is still 50%. And what's really great about AV is it's not only the winner who gets more than 50% of the vote, so might some of the losers.

Let's return to the constituency I mentioned in my last post, but give it a more definite result. First-choice votes go:

Labour: 15,000
Liberal: 11,000
Tory: 9,000

The Tory drops out and let's suppose 7000 of his second-choice votes were for the Liberal. The Liberal now has 18000, which is more than 50%, and wins. First-choice votes of one candidate plus second choice votes of another gives greater than 50%.

The trouble is, the Labour candidate also got more than 50% if you add the first choice votes of one candidate and the second choice of another. The Labour first-choices plus the Liberal second-choices (assuming a reasonable split) also come to greater than 50% . (It's actually mathematically possible that the Tory could get more than 50% too if the Labour candidate's second choice votes were largely for the Tory - but that's unlikely).

So 50% is a necessary but not sufficient condition to win an AV election. How then is the winner decided? We could go back to a FPTP-style approach, where it's the candidate with the highest number of votes (first and second choice combined), but in this example, that would probably be Labour. Here the Liberal wins with more than 50% of a particular set of first and second choice votes, but still fewer than the combination of votes that Labour got, but didn't get counted.

So the total number of votes cast is more than 100%?

Let's consider the definition of percentage. The percentage of votes for a candidate is:

(n / T) * 100

where n is the number of votes received by the candidate and T is the total number of votes overall.

But that term T could actually have several meanings. It could be the total number of voters, or the total number of votes cast or the total number of votes counted. (It could also be the total number of eligible voters, but turnout is a problem under any system.) Under FPTP those three definitions of T are all the same thing, because FPTP is one person-one vote.

But AV gives several votes to each voter. (You can argue that it's a good thing, but you can't deny it - the voter gets to indicate support for more than one candidate. It may not be that multiple votes are counted, but multiple votes are cast.) If you go with T being the number of voters, then it's clearly true - the winning candidate has more than 50% of the vote. In this case, 51%.

But if T is the total number of votes cast, with three votes per voter (assuming that there were no other candidates and that every voter used all their votes) that gives a total of 105,000 votes - and the winning Liberal candidate gets (18,000 / 105,000) * 100 = 17%.

On the other hand, if T is the number of votes counted, things get more complicated. In the first round, 35,000 votes were counted. The Tory dropped out and so in the second round a further 9000 votes were counted, giving a T of 44,000. So now the winning Liberal candidate's percentage is (18,000 / 44,000) * 100 = 41%. Not a bad result for a winner under FPTP, but this is AV, which supposedly guarantees the winner gets more than 50%. You may think I'm being unfair, counting those second-choice Tory votes into T with the same weight as first-choice votes, but if you do, then you must surely also object to them being counted into n (the votes for the winning candidate) with equal weight - that's one of the main objections that many people have to AV.

So the question that we must all answer, whether we are pro or anti or could not care less, is:

What is your definition of T?

Wednesday, 20 April 2011

Electile Disfunction

I won't deny it; I've been mildly opposed to the Alternative Vote ever since Gordon Brown first mooted it as an opening gambit in his attempts to form a coalition government with the Liberal Democrats back in 2009, over six months before the general election.

But one objection that I'd never really held with was the idea that AV is complicated. All the voter has to do, so they say, is write down the numbers 1 to 9 (or whatever) in order of preference. It's simple - in much the same way that solving a Sudoku is simple (or indeed that playing the flute is simple, according to Monty Python).

But the more I think about it, the more complicated it gets. Problems can occur in many areas, but the one I'd like to focus on is the phenomenon of the second choice marginal.

Let's consider an imaginary constituency with just Tory, Liberal and Labour candidates (or one in which other, less popular candidates have already been knocked out in earlier rounds of AV). I use real party names rather than abstractions such as A, B and C not to express any party bias, but because it's easier to follow and easier to decide whether such a scenario could really happen. Suppose the first choice votes are roughly:

Labour: 15,000
Liberal: 10,000
Tory: 10,000

'Roughly?' I hear you bellow. 'Surely we must be accurate here.' Well, yes and no. If this were a First Past the Post election, then those kind of round numbers are quite clear enough to show that Labour wins. Of course, even under FPTP we have marginals if the two leading candidates are close, and then accuracy matters, but under AV we also have the possibility of this kind of second choice marginal (or, indeed, third, fourth of fifth choice) where precise counting even for second place really matters.

More marginals? Isn't that one of the key aims of AV; to force parties to genuinely campaign in more seats, rather than just focussing on the few marginals that matter so much under FPTP? True enough, but you might find that what they're campaigning for isn't quite what you'd expected.

Under AV the winning candidate needs to get more than half the votes cast, so in this case the winning post is 17,500. (Odd, isn't it, that it's AV that actually has the fixed finishing post, and so-called First Past the Post that doesn't?) No one here has 17,500, so we have to consider those second choice votes.

We can ignore the Labour second choices, because they're never going to be counted, though they'll probably be mostly for the Liberals. The Tory second choices are likely to be mostly Liberal too. Admittedly there may be a lot of support from Tories for, say, UKIP, but we're assuming they've been eliminated by now. At this stage, a Tory's second choice can only be Labour or Liberal (or nothing, but that's another story).

As for the Liberal voters, let's assume they spilt 50-50 amongst Tory and Labour. In reality, there might well be more of a bias towards Labour, but it doesn't much matter. With Labour only needing around 2,500 to win, the Liberal spilt could be up to 75% pro Tory, and the mathematics would still be much the same.

So, we had Liberals and Tories on about 10,000 each. Time to be specific. Let's suppose that the Liberal got 10,005 and the Tory 10,000. The Tory drops out and his second choice votes get allocated. We've assumed they're mostly Liberal and very few Labour, and so it seems reasonable that the Liberals will pick up the extra 7,500 they need and will win. This is exactly the sort of result that AV is supposed to achieve. The Liberals come second in the first round, but win on the second round.

But just suppose it goes the other way. Suppose it's the Tory who gets 10,005 and the Liberal 10,000. Then the Liberal drops out and his second choice votes get reallocated. We've assumed it's a 50-50 spilt, so Labour gets 5,000 more votes and wins. Just read that again:

The Tory is more popular than the Liberal and therefore Labour wins.

And on top of that, the difference is the matter of just a few votes. Under FPTP a few votes will matter in a marginal, but at least there a vote for Labour will help Labour, a vote for the Liberals will help the Liberals. Here it's the swing between Tories and Liberals that determines a result between the Liberals and Labour.

So what's a Tory voter to do? In this particular constituency, they know that their favoured candidate has no chance of winning, so the next best option is for the Liberal to win. But if they vote Tory first and Liberal second, that actually increases the chance of Labour winning, by pushing out the Liberal on the first round and thereby getting his second choices counted. It's a better bet for the Tory to vote Liberal first and Tory second, so that the Tory drops out and his second choice votes go to the Liberal. It's classic tactical voting; if you're a Tory afraid of Labour, vote Liberal.

In fact, it's better than tactical voting under FPTP. Not only does the Tory vote for the Liberal mean one more vote for the Liberals; if it makes the Tory candidate drop to third, it means thousands more votes for the Liberal as all those second choices get counted.

And it's not just Tories who can vote tactically. Remember the set up: Liberal second place is good for the Liberals; a Tory second place is good for Labour. So why don't a few hundred Labour supporters tactically vote Tory? It costs a few hundred votes, but if it pushes the Liberals into third and reaps a few thousand Labour votes it's a worthwhile reward.

In both styles of tactical voting, there is a powerful psephological lever in operation. Switching a small number of votes away from your first choice party can actually liberate a huge number in favour of the result you want. It may take a fair deal of voter management from the political parties, but guess what? - they're good at that.

And what about recounts? Let's go back to that scenario where the Tory gets 10,005 and the Liberal 10,000. That means Labour wins. The Liberal isn't happy and there's only five votes in it, so it's worth asking for a recount. But the thing is, the Tory (with Nick Berry's Every Loser Wins ringing in his ears) isn't happy with it either - because Labour wins. So both the winning and the losing candidate (in the second and third place play-off) will be asking for a recount. At least under FPTP it tends to be the loser who wants a recount and the winner who doesn't. It puts the returning officer in a difficult position of perhaps having to act against the requests of both candidates.

And will those candidates have enough information to decide whether a recount is worthwhile. With different second choice voting patterns, it's quite possible that the Labour candidate would win regardless of who comes second. Would the Tory and Liberal candidates know that before deciding whether it's worth bickering over the few votes that determine second and third place?

Of course, we've been looking at a specific example which won't occur everywhere. But with 650 constituencies, this sort of thing could crop up more than once, along with other permutations that aren't even dreamt of here.

Putting the numbers 1 to 9 nine in order has never been more of a challenge.