A candidate can beat every rival in pairwise comparisons, or majority preferences can cycle.
By the end: Identify a Condorcet winner and explain why individually coherent rankings need not produce a coherent majority ranking.
Count each comparison separately
A pairwise comparison asks how many voters rank one candidate above another. A Condorcet winner strictly defeats every other candidate head-to-head. In the complete 40/35/25 profile, C beats A by sixty to forty and B by sixty-five to thirty-five, even though C has the fewest first preferences.
Instant-runoff eliminates C first in that profile. Thus winning under instant-runoff does not guarantee being a Condorcet winner. This is a concrete distinction between criteria, not an assertion that the pairwise criterion must outweigh every other objective.
Build a comparison from individual orders
A pairwise count asks one question at a time. To compare A with B, ignore C's position except insofar as the complete order tells you which of A and B comes first. A>C>B contributes to A over B even though B is not immediately below A. B>C>A and C>B>A both contribute to B over A. The compromise profile therefore gives A forty and B sixty. The two counts sum to one hundred.
Repeat the operation for A against C, starting again from the original orders. The forty A>C>B papers favor A. The thirty-five B>C>A and twenty-five C>B>A papers favor C, giving C sixty against A forty. These sixty happen to be the same two groups that favored B over A, but their reasoning need not be identical. A complete ranking allows a different coalition in each comparison. The calculation does not remove any voter or transfer any ballot value between comparisons; it reads another relation from the same submitted orders.
For B against C, the forty A>C>B papers favor C, the thirty-five B>C>A papers favor B, and the twenty-five C>B>A papers favor C. C therefore wins sixty-five against thirty-five. C strictly defeats both rivals and is the Condorcet winner in this profile. The definition requires a victory over every other candidate, not merely the highest sum of pairwise support. With three candidates, two victories suffice.
Read the same orders separately for each pair
Question
Groups favoring winner
Opposing group
Result
A against B
35 B>C>A + 25 C>B>A
40 A>C>B
B wins 60–40
A against C
35 B>C>A + 25 C>B>A
40 A>C>B
C wins 60–40
B against C
40 A>C>B + 25 C>B>A
35 B>C>A
C wins 65–35
Each row uses all 100 voters once. Reusing voters across different questions is legitimate; adding the rows as distinct people is not.
Compare that finding with first choices. Only twenty-five people place C first, fewer than the forty for A or thirty-five for B. Nevertheless, sixty prefer C to A and sixty-five prefer C to B. There is no contradiction because these are different questions. First-choice counts ask which name is highest among all candidates; a pairwise comparison asks which is higher within one named pair. A candidate can receive broad support against each rival without leading the original first-choice count or being everyone's first choice at once.
Instant-runoff and pairwise evaluation consequently disagree here for a precise reason. The former eliminates C on the first-choice count before testing the remaining A/B contest. The latter tests C directly against each rival without eliminating it first. B wins the ranked elimination count sixty to forty, while C wins both comparisons involving C. This is evidence that instant-runoff does not always elect a Condorcet winner. It does not determine which criterion a community should prioritize, nor establish a frequency of such disagreements in observed elections across different institutions.
Three rules inspect different features of one profile
Criterion or rule
Relevant operation
Result
Plurality
Compare first choices 40 / 35 / 25
A wins
Instant-runoff
Eliminate C; transfer its 25 to B
B wins 60–40
Strict Condorcet criterion
C defeats A and B separately
C is Condorcet winner
The complete compromise rankings are held fixed. This demonstrates disagreement, not an empirical frequency or a judgment about which criterion should dominate.
These calculations assume complete strict rankings of the invented candidates. Strict means no equal ranks, and complete means every candidate appears. The supplied pairwise function requires those orders; it does not fill gaps in the optional-exhaustion profile. Approval and score reports need additional conventions as well. A voter who places C second might approve C or reject C under different acceptance thresholds. A rank also does not quantify how much better C seems than another candidate. Pairwise support measures ordered comparisons, not approval rates or interpersonal welfare totals.
A majority can cycle
Consider thirty-five A>B>C voters, thirty-three B>C>A voters, and thirty-two C>A>B voters. A beats B by sixty-seven to thirty-three. B beats C by sixty-eight to thirty-two. C beats A by sixty-five to thirty-five. There is no candidate who defeats every rival.
Every individual ranking is consistent, but the majority relation cycles. Transitivity would require A above C whenever A is above B and B is above C. A sequence of pairwise elimination votes can then depend on the agenda. More elaborate Condorcet methods specify how to resolve cycles; the word Condorcet alone does not name a complete counting algorithm.
Find the changing majority coalition
Now use thirty-five A>B>C voters, thirty-three B>C>A voters, and thirty-two C>A>B voters. Every individual order is transitive. Someone with A>B>C ranks A above C as well as A above B and B above C. Nothing in these papers gives an individual a circular order. A cycle can nevertheless appear when different groups form a majority on different questions. To see why, retain the group identities while adding their counts.
Against B, A receives the thirty-five A>B>C voters and the thirty-two C>A>B voters. Together they supply sixty-seven, defeating B's thirty-three. Against C, B receives the thirty-five A>B>C voters and the thirty-three B>C>A voters. Their sixty-eight defeat C's thirty-two. If society were one person with a transitive ranking, these first two relations would require A above C. But the final comparison uses another majority. The thirty-three B>C>A voters and thirty-two C>A>B voters prefer C to A, giving C sixty-five against A thirty-five.
No candidate defeats every rival. A loses to C, B loses to A, and C loses to B. The majorities overlap, but no single coalition of voters supports every winning direction. Each group appears in two of the three winning coalitions and opposes the third. Adding sixty-seven, sixty-eight, and sixty-five would produce two hundred winning comparison entries, not two hundred distinct supporters. There are still one hundred voters.
Different groups form the three majorities
Strict victory
Supporting groups
Opposing group
Winning count
A over B
35 A>B>C + 32 C>A>B
33 B>C>A
67 of 100
B over C
35 A>B>C + 33 B>C>A
32 C>A>B
68 of 100
C over A
33 B>C>A + 32 C>A>B
35 A>B>C
65 of 100
Every individual order is transitive. Each group joins two winning coalitions and opposes the third. The sum 200 is winning comparison entries, not distinct people.
A sequential agenda can turn that cycle into different final choices. Suppose A first faces B, the winner then faces C, and every comparison follows the stipulated rankings sincerely. A defeats B sixty-seven to thirty-three, then C defeats A sixty-five to thirty-five. C finishes as the survivor. Begin instead with B against C. B wins sixty-eight to thirty-two, then loses to A sixty-seven to thirty-three. A survives that agenda. Starting with C against A would leave C to face B, so B would survive the third possible opening pair.
These are paper agendas, not controls in the pairwise lab and not predictions about a legislature's actual procedure. They assume the same participants, sincere votes, a fixed candidate set, and no strategic anticipation of later rounds. Real participants might respond to the agenda or have rules about reconsideration. The mechanical example establishes that the final survivor can depend on comparison order under those assumptions. Calling an agenda fair or unfair would require criteria about who chooses it and which alternatives deserve consideration, as well as a description of the institution's authority.
Change only the order of sincere pairwise contests
Opening contest
Opening survivor
Final contest
Final survivor
A versus B
A, 67–33
A versus C
C, 65–35
B versus C
B, 68–32
B versus A
A, 67–33
C versus A
C, 65–35
C versus B
B, 68–32
Paper agendas for the cycle profile, not lab controls. Every contest uses the same 100 voters and sincere stipulated rankings; strategic anticipation is not modeled.
A method described as Condorcet needs a cycle-resolution rule if it must always select an officeholder. Requiring election of a Condorcet winner whenever one exists does not specify what happens here, because none exists. Different complete methods can share that requirement and handle this profile differently. The graph is therefore a diagnosis of the majority relation, not a complete election algorithm. When reading a result, distinguish no strict Condorcet winner from no possible decision. The institution can still decide, but must supply a procedure beyond the strict-winner definition.
Work through the mechanism
Which candidate defeats which?
Changing the profile replaces all rankings and pairwise counts immediately. It does not animate a hypothetical sequence of elections.
Stipulated ballots, not inferred from first-choice totals
Voters
Ranking, best first
40
A > C > B
35
B > C > A
25
C > B > A
Voters ranking the row above the column
Row over column
A
B
C
A
Not a contest
40
40
B
60
Not a contest
35
C
60
65
Not a contest
Condorcet winner: C, with a strict victory over each rival.
Each arrow points from the winner to the loser of one pairwise contest. Its label gives the winning count out of 100. The same voter participates in several hypothetical comparisons, so matrix totals must not be added as if they were distinct people.
Profile
A versus B
B versus C
C versus A
Complete compromise profile
A loses 40–60
B loses 35–65
C wins 60–40
Cycle profile
A wins 67–33
B wins 68–32
C wins 65–35
Pairwise totals use the same voters separately in each hypothetical contest; they are not summed across contests.
Read the matrix without double counting
Select the compromise preset in the pairwise lab. The table counts voters ranking the row candidate above the column candidate. Read the cell in row C and column A as sixty voters preferring C to A. The opposite cell, row A and column C, is forty. Read the row label first each time; reversing that convention reverses every claimed victory. Diagonal cells are not contests because comparing a candidate with itself cannot split the electorate into two strict preference groups. They should not enter a candidate's win count.
Check every opposite pair before interpreting the diagram. In the compromise matrix, A/B is forty against sixty, A/C is forty against sixty, and B/C is thirty-five against sixty-five. Each pair totals one hundred because each complete strict ranking puts exactly one member of that pair first. The three comparisons together contain three hundred directed support entries when both sides are included. That is three answers per voter across three questions, not three hundred people. The matrix's unit is voters per named comparison, and the comparison must remain attached to the number.
The directed graph summarizes strict victories rather than transfers. In the compromise profile, its directions are B toward A, C toward A, and C toward B. C has a winning direction to each rival and none returning to it. The labels give the winning counts out of one hundred. Their placement is schematic; an edge's length does not measure winning margin or ideological distance. For a margin, subtract the opposite count. C's sixty against A forty is a twenty-vote margin, while C's sixty-five against B thirty-five is a thirty-vote margin.
Winning counts and margins have different meanings
Compromise pair
Support in first direction
Support in reverse
First direction margin
B over A
60
40
+20 votes
C over A
60
40
+20 votes
C over B
65
35
+30 votes
Each support pair sums to 100. A margin subtracts the opposing count; it is not the winning count and not a percentage of a new electorate.
Switch to the cycle preset and predict the matrix before looking at its graph. A's row contains sixty-seven against B and thirty-five against C. B's row contains thirty-three against A and sixty-eight against C. C's row contains sixty-five against A and thirty-two against B. The opposite cells still sum to one hundred, so conservation of comparison counts does not imply a transitive social order. The graph now forms a directed cycle. The preset replaces all rankings and results immediately; there is no transfer stage or hypothetical election sequence to advance.
For a numerical cross-check, calculate the cycle margins from both sides rather than reading only the winners. A's margin over B is sixty-seven minus thirty-three, or thirty-four. B's margin over C is sixty-eight minus thirty-two, or thirty-six. C's margin over A is sixty-five minus thirty-five, or thirty. All three victories are strict even though there is no candidate who wins all contests. Selecting the largest margin alone would add a new decision rule, not discover a missing Condorcet winner.
The cycle matrix preserves every pair total
Row preferred over column
A
B
C
A
Not a contest
67
35
B
33
Not a contest
68
C
65
32
Not a contest
Units are voters per comparison. A beats B by 34 votes, B beats C by 36, and C beats A by 30. Opposite entries sum to 100; the six off-diagonal entries sum to 300 comparison answers.
Keep the matrix available while explaining the graph aloud. For each edge, identify a supporting ballot group, an opposing group, and the pair's denominator. Then explain why C is a strict winner in one preset but not the other without appealing to color or node position. This practice separates a representation from the operation it represents. If a copied diagram and its matrix disagree, reconstruct the counts from the ranking table.
Keep the information requirements visible
The matrix uses complete strict rankings. Each off-diagonal entry counts voters who put the row candidate above the column candidate. Opposite entries sum to the number of voters. Diagonal cells are not contests, and the model does not turn a tie into a strict victory.
For incomplete rankings or indifference, a real rule must say how missing or tied preferences are treated. Nor does ordinal rank measure how strongly someone prefers one outcome. Pairwise majorities answer a particular preference question; they do not directly measure welfare, legitimacy, or willingness to compromise.
Separate criteria from theorem claims
A strict Condorcet winner must win every comparison, so a tie can prevent that status without creating a cycle. Consider fifty A>B>C papers and fifty B>A>C papers. A and B tie fifty to fifty, and each defeats C one hundred to zero. Neither strictly defeats every rival. The example has no three-way cycle; it has an unresolved top comparison. A complete voting method would need a tie procedure or another criterion. Do not silently import the instant-runoff elimination priority into the definition of a strict pairwise victory.
This tied profile is a paper contrast rather than a third lab preset. The matrix uses complete rankings, and both ranking groups supply a definite answer to every comparison. The tie arises from equal opposing numbers, not from individual indifference or missing information. If ballots allowed tied ranks or omitted candidates, the counting convention would need to specify whether those reports contribute to either side or change the comparison denominator.
No strict winner can mean a tie, not a cycle
Comparison
First side
Second side
Relation
A versus B
50
50
Tie
A versus C
100
0
A strictly wins
B versus C
100
0
B strictly wins
Paper profile: 50 A>B>C and 50 B>A>C. No individual is indifferent. Neither A nor B strictly defeats every rival. No alphabetical elimination rule is part of this criterion.
Arrow's theorem concerns a different object from choosing a single officeholder. It studies a rule that maps profiles of individual rankings into a complete transitive social ranking. With at least two voters, at least three alternatives, and an unrestricted domain of individual orders, no such rule can simultaneously satisfy unanimity, independence of irrelevant alternatives, and non-dictatorship. Independence concerns each social pair comparison depending only on individuals' rankings of that pair. The theorem is a universal incompatibility claim under its assumptions, not a claim that every submitted profile has a majority cycle or that all rules always fail.
The Gibbard-Satterthwaite theorem instead concerns deterministic single-winner choice with at least three attainable outcomes and unrestricted strict preferences. Under those conditions, a strategy-proof rule must be dictatorial. Strategy-proof means no individual can ever obtain a strictly preferred outcome by misreporting while others' reports stay fixed. This does not mean every voter can always gain by lying. Arrow addresses properties of an aggregate order; Gibbard-Satterthwaite addresses incentives for a choice rule. Neither theorem measures manipulation frequency or identifies a preferable system.
The majority cycle explains one obstacle to treating society as a single ranked chooser, but it is not a complete proof of either theorem. Showing one cyclic profile disproves the claim that pairwise majority always produces a transitive order on unrestricted rankings. It does not establish every assumption or every quantified statement in a general impossibility result. Likewise, agreement between plurality and pairwise choice in one profile would not prove that those methods always agree. A counterexample can defeat a universal assertion; a favorable example cannot establish one without a general argument.
An aggregate order is not a single-winner choice
Result
Object and domain
Incompatible requirements or implication
Arrow
Complete transitive social ranking; unrestricted orders; at least 3 alternatives
Unanimity, independence of irrelevant alternatives, and no dictator cannot all hold
Gibbard-Satterthwaite
Deterministic single winner; unrestricted strict preferences; at least 3 attainable outcomes
Strategy-proofness implies dictatorship
These are universal statements about rules under their assumptions. Neither says every profile cycles, every voter can gain by lying in every profile, or every institutional objective is equivalent.
For an unfamiliar claim, first identify the requested output. Is it a candidate for one office, an ordering of alternatives, or a distribution of legislative seats? Then identify the evidence. A matrix establishes conditional comparison counts, a behavioral study may address strategic reporting, and a normative argument must name its objectives. The selected Stanford discussion supplies formal context for cycles and theorems, not a current statutory instruction or a full empirical comparison of reforms. Use the two lab profiles as reproducible counterexamples while keeping those broader questions separate and their assumptions visible.
Exercises
Find the compromise winner
In the complete 40/35/25 profile, C defeats A 60–40 and B 65–35. What does that establish, and why does instant-runoff still eliminate C?
Hint
Pairwise victories and first-choice totals are different quantities.
Worked answer
Criterion
Relevant count
Outcome
Condorcet
C beats both rivals
C
Initial IRV elimination
C has fewest first choices
Eliminate C
The disagreement is produced by distinct rules on the same rankings.
C is the Condorcet winner because it strictly defeats each rival. Instant-runoff initially compares first choices, where C has twenty-five against A’s forty and B’s thirty-five. Its elimination rule does not search for the pairwise winner.
Does a cycle imply irrational voters?
A beats B, B beats C, and C beats A in the cycle profile. Must some individual have submitted a cyclic ranking?
Hint
Inspect the three complete individual ranking types.
Worked answer
Comparison
Supporting groups
Total
A over B
35 + 32
67
B over C
35 + 33
68
C over A
33 + 32
65
The coalition forming the majority changes between comparisons.
No. A>B>C, B>C>A, and C>A>B are each transitive individual orders. Different majorities support different comparisons, so the aggregate relation cycles. Treating the electorate as a single person with a single ranking would hide the mechanism.
Sources and further reading
The explanations, counts, and worked answers are complete without these optional links. Reference annotations identify the selected material reviewed. Historical country examples are not statements of current law, and the bounded models are not certified election-counting software. Mechanical results, behavioral evidence, and judgments about desirable representation remain separate.
Selected discussions of majority cycles, preference aggregation, Arrow’s theorem, approval/cardinal information, and the Gibbard-Satterthwaite theorem. The lessons state assumptions rather than treating impossibility results as a universal ranking of systems. Its choice-set discussion allows weak pairwise winners; this course uses strict victories for a Condorcet winner.
Selected explanations of system families, district magnitude, thresholds, proportional formulas, mixed systems, and design objectives. Country examples and legal thresholds are historical, not statements of current law.