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Julian D. A. Wiseman
Contents: The Problem; Single Transferable Vote; Borda Count; First Past The Post; Proportional Representation; Solution.
Publication history: Only here. Usual disclaimer and copyright terms apply.
Twelve members of a democracy wish to elect a leader from amongst six candidates called A, B, C, D, E and F. The voters’ preferences are:
This example started with one derived by Joseph Malkevitch of York College CUNY, being an example with 55 voters of 6 types that did not consider approval voting: 18 × A>D>E>C>B; 12 × B>E>D>C>A; 10 × C>B>E>D>A; 9 × D>C>E>B>A; 4 × E>B>D>C>A; 2 × E>C>D>B>A. This author then shrank the number of voters and number of types of voters, again without approval voting. There can be as few as 11 voters, albeit at the price of having a draw in the first round of STV: 4 × A>D>E>C>B; 3 × B>E>D>C>A; 2 × C>E>D>B>A; 1 × D>C>B>E>A; 1 × E>D>C>B>A. In November 2008 approval voting was added to the list, without the STV draw.
Who should be elected?
A? British-style first-past-the-post (single-member plurality) elects A.
B? If, French style, there is a run off between those two candidates with the most first place votes, then B defeats A on the second round 8:4.
C? Single Transferable Vote gives the victory to C. First E is eliminated, having no first-place preferences. Then D, that vote going to C, then F, leaving A:B:C with votes split 4:3:5, thus eliminating B; C winning the final round 8:4.
D? If we assign 6 points to each first place vote, 5 to each second, etc, thus giving each candidate a Borda count, D would win, the candidates scoring 35:39:33:53:42:50.
E? Approval voting allows voters to choose multiple approved candidates. For these purposes let’s say that voters would approve their first three choices, and not approve the last three. E wins, the candidates’ approval-vote totals being 4:3:3:9:10:7.
F? Consider a head-to-head between each pair of candidates. F would beat A 7:5, F would beat B by 8:4, F beats C by 9:3, F beats D by 7:5, and F beats E 7:5. F is called the Condorcet winner. (Often voting is so muddled that there is no Condorcet winner.)
Of course, this could be still more confusing. Instead imagine that twelve members of a democracy wish to choose an electoral system from amongst six possibilities called A, B, C, D, E and F… .
The French President is elected by a system called ‘Run-Off Voting’. A first round of voting is held, after which there a run off between those two candidates with the most votes. Both this system, and the Single Transferable Vote, suffer from being non-monotonic, and hence (in effect) random.
What is meant by ‘monotonic’? Simply that voting for a candidate doesn’t harm that candidate.
Consider the case in which seventeen voters have preferences amongst three candidates as follows:
In either of the two systems under consideration, C is knocked out at the first round. B acquires C’s votes, and so defeats A by a margin of 9 to 8.
But what if two of the voters who favour A (and are indifferent between B and C) instead vote for C (in the first round of a 2-round system) or vote for C then A then B (in STV). Then votes would be as follows:
October 2016: Nate Silver explained How Evan McMullin Could Win Utah And The Presidency, except that “if McMullin gets too strong, he could literally cost himself the election.” It’s a neat example of non-monotonicity in the USA.
June 2019: the UK’s Conservative Party, despite rightly opposing the use of Single Transferable Vote in general elections, uses it as part of the system for the election of its own leader. MPs vote using STV until only two candidates remain, and then it goes to a run-off in which the electorate comprises all party members. The BBC, in Tactical voting claims over Johnson and Hunt win, suggested that some supporters of Boris Johnson voted for Jeremy Hunt with the aim, achieved, of eliminating Michael Gove. Presumably the Johnson supporters (or Johnson campaign) believed that members’ preferences might be Gove > Johnson > Hunt.
First round preferences for A : B : C now total 6 : 5 : 6, so B is eliminated in the first round, and in the second round A wins by 9 to 8.
So (for two voters at least) voting for A meant that A lost, whereas voting for C then A meant that A won. When voting for (or giving a higher preference to) a candidate never harms that candidate, voting is said to be monotonic. When this can hurt, voting is non-monotonic.
However, this non-monotonicity is not as bad a problem as it seems.
The tactical ‘dishonesty’ relies on excellent information about voters’ true intentions. In practice, the required information is only likely to be known if there are few voters: seventeen rather than seventeen thousand. So, for a national election, in which constituency electorates can vary from a few thousands to many millions, a political party would have neither the precision of information nor the precision of command to carry out such a manoeuvre. (For an excellent summary of this argument see ¶s 145 to 150 of Lord Jenkins’ Report Of The Independent Commission On The Voting System.)
However, as argued by Lord Alexander in the same report, the ‘randomness’ of such systems is unavoidable:
In addition, as all experts on electoral systems have acknowledged, AV [and STV and similar systems] can operate haphazardly depending upon the ranking of candidates on first preference votes …
Suppose within a constituency, Conservatives receive 40% of first preferences. Labour are second on 31% and Lib Dems third on 29%. Lib Dems second preferences happen to be split 15/14 in favour of Labour. The Conservatives are therefore elected with 54% of the total vote (i.e. 40% + 14%).
But now suppose the position of Labour and Lib Dems had been reversed on first preferences, with Lib Dems 31% and Labour 29%. If Labour second preferences were split 20/9 in favour of Lib Dems, the Lib Dems would be elected with 51% of the total vote (i.e. 31% + 20%).
So the result would be different depending on which horse was second and which third over Becher’s Brook first time round. This seems to me too random to be acceptable.
Unfortunately, this ‘haphazardness’ and this ‘non-monotonicity’ go hand-in-hand: the former is the large-scale manifestation of the latter.
As a second example of the pathological behaviour of ‘elimination’ methods, consider the following:
First round votes are 4 : 2 : 3 for A : B : C, so B is eliminated and A wins 5 : 4. But note that in pairwise contests, B would beat A by 5 to 4, and B would beat C by the same margin. Yet despite the fact that B was preferred to both the others, B was eliminated in the first round.
STV can behave strangely in other ways. Consider a constituency in which:
|3||people vote||C then B then A.|
STV would eliminate C, and B wins the second round 7 : 6. Also assume that there is a neighbouring constituency in which the roles of A and C are reversed:
|3||people vote||A then B then C.|
This constituency is also won by B. Now imagine that these two constituencies are merged: B has 8 first-preference votes, A and C have 9 each, so B is eliminated and the result depends on the second preferences of those preferring B. Thus merging two constituencies both won by party B can produce a larger constituency not won by party B!
As we saw earlier, the Borda count assigns 1 point to the last preference, 2 to the second-last preference, and so on. If some lower-order preferences are omitted, then the points are assigned equally between the remaining candidates.
So, consider an election between A and B, in which A receives 60% of the first preference votes (and therefore, implicitly or explicitly, 40% of the second preference votes), and B receives 40% of the first preferences and 60% of the second preferences. Therefore A scores 60% × 2 + 40% × 1 = 160%, whilst B scores 40% × 2 + 60% × 1 = 140%. A is therefore a clear winner.
But the B team has a counter-strategy: B fields two candidates, B1 and B2, of which B1 is clearly superior. Now 60% of the votes are for A then B1 then B2, and 40% are for B1 then B2 then A. A scores 60% × 3 + 40% × 1 = 220%; B1 scores 60% × 2 + 40% × 3 = 240%; and B2 scores 60% × 1 + 40% × 2 = 140%. The winner is therefore B1.
A has two possible counter-counter-strategies. Either it could field lots of candidates, or it could ask its supporters to vote for the worse B candidate (B2) in second place, with B1 into third place. Either way, the election has become one of tactical manipulation of the rules.
First Past The Post is the system that has been in longest continuous use, in the United Kingdom, and in the United States.
But is also subject to non-intuitive behaviour. Imagine that there three constituencies, each containing 5 voters. Nationwide, the votes split 9 : 6. If these were split equally amongst the three, one party would win all three seats by a majority of 3 : 2. But if the voters are split into one constituency of 5 : 0, and two of 2 : 3, then the other party will have won a majority of the seats. The result of the election will have been determined by the geographical split of the constituencies, rather than by the votes themselves.
First-past-the-post can behave worse when there are three parties. Assume that there are 5 seats, each with 6 voters. In three seats the votes split 3 : 1 : 2, in one seat 1 : 3 : 2, and in the last seat 0 : 4 : 2. The first party will have won three seats, the second two seats, and the third none, despite the national vote having split equally. This dependency on the geographic split of the vote is widely deemed to be unfair.
Proportional electoral systems (including the various Additional Members Systems), can put hugely disproportionate power into the hands of tiny minorities. In doing so, such systems can cause frequent turnover of short-lived governments, most famously in Italy.
Why? Imagine that, in a proportional system, three parties split the vote equally. Each would have an equal number of seats. A bill would then require the consent of any two parties, and any two parties can prevent the passage of a bill. So far so fair.
But now imagine that the votes split 49% : 48% : 3%. Exactly the same applies: a bill would require the consent of any two parties, and any two parties can prevent the passage of a bill. All three parties would have equal power: less fair.
If there are more than three parties the situation can be much worse. If (for example) any likely coalition will involve three or more parties, then power is effectively moved from the ballot box to the post-election negotiating table, as large parties haggle over the ‘price’ of the small parties (a price normally measured in cabinet seats and special-interest policies).
So what’s the answer? In the opinion of this writer, the correct answer must have several features:
Don’t ask questions that you can’t answer. If voters are asked for their full preferences (… prefers A to B to C to D to …) there is no sensible consistent thing to do with the preferences. So don’t ask.
Don’t even ask voters for a second preference. It allows extremist parties to get started. If those who would prefer to vote fascist first and mainstream-right second (or communist first and mainstream-left second) have only one choice, and the extremist vote would be a wasted vote, then the extremists will not be able to get started. (See ¶146 of Lord Jenkins’ Report: “In many situations of life a decision has to be made in favour of a second or third best choice and there is no inherent reason why what has often to be applied to jobs, houses, even husbands and wives should be regarded as illegitimate when it comes to voting”.)
Avoid proportionality. Give parties an incentive to negotiate coalitions before the election, and to reveal to the electorate the results of that negotiation.
Avoid systems in which the answer depends on the geographic distribution of the votes: they politicise Boundary Commissions.
Unfortunately, these conditions are widely believed to be impossible to satisfy. “Unfortunately”, because this belief is wrong. PR-Squared satisfies all these criteria.
Julian D. A. Wiseman
January 2000, October 2001, November 2008, and later amendments
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