How often the majority goes in a circle

The majority that goes in a circle built every profile of three voters over three candidates — 216 of them — and found 12 in which pairwise majority runs round a cycle. It quoted, without establishing, the number that everyone who works on voting knows: as the electorate grows, the share of cyclic profiles approaches about 8.8%. It also named the assumption hiding in that number, which is that every ordering of the candidates is equally likely for every voter, independently.

This essay establishes the number, in two ways that share no arithmetic, and then does the more useful thing, which is to move the assumption and watch the number move with it. The count can be carried exactly to 41 voters, where it stands at $8.57\%$. The limit follows from a picture of three correlated bell curves, and it comes out as a solid angle on a sphere. More candidates make things much worse. And a single shared axis among the voters makes cycles disappear entirely — so the famous $8.77\%$ is a fact about one model of ballots, not about majority rule.

Where the count starts #

With three candidates there are six rankings, so $n$ voters produce $6^n$ ordered profiles, and each one either cycles or does not. That is a finite count for every $n$, and in principle it answers everything. In practice $6^n$ outgrows any computer by $n = 20$. But the verdict does not depend on which voter holds which ranking — only on how many voters hold each of the six — so the profiles can be grouped by their six counts. An electorate of 41 voters has only $\binom{46}{5} = 1{,}370{,}754$ distinct count vectors, and each one stands for a known number of ordered profiles: the multinomial coefficient, $41!$ divided by the factorials of its six counts. Summing over the cyclic ones gives the share exactly.

There are two natural ways to weight the count vectors, and they are two different models of an electorate. Under the first, which is the one Guilbaud used and the one the 216-cell grid illustrates, each voter independently picks one of the six rankings at random; a count vector then carries the weight of all the ordered profiles it stands for. Under the second, every count vector is equally likely — every way the ballots could come out, as a tally, is as likely as every other. The first model is called the impartial culture, the second the impartial anonymous culture, and the names are more solemn than the difference, which is just a choice of how to weight a tally.

Every electorate to forty-one, counted #

Both curves rise and flatten. Under the first model the share at three voters is $12/216 = 5.56\%$, exactly the grid’s count, and it climbs through $6.94\%$ at five and $7.50\%$ at seven to $8.57\%$ at forty-one, always from below. Under the second it starts at $2/56 = 3.57\%$ — of the 56 possible tallies of three ballots, only the two made of one each of the three rankings in a cycle are cyclic — and climbs to $6.20\%$.

Two things are worth noticing before the limit is derived. The first is that the curves really do flatten, which a reader of the 216-cell grid had no way to know: the share could in principle have kept climbing towards a half. The second is that the two models disagree by about a quarter, at every size. Neither is more correct than the other. They are two ways of saying “nothing is known about the voters”, and they do not say it the same way, which is a first warning that the headline number belongs to a model.

Why the limit is a solid angle #

The limit comes from what happens to the three pairwise margins when the electorate is large. Write $X$ for A’s margin over B, $Y$ for B’s margin over C, and $Z$ for C’s margin over A. Each voter adds $+1$ or $-1$ to each of them. A cycle A → B → C → A is the event that all three are positive; the other cycle is all three negative.

Each voter’s contribution to $(X, Y, Z)$ is one of six vectors, and a quick check on the six rankings gives the correlations: any two of the three contributions agree in sign exactly a third of the time, so each pair of margins has correlation $-\tfrac13$. By the same averaging that turns coin flips into a bell curve, the margins divided by $\sqrt n$ become, for a large electorate, three normal variables with those correlations. So the cycle probability tends to twice the chance that three standard normal variables with pairwise correlation $-\tfrac13$ are all positive.

The correlation is worth reading off the ballots rather than taking on trust, because it is the whole of the argument. List the six rankings with their three verdicts. A ≻ B ≻ C says A over B, B over C, and A over C, so its contributions to $(X, Y, Z)$ are $(+1, +1, -1)$. A ≻ C ≻ B gives $(+1, -1, -1)$; B ≻ A ≻ C gives $(-1, +1, -1)$; B ≻ C ≻ A gives $(-1, +1, +1)$; C ≻ A ≻ B gives $(+1, -1, +1)$; C ≻ B ≻ A gives $(-1, -1, +1)$. Across the six, each coordinate is $+1$ three times and $-1$ three times, so every margin is fair. But the first two coordinates agree in only two of the six rankings, and the same is true of every pair of coordinates. The two missing patterns, $(+1, +1, +1)$ and $(-1, -1, -1)$, are exactly the cycles — no single voter’s ranking is cyclic, and that one fact, spread over a large electorate, is the negative correlation. A cycle has to be assembled from voters none of whom holds it, and the correlation measures how much the voters resist being assembled that way.

That chance has a closed form, and a picture. Three such variables can be written as the projections of one random direction in space onto three unit arrows, and the correlation between two of them is the cosine of the angle between their arrows. An angle whose cosine is $-\tfrac13$ is $109.47°$ — the angle between two arms of a methane molecule, from the centre of a regular tetrahedron to two of its corners. All three positive means the random direction lies within ninety degrees of all three arrows, which cuts out a spherical triangle; its share of the sphere is its area over $4\pi$, and the area of a spherical triangle is its angle excess. The result is Sheppard’s orthant formula,

$$P(X, Y, Z > 0) = \frac18 + \frac{3}{4\pi}\arcsin\!\left(-\tfrac13\right) = 0.04387,$$

and doubled, for the two directions round the cycle, it is Guilbaud’s $8.774\%$.

It helps to know what the number would have been without the correlation. If the three pairwise contests were independent fair coins, all three would point the same way round the circle a quarter of the time, and a quarter of electorates would cycle. The structure of rankings — the fact that no voter holds a cycle — cuts that quarter to under a tenth. In the picture, the three arrows are pushed apart from ninety degrees to the tetrahedral angle, and the spherical triangle where all three half-spaces overlap shrinks from an eighth of the sphere to four and a half hundredths of it. The exact counts to 41 voters approach it from below, and the figure checks that they never pass it.

Cycles are close contests #

The scatter shows the geometry at work. A cycle needs the third margin to agree in sign with the first two, and the negative correlation says that when the first two are both large and positive, the third is pushed towards negative. So cycles live near the centre of the cloud, among electorates in which every pair is fairly close, and become rare in lopsided ones. The simulated share, $8.6\%$ from sixteen hundred electorates, sits within sampling error of the exact limit, which is the only thing a simulation is used for here — every other number in the essay is counted.

This is the most useful qualitative fact the model offers: under it, a cycle is a phenomenon of close elections. An electorate with a clear favourite in any pair is unlikely to cycle, whatever the other pairs do, because a cycle requires every pair’s margin to point round the circle, and a large margin in one pair drags the others away from that alignment.

More candidates, and the winner disappears #

With three candidates, the absence of a Condorcet winner and the presence of a cycle are the same event. With more, they are not — several different structures can fail at once — and the natural question becomes simply how often no candidate beats every other.

The same limiting picture applies. Every pair of candidates has a margin, the margins are jointly normal in a large electorate, and the correlations are fixed by the geometry of rankings: two pairs sharing a candidate on the same side are correlated at $+\tfrac13$, two pairs chained through a candidate at $-\tfrac13$, two disjoint pairs not at all. A Condorcet winner is a candidate all of whose margins point outwards. Sampling that normal vector answers the question, and the answer climbs fast: one chance in six at four candidates, one in four at five, nearly one in two at ten. As the number of candidates grows the chance of a Condorcet winner goes to zero.

Here too the correlations are doing most of the work, in the other direction. If every pairwise contest were an independent coin, a given candidate would win all of its $m - 1$ contests with chance $2^{-(m-1)}$, and a Condorcet winner would exist only about $m/2^{m-1}$ of the time — one electorate in fifty at ten candidates. The true figure is one in two. Pairs that share a candidate are positively correlated when the shared candidate sits on the same side, so a candidate who is strong against one rival tends to be strong against the others, and the correlations keep winners far commoner than independence would.

The reason winners still become rare is not mysterious. A candidate needs to win $m - 1$ contests, and although the contests are correlated, they are not correlated strongly enough to make winning all of them anything but a long shot as $m$ grows. Under this model, a majority relation with many candidates and no structure in the electorate is almost never a ranking with a top.

What a missing winner does to every other rule #

A cycle is not only a curiosity of the majority relation. It is the situation in which every other rule has to make a choice that no pairwise argument can defend, and so the numbers above are also a measure of how often the disagreements of five reasonable rules are forced rather than accidental.

When a Condorcet winner exists, any rule that respects it — Copeland’s, Black’s, the minimax rule, Kemeny’s ranking — returns it, and a rule that does not respect it can be told, on the numbers, that some other candidate beats its choice head to head. When no Condorcet winner exists, every candidate is beaten by somebody, and whichever candidate a rule returns can be challenged by the one that beats it. The rules then diverge because they break the cycle in different places: minimax chooses the candidate whose worst defeat is smallest, Borda the one with the most total support, instant runoff the one who survives a sequence of eliminations. None of them is wrong. They are answering a question the majority relation left open.

It is also where the incentive to misreport lives most comfortably. A voter who can tip one pairwise margin in a cyclic profile can change which pair is the weakest link of the cycle, and so change which candidate every cycle-breaking rule returns. So a figure that says eight or nine electorates in a hundred are cyclic under the random model is also saying how often the choice among rules genuinely matters under that model — and, by the scatter, that those are precisely the close contests in which it would be most keenly felt.

A shared axis, and the cycles vanish #

Everything so far has assumed the same thing about voters: that each ranking is as likely as any other, independently. The model is chosen because it says nothing, and saying nothing turns out to be a strong statement. A cycle among A, B and C needs some voters who rank each candidate last — someone must put A at the bottom, someone B, someone C. If voters share any structure at all, that requirement is the first thing to fail.

The cleanest structure is a shared axis. Arrange the candidates on a line, A — B — C, and suppose voters have single-peaked preferences on it: a favourite, and a dislike that grows with distance from it. Then nobody ranks B, the middle candidate, last, and Black’s median voter theorem says pairwise majority can never cycle. Four of the six rankings are single-peaked on this axis.

The fall is steep, and it is steeper for larger electorates. With half the voters on the shared axis and the other half random, 41-voter electorates cycle only $1.19\%$ of the time — a seventh of the random model’s rate — and the decline accelerates with size, because a large electorate’s margins reflect the average voter, and the average voter leans towards the axis. In the limit of a large electorate, any positive share of axis-bound voters is enough to make cycles vanishingly rare under this mixture: the shared axis gives B a steady advantage over both A and C, so two of the three margins grow in proportion to the electorate, and they grow in opposite directions round the circle, which no cycle can survive.

That is the real content of the $8.77\%$, and it cuts in both directions. The number is a precise answer to a question about a model of complete ignorance, and it is often read as an estimate of how often majorities misbehave in practice. It is not one. It is the rate at which majority cycles under the least structured electorate imaginable, and this field makes no claim about which model any actual electorate resembles — that is a question about observed preferences, and voting rules are studied here as permutations, not as descriptions of anyone.

What the counts cannot show #

They cannot reach the limit. The exact counts stop at 41 voters, where the anonymous profiles number 1,370,754, and the limit is established by the normal approximation, not by counting. The counts approach $8.774\%$ from below and are within $0.21$ percentage points of it at 41; they are evidence for the derivation, and the derivation is what proves the limit, through the central limit theorem applied to the three margins.

They cannot say which model is right. Both models in the first figure are defensible descriptions of ignorance, and they disagree by a quarter. Neither is a description of any electorate, and the mixture model is one family of departures among infinitely many. The figures show how strongly the answer depends on the model; they do not choose one.

And the simulated figures are samples. The scatter and the many-candidate bars are drawn from seeded random draws, not from exhaustion, and each one is checked against an exact value — the scatter’s share against $8.774\%$, the three-candidate bar against the same number — within four standard errors. Their other values carry sampling error of about a quarter of a percentage point, which is smaller than any difference the text relies on.

Where the count goes next #

Two directions are natural from here. One is to replace the probability model with a worst case: not how often a cycle occurs, but how many voters it takes to produce every possible majority tournament, which is McGarvey’s theorem made quantitative — a question about the smallest profiles, answerable by search. The other is the question the jury theorem asks, which turns the whole subject round: instead of voters with preferences, voters with judgements about a single fact, each more likely right than wrong, and the question of how often the majority is correct. There, the same law of large numbers that produced Guilbaud’s limit produces something much more reassuring.

The share of cycles is a property of the model #

Under the model in which every voter independently picks a ranking at random, the share of three-candidate electorates whose majority cycles rises from $5.56\%$ at three voters to $8.57\%$ at forty-one, counted exactly, and tends to Guilbaud’s $8.774\%$ — twice the chance that three normal margins correlated at $-\tfrac13$ are all positive, a solid angle cut out by three half-spaces at the tetrahedral angle. Under the model in which every tally is equally likely, the limit is $6.25\%$.

Cycles are concentrated among close contests, and with more candidates a Condorcet winner goes missing with rising likelihood — nearly half the time at ten. But the whole calculation rests on voters sharing no structure. Let some of them share a single axis and cycles become rare; let all of them share it and they cannot occur.

A probability of misbehaviour computed under complete ignorance is a fact about ignorance — before it is read as a fact about the system, the model has to be named.