A party that loses every issue can win

The majority that goes in a circle found that majority voting over three candidates can prefer A to B, B to C and C to A, so that “what the majority wants” is not a ranking at all. That paradox arises when there are several candidates and one question. This essay is about the other way an election can be several decisions at once: one choice between two parties, but parties that stand for positions on several issues. A voter who agrees with one party on most issues votes for it, and the question is whether the party that wins the election is also the party whose positions win on the issues.

It need not be, and the gap can be as wide as it could possibly be. The party that wins the vote can be on the losing side of every single issue. Moisei Ostrogorski, a Russian political scientist writing about party machines in 1902, noticed the possibility; Douglas Rae and Hans Daudt turned it into a precise paradox in 1976; and in the same year the philosopher Elizabeth Anscombe found a second paradox in the same arithmetic, about voters rather than parties. Both can be computed exactly for small electorates, which settles how often they happen, and both are fenced in by a theorem of Carl Wagner’s about how large majorities must be before neither can.

A party that loses every issue wins the vote #

The smallest clean example has five equal groups of voters and three issues. Party X takes the “yes” side on all three, party Y the “no” side. Groups A, B and C each agree with X on two of the issues and with Y on the third, a different third for each group; groups D and E agree with Y on everything.

Every member of A, B and C votes for X, because X is right about two of the three issues by their lights, and X wins three groups to two. On each issue, though, exactly one of A, B and C sides with Y, joining D and E: Y’s position wins issue one three groups to two, and issue two, and issue three. So the electorate prefers Y’s position on every question and elects X.

Nothing is inconsistent about any voter. Each holds a definite view on each issue and votes for the party closer to those views, counting issues equally. What has happened is that two different majorities are being taken in two different orders. Taking the majority within each voter first (which party is closer) and then across voters gives X; taking the majority across voters on each issue first, and then seeing which party those majorities match, gives Y. The two orders of aggregation disagree, and the five groups are arranged so that they disagree as completely as possible. It is the same structure as the court that contradicts itself, where judges voting on premises and judges voting on the verdict reach opposite results, and which deciding the premises or the conclusion turned into a choice every deliberating body has to make; there the premises were combined by and, and here the issues are combined by majority, which is what makes the reversal total rather than partial.

How often the winner loses most of the issues #

A single constructed example says the paradox is possible. How common it is depends on what electorates are like, and the simplest model to compute with is one in which every voter’s opinion on every issue is a fair coin, independent of everything else. With three issues there are eight kinds of voter, an electorate of $n$ voters is a count of how many of each kind it has, and the probability of every count is a multinomial coefficient. Enumerating every count settles the frequency exactly.

With three issues and 21 voters the winning party loses a majority of the issues in 21.9 per cent of electorates. The share rises with the number of voters and is still rising at 21, towards a limit near 23 per cent. Losing every issue, the extreme case of the five groups, happens in 0.88 per cent of electorates of 21 voters. It is impossible with three voters, which the enumeration confirms over all 120 kinds of three-voter electorate, and it first becomes possible at five. With five issues the winner loses most of them in about a quarter of electorates. A party winning most of the issues it stood for is the usual outcome, then, but in roughly one election in five it is not, and nothing in the result of the vote reveals which kind of election has just been held.

Three voters cannot, five must look like this #

The enumeration found the extreme case impossible with three voters and possible with five, and a count explains both — the same count that will reappear in Wagner’s theorem. Suppose party X wins and loses every one of three issues. Count the agreements with X, one for each voter and each issue on which they side with X.

With three voters, X’s win needs two of them, and each of those agrees with X on at least two issues, so there are at least four agreements. But X loses every issue, so on each issue at most one of the three voters sides with X, which allows at most three agreements in all. Four cannot fit into three, and the case is impossible. With five voters, X’s win needs three of them, each with at least two agreements: at least six. Losing every issue allows at most two agreements per issue: at most six. So the count fits only exactly — each X voter agrees with X on precisely two issues, the other two voters agree with X on nothing, and each issue has precisely two X agreements. Spreading three pairs of issues over three voters so that each issue is covered twice leaves only one arrangement up to relabelling, and it is the five groups in the first figure. The smallest example is not merely an example; it is the only one of its size.

The pattern resembles the construction in how few voters any majority needs, which found the smallest electorates that can produce a given pattern of head-to-head majorities among candidates. There, as here, small electorates are pinned down by counting, and paradoxes that look like accidents of large crowds turn out to need only a handful of voters arranged just so.

Most voters outvoted on most issues #

Anscombe’s paradox looks at the same electorates from the voters’ side. Settle each issue by majority, and then ask each voter on how many issues they ended up on the losing side. Each issue was decided by a majority, so on each issue most voters won. It does not follow that most voters won on most issues.

The paradox cannot happen with three voters and three issues, appears with five, and occurs in 3.5 per cent of electorates of 21 voters: about one election in 28 leaves a majority of voters outvoted on a majority of the questions it settled. With more issues it becomes rarer in this model, 1.3 per cent with five issues and 0.24 per cent with nine, because the number of issues on which a voter loses spreads out like a sum of coin tosses and concentrates as there are more of them. Anscombe’s own concern was referendums, where a series of questions is put separately to the same electorate, and her point was that “the will of the majority” on each question can add up to the frustration of the majority overall.

How many issues each voter loses #

The reason the paradox is possible at all is visible in a single large electorate. When opinions are coin tosses, each issue is decided by a narrow majority, barely more than half, so being on the winning side of any one issue is itself close to a coin toss.

The counts spread like the number of heads in seven tosses, and 466 of the 1,001 voters — 46.6 per cent — are outvoted on a majority of the issues. That is not quite the paradox; the paradox needs more than half. But it shows how close an ordinary electorate sits to it. Everything turns on how far above half the issue majorities are: a narrow majority on each issue leaves a large minority losing on most issues, and pushing that minority past half takes only a little arrangement.

Three-to-one majorities rule the paradox out #

The arrangement that does it can be built explicitly, and it shows exactly how far above half the majorities can be while the paradox still happens. Take $2m + 1$ issues and $4m + 1$ voters. Give $2m + 1$ of them, one for each issue, a block of $m + 1$ consecutive issues (counting round in a circle) on which they oppose the outcome, and let the other $2m$ voters agree with the outcome everywhere. Each issue is opposed by exactly $m + 1$ voters, a minority, so the outcome stands; and the $2m + 1$ opposing voters, a majority, each lose on $m + 1$ issues, a majority of the issues.

The average majority in this family is $3m/(4m + 1)$: 0.600 with three issues, 0.692 with seven, 0.735 with twenty-five, creeping towards three quarters and never reaching it. Random electorates showing the paradox come nowhere near the family: among sixteen random seven-voter cases the largest average majority was 0.619. The ceiling the family approaches is a theorem. Carl Wagner proved in 1983 that if the average share of voters on the winning side, over all the issues, is at least three quarters, Anscombe’s paradox cannot occur. The proof is a count. If the average majority is at least three quarters, the total number of losing votes, summed over issues, is at most a quarter of the number of voters times the number of issues; a voter losing on a majority of issues accounts for more than half the issues’ worth of losses; and more than half the voters each losing more than half would need more than a quarter in total. The family shows the bound cannot be lowered: any average majority short of three quarters can be beaten.

The same rule bounds the extreme form of Ostrogorski’s paradox. If the winning party loses every issue, each of its voters — a majority — agrees with it on most issues, and so disagrees with the outcome on most issues: that is Anscombe’s paradox, and three-to-one majorities forbid it. The milder form, a winner that loses most but not all of its issues, escapes this argument, which is why it is so much more common. Wagner’s rule gives the practical moral: a decision taken issue by issue cannot be badly at odds with a decision taken in one vote unless the issues themselves are close.

Paradoxes that need close votes #

Every paradox on this page lives on narrow margins. In the five groups each issue is decided three to two; in the coin-toss electorates every issue is close to even; in Wagner’s family the majorities are as large as they can be while the paradox survives, and they stay below three to one. Where opinion on every issue is lopsided, the order in which majorities are taken does not matter, and the party with the popular positions wins.

This is a close relative of a theme that runs through how often the majority goes in a circle, where cycles among candidates turned out to be common in random electorates and rare in electorates with structure — voters arranged along a single left-to-right line, as in Duncan Black’s theorem in a lie that pays. Structure suppresses Ostrogorski’s paradox too. If voters’ opinions are correlated across issues, so that a voter who is left on one issue tends to be left on the others, the party vote and the issue votes tend to agree, and in the extreme where every voter is either all-yes or all-no the two coincide. The paradox is the price of issues that cut across one another.

The paradox also sharpens a question that a majority wiser than its members raised for a different purpose. Condorcet’s jury theorem says that when each voter is a little more likely right than wrong about a question of fact, the majority is far more likely right. If the issues have right answers and the voters are such jurors, the issue-by-issue majorities are each likely to be correct — and Ostrogorski’s paradox says the party vote can overturn all of them at once. Which of the two verdicts deserves trust is then not a matter of taste, and it depends on whether the questions really are separate questions with answers, or whether voters are right to judge a party as a package. Like the five reasonable counts in five rules and five winners, both procedures are defensible, and they give different answers on the same opinions.

Real electorates are neither coin tosses nor perfectly ordered, and how often the paradox occurs in them is an empirical question that the model cannot answer. What the model shows is that the paradox needs no unusual voters and no strategic behaviour: honest voters weighing issues equally, in electorates with nothing special about them, produce it about one time in five.

What the pictures cannot show #

The frequencies are exact for three issues, and only for the fair-coin model: every opinion independent, every issue equally divisive. Change the model — correlated opinions, issues of different importance, voters who weigh some issues more than others — and the numbers change, and nothing on this page says by how much. The five-issue curve is sampled, so its values carry an error of about a fifth of a percentage point.

The figures also cannot show Wagner’s theorem. The family approaching three quarters and the random cases staying below it are consistent with the theorem and show that its bound cannot be improved; the theorem itself, that no electorate whatever with average majority three quarters or more shows the paradox, is the counting argument above, and no computation over particular electorates could establish it. Nor do the pictures show what any party should do about it. The paradox is a fact about aggregation, and it does not say whether the issue-by-issue verdict or the election is the truer expression of what the electorate wants.

Still open: how often real elections do it #

The models on this page say how often the paradoxes occur when opinions are random, and the theorems say when they cannot occur at all. Neither says how often they occur in actual elections, and that is the question that matters for how far a vote between two parties can be read as a verdict on the issues they stood for. Answering it needs something elections never record: every voter’s position on every issue, alongside their vote. Opinion surveys can supply an approximation for a sample of voters and a short list of issues, and the paradox has been looked for in such data, but the issues a survey asks about are chosen by the survey, voters do not weigh them equally, and a party’s platform is not a list of yes-or-no positions.

How often does the winner of a real two-party election stand on the losing side of most of the issues that divided the electorate? The fair-coin answer, about one time in five, is surely too high for electorates whose opinions line up along a left–right spectrum, and the exact answer for any real country is unknown. The arithmetic of the five groups shows that the question is not idle; the data to settle it do not exist in the form the arithmetic needs.