FPTP Double Counts Some Votes
In my last piece, I argued that what matters when we’re thinking about a fair and democratic electoral system is not how many votes each person casts, but how many votes they have counted. I argued that, on this understanding, AV respects the maxim ‘one person, one vote.’ I didn’t criticize FPTP as such, allowing that it also satisfied this requirement. In this post, however, I want to show that FPTP actually double counts some votes and is therefore unfair.
For simplicity, I’m going to focus on an example where we have only three candidates, who I will call A, B, and C, and twelve voters. Let’s assume that (first preference) votes are allocated as follows:
A: 5 votes
B: 4 votes
C: 3 votes
Under FPTP, A is declared the winner, because A has more votes than anyone else. But what does this mean? Does it mean that A has a majority? The answer, evidently, is not in the strict sense, since a majority means ‘more than half’ and A only has five of the twelve votes, which is less than half.
It is sometimes said that A has a ‘relative majority.’ That is, A has a majority over B (five to four) and also a majority over C (five to three). Note, however, that when we compare A only with B, we ignore those who voted for C. In saying A has a majority of five to four (over B), we are only counting nine votes. Those three votes for C are excluded here (though, of course, they come into play when A’s votes are compared to C’s – there it is B’s votes that are ignored).
Saying that A has a majority – in this sense – over each of B and C taken separately does not show that A has a majority over the two of them together. It might be that all those who voted B would also prefer C to A (i.e. their full preferences were B > C > A) and similarly those who voted C prefer B to A (i.e. C > B > A). In this case, a majority (seven people) prefer B to A and a majority (seven people) prefer C to A.
How then does FPTP declare A the winner? The problem, it seems, is that FPTP either ignores some people’s votes – as votes for C are ignored when simply comparing A and B – or it double counts some. Think again of the comparison between A and B. We might imagine the votes ‘cancelling out’ until, after four votes on each side have been cancelled, A wins because there is still a remainder (one) in favour of A. To think that A also defeats C, however, we have to allow that each vote for A can cancel not only one of B’s votes but also one of C’s.
Someone who voted B doesn’t, I think, have good reason to accept defeat when only first preferences are known. They might acknowledge that B trails A by five to four at this stage, but they can rightly reply that they are not yet shown to be in a minority when we do not know how the other three (C voters) would rank A and B.
Now, it might be that C voters would prefer A to B. If this is the case, then AV would declare A to be the winner. This is a case, however, where AV and FPTP would agree. Most cases would probably be like this, so there’s little need to argue between them. If A is picked by both FPTP and AV then, uncontroversially, A should be the winner. (Remember, what we’re concerned with is who should win.)
To bring out the difference between FPTP and AV, we need a case where they come apart. Therefore let us assume that all C voters prefer B to A. In this case, B can reasonably complain about any electoral rule (such as FPTP) that awards the election to A. Here B is preferred to A by a majority of the electorate (seven of twelve), so surely the idea of equal votes and majority rule tells us that B should win.
The only way we can say that A ought to win is if we illegitimately infer that because A has a majority over each of B and C taken separately (five to four and five to three, respectively) then A also has a majority over the two of them together – but this is not so, since in this example we have assumed that a majority would actually prefer either B or C to A in a two-horse race (seven to five in either case). A only wins if votes for A are counted twice, first as defeating votes for B and then again as defeating votes for C.
As I argued last time, there’s nothing unfair about counting second preferences. Under AV, each person has only one vote counted (the three C supporters have their second preference for B counted instead of a vote for C, only once C has been eliminated). This is in stark contrast to FPTP, where as I’ve just argued A only has a majority if either some votes (those for B or C) are ignored or if those for A are counted twice.
Democracy is about responding to the people’s preferences, so surely it’s more democratic to have full information about people’s preferences. Imagine that A and B had tied in the first round (say, four against four, with three for C and one abstention). How could this be resolved?
One possibility would be some random device – such as the drawing of straws or flip of a coin. That procedure is actually deployed in the case of at least some tied elections. An alternative, however, is to break the tie by appealing to voters’ preferences – we already know those of the eight people who voted for either A or B, but we could ask either the one abstainer and/or the three who had voted for C which, out of A and B, they would have voted for had they had to. This is what AV does and surely, since it responds to people’s preferences, that is a more democratic way to break the tie between A and B.
But, if the second preferences of C voters (between A and B) are the most democratic way to break a tie between A and B, why shouldn’t they also come into a close contest? Once again, supporters of B have no reason to accept five to four as a defeat, if three people’s preferences between A and B have not yet been considered.
If C voters prefer A then, fair enough, B is in the minority (eight to four). But that is never shown under FPTP. AV will establish, once and for all, whether it is A or B that has a majority. If A, then it agrees with FPTP. But if B is preferred to A by seven of the twelve voters, then surely it’s more in keeping with democracy, majority rule, and equal votes to declare B the winner.