Tournament
Named by 5 essays across 2 fields — each of them below, with the objects they name alongside it.
The majority that goes in a circle
Every voter hands in a ranking, and a ranking is transitive by construction. Compare the candidates two at a time and let the majority decide each pair, and the verdicts need not fit together into a ranking at all.
How often the majority goes in a circle
Three voters and three candidates give 216 profiles, and 12 of them are cycles. Count every electorate up to 41 voters exactly and the share climbs towards 8.77%, a number Guilbaud found in 1952 as the solid angle where three half-spaces at the tetrahedral angle overlap. Add candidates and a winner goes missing half the time; let voters share one axis and cycles vanish. The number is always a property of the model of how ballots are drawn.
How few voters any majority needs
Any pattern of head-to-head majorities whatever — cycles within cycles, a candidate who beats the winner of every other contest and loses to its loser — can be produced by voters who each rank the candidates sensibly. McGarvey's recipe needs n(n − 1) of them for n candidates. The truth is far fewer: every pattern on five candidates takes three voters at most, a counting argument shows the number must eventually grow, and it grows only like n divided by its logarithm.
The sizes at which a sentence can come true
Write a sentence about a relation and ask for which numbers of points some relation makes it true. "Every two points have exactly one common neighbour" comes true at every odd size and no even one, and every graph that satisfies it is a windmill of triangles. "Every two players are both beaten by some third" fails for every tournament of two to six players and holds from seven on. A search can settle one size at a time; only a proof settles all of them, and the two sentences need different kinds.
Six choices make a seventh
Bertrand Russell's millionaire needs the axiom of choice to pick one sock from each of infinitely many pairs. Does having that power for pairs give the same power for sets of four? It does, by a short trick that Tarski found: choosing from the six pairs inside a set of four makes a tournament, and all 64 tournaments single out something definite. For sets of three it fails, and whether a size follows from smaller ones is settled almost entirely by arithmetic: from pairs, 4 is the only other size that can follow.
Named alongside it
The objects these essays reach for when they reach for this one.
Exhaustive searchCondorcet cyclePairwise majorityPreference profileVoting ruleAxiom of choiceBacktrackingChoice functionCondorcet paradoxCounterexampleCounting argumentEigenvalue