The minimal tilted cycle — five voters, and already the methods disagree¶
Five voters. Three candidates. Every voter perfectly rational, the electorate not: Ada beats Ben 4–1, Ben beats Cara 3–2, Cara beats Ada 3–2. This is the smallest election in existence in which a majority cycle is lopsided — and that lopsidedness is exactly what pries the Condorcet family apart. Ranked Robin's Copeland count still ties all three; the maximin family drops Ben. Five ballots is all it takes. This page proves the minimality — by hand and by brute force — and runs every rule on it.
→ The theorem behind it: Condorcet-Consistent Choice Among Three Candidates · the collapse it complicates: three-candidate collapse · the symmetric sibling: reinforcement paradox — North district · cycle resolution at four candidates · the tie-break shape: three-way dead rung.
The ballots¶
Each row is one voter's ranking, most-preferred first (N: prefix = N identical ballots).
2:Ada>Ben>Cara
1:Ben>Cara>Ada
2:Cara>Ada>Ben
Same cast as the reinforcement pair on purpose — it is the same paper's three-candidate world, with one thing changed: the cycle is tilted instead of symmetric.
What the cycle looks like¶
full report → cases/cases_pages/tilted_cycle_c3_b5_rr.md · _tabulated mirror
Round-Robin — every pair, head-to-head (For – Against):
Ada beats Ben 4 – 1
Cara beats Ada 3 – 2
Ben beats Cara 3 – 2
Win–loss record — Copeland score = wins + ½·ties:
# Candidate W–L–T Copeland Margin Beats
1 Ada 1–1–0 1 +2 Ben
2 Cara 1–1–0 1 +0 Ada
3 Ben 1–1–0 1 -2 Cara
Winner — Ranked Robin (RCV-RR): Ada
*** 3 candidates tie for the most wins — a Condorcet cycle.
Nobody is a Condorcet winner; nobody is a Condorcet loser. Every individual ballot is transitive. The intransitivity is manufactured purely by aggregation — that is the whole content of the word cycle.
Why five voters, and why 4–1 / 3–2 / 3–2¶
The numbers are not chosen for teaching convenience. They are the only numbers available. Here is the argument in full.
Fix the cycle direction A→B→C→A and look at the three margins along those arcs. Any single voter's ranking is transitive, so it agrees with either two of the three arcs or one of them (never three — that would be a cyclic ranking, and never zero — that would be the reverse cycle). So each voter contributes to the sum of the three cyclic margins:
- agrees with 2 arcs, disagrees with 1 →
(+1) + (+1) + (−1) = +1 - agrees with 1 arc, disagrees with 2 →
(+1) + (−1) + (−1) = −1
Therefore the three cyclic margins always sum to at most n (with equality only when every voter agrees with two arcs). And a second constraint comes free: with n voters each pairwise margin is (votes for) − (votes against) out of n, so every margin shares n's parity. Now walk up the electorate sizes, requiring all three margins strictly positive (that is what "cycle" means):
| n | margins must be… | and sum ≤ n, so… | result |
|---|---|---|---|
| 1, 2 | ≥ 1 each, sum ≥ 3 | 3 > 2 | no cycle possible |
| 3 | odd, ≥ 1 each, sum ≥ 3 | sum = 3 exactly | (1,1,1) only — forced symmetric |
| 4 | even, ≥ 2 each, sum ≥ 6 | 6 > 4 | no cycle exists at all |
| 5 | odd, ≥ 1 each, sum ≤ 5 | (1,1,1) or (3,1,1) | first tilted cycle — and only one shape |
| 6 | even, ≥ 2 each, sum ≤ 6 | (2,2,2) only | forced symmetric again |
So: three voters give you a cycle but a perfectly symmetric one, in which no rule can prefer anyone. Four voters give you no cycle whatsoever. Five voters is the first electorate that can tilt a cycle, and (3,1,1) — margins of 3, 1, 1, i.e. 4–1, 3–2, 3–2 — is the only tilted shape it can produce. That is why the paper reaches for exactly this profile: it is not a small example, it is the minimal one, uniquely determined.
The charming coda is n = 6: still forced symmetric, (2,2,2) — which is precisely the North district's rock-paper-scissors tie. The first even electorate that can tilt is eight.
Machine-checked¶
minimality_check.py in this folder enumerates every profile of strict rankings over three candidates for n = 1…8 — no engine, no library, pure standard library — and reports the margin shapes that actually occur:
n cycles found distinct margin shapes (largest first)
-- ------------ ---------------------------------------
1 0 —
2 0 —
3 2 (1, 1, 1)
4 0 —
5 12 (3, 1, 1), (1, 1, 1)
6 2 (2, 2, 2)
7 42 (5, 1, 1), (3, 3, 1), (3, 1, 1), (1, 1, 1)
8 12 (4, 2, 2), (2, 2, 2)
python3 method_comparisons/minimal_tilted_cycle/minimality_check.py
Exhaustive search agrees with the parity argument line for line.
What the tilt buys: the methods split¶
Every winner below is computed, not asserted — Ranked Robin and RCV-IRV by the LH engine (the two cases/ files), the rest by pref_voting as an independent witness.
| Rule | Winner(s) | Reads | Why |
|---|---|---|---|
| Copeland / Ranked Robin | Ada, Ben, Cara — a three-way tie | wins | every record is 1–1 |
| ↳ LH's Ranked Robin, after tie-break | Ada | +margins | total margin +2 / 0 / −2 |
| Minimax (maximin) | Ada, Cara | margins | worst loss −1 each; Ben's is −3 |
| Ranked Pairs | Ada, Cara | margins | the three-candidate collapse — |
| Schulze (beat path) | Ada, Cara | margins | at m = 3 these are |
| Split Cycle | Ada, Cara | margins | one and the same rule |
| Stable Voting | Ada, Cara | margins | as maximin |
| Kemeny–Young | Ada, Cara | margins | ditto |
| Dodgson | Ada, Cara | margins | ditto |
| Leximax (leximin) | Ada | margins | tied at −1, then +3 beats +1 |
| Nanson — strict | Cara | Borda | Ben out (4 < avg 5), then Cara beats Ada 3–2 |
| Nanson — weak | Ada | Borda | Ben and Cara out (≤ avg), Ada left standing |
| Borda | Ada | points | 6 / 5 / 4 |
| Bucklin · Coombs | Ada | ranks | — |
| Plurality | Ada, Cara | 1st choices | 2 / 2 / 1 |
| RCV-IRV | Cara | 1st choices | Ben out on 1, transfers to Cara → 3–2 |
Two things to take from that table.
First, the headline: Copeland is not in the maximin family, and five voters proves it. On the symmetric cycle every one of these Condorcet rules returns all three names — indistinguishable. Tilt by a single voter and the margin-reading rules immediately see that Ben's worst defeat (−3) is far worse than anybody else's (−1) and drop him, while Copeland — which counts only whether you won, never by how much — still reports 1–1, 1–1, 1–1. That is the practical content of "Ranked Robin doesn't inherit maximin's no-show immunity": it is a different rule, and this is the smallest election where you can watch it be different.
Second, and fairly: nobody here is wrong. Ada, Cara, and the {Ada, Cara} set are all defensible answers, and IRV's Cara is a perfectly legitimate reading — in a cycle there is no Condorcet winner to miss, so none of these rules can be accused of missing one. What the profile shows is the cost of a cycle: five ballots, and reasonable rules already land in different places. Note too that strict and weak Nanson — two spellings of the same rule — disagree with each other (Cara vs Ada), though both stay inside maximin's {Ada, Cara}. Precision about tie-breaks is not pedantry; it changes the winner.
The "three-way dead rung" connection¶
Ranked Robin's tie here has the shape of the three-way dead rung — every candidate tied on the deciding measure — but only one rung deep:
| Rung | Symmetric 6-voter cycle | This tilted 5-voter cycle |
|---|---|---|
| Copeland (wins) | 1–1, 1–1, 1–1 — dead | 1–1, 1–1, 1–1 — dead |
| Total margin | 0 / 0 / 0 — dead | +2 / 0 / −2 — alive → Ada |
| Lot | decides the winner | never reached |
So the tilt is doing double duty: it is what lets maximin separate the candidates, and it is what lets Ranked Robin finish deterministically instead of falling to a coin. The dead rung is real but shallow.
And there's a theorem under that table. The six-voter tie isn't merely deeper — it is mathematically forced: no anonymous, neutral, Pareto rule can name a single winner whenever the voter count n has a divisor r with 1 < r ≤ m candidates (Moulin, 1983). At n = 6, m = 3 both 2 and 3 qualify, so every defensible rule ties there and the lot isn't a shortcut — it's the only thing left. At n = 5, m = 3 nothing is forced (5 is prime and exceeds 3), which is exactly why five voters is the interesting number: the ballots still carry an asymmetry, and a rule that can't find it is discarding information rather than running out of it. That's the difference between the dead margin row and the live one. Full treatment: Ties Are Forced.
Why this case is LH-only¶
No BetterVoting election backs it, deliberately. A Copeland three-way tie is precisely the case where BV breaks the tie at random (head-to-head, then random) while LH breaks it by margin, then lot. A random BV result cannot be frozen into a stable teaching artifact, so minting a permanent public election here would produce an unreproducible page. The tie-break divergence page covers that difference on its own terms.
The cases¶
| Case | Method | Winner | Page | Source |
|---|---|---|---|---|
| Tilted cycle, Ranked Robin | Ranked Robin (Copeland) | Ada (after margins tie-break) | page | yaml |
| Tilted cycle, RCV-IRV | RCV-IRV | Cara | page | yaml |
Reproduce the wider table:
uv run STARVote_LH_tabulation_engine/tools_adam/pref_voting_tabulation_engine/cycle_resolution_report.py method_comparisons/minimal_tilted_cycle/cases/tilted_cycle_c3_b5_rr.yaml
Source: Felix Brandt, Chris Dong & Dominik Peters, "Condorcet-Consistent Choice Among Three Candidates" (arXiv:2411.19857, 2024; author PDF) — neutral academic social-choice theory, summarized here. Cross-checks by pref_voting (Eric Pacuit & Wesley Holliday). Glossary: Condorcet cycle · Copeland.