Reinforcement paradox — when both halves pick Ada, but the whole picks Cara¶

Two districts. Ada wins both. Merge them, and Cara wins — the one thing a voting rule's "reinforcement" (a.k.a. consistency) promise says can't happen. This is a live, countable demonstration of a theorem from Brandt, Dong & Peters, "Condorcet-Consistent Choice Among Three Candidates" (2024): every Condorcet method must show this paradox once there are ≥ 8 voters. The lesson is fair to a fault — it cuts against STAR too, and shows exactly which methods keep the promise and which don't.

▶ Live on BetterVoting: vote · results ↗ (election t4by6x, BV2254) — the combined 9-voter electorate as two races (STAR + Ranked Robin); both elect Cara, matching the counts below.

→ The theorem behind it: Condorcet-Consistent Choice Among Three Candidates · related: the No-Show paradox · multiple-districts / consistency · cycle resolution · Runoff Reversal.

→ The same event with a price tag instead of a checkmark: distributed voting measures what slicing an electorate into districts costs in voter welfare (answer: a factor of k, the number of districts — and it survives even when every district counts perfect utilities). Consistency asks did the promise break; distortion asks how much did the break cost.

The whole story in one table¶

Nine voters in two towns choosing among Ada, Ben, and Cara. Every winner below is engine-verified (the four cases/ files):

Electorate	Score/Range	Approval	Plurality	STAR	Ranked Robin (Condorcet)
South (3 voters)	Ada	Ada	Ada	Ada	Ada
North (6 voters)	tie	tie	tie	tie	tie (a 3-way cycle)
Combined (9)	Ada ✅	Ada ✅	Ada ✅	Cara ⚠️	Cara ⚠️

Reinforcement (consistency) says: if a candidate wins two separate electorates, she must win their union. Ada wins South outright and is a co-winner of North's dead heat — she is a winner in both. So reinforcement demands Ada win the combined election. Three methods keep that promise; two break it. The rest of this page is why.

The three counts¶

South (3 voters) — Ada is the clear Condorcet winner. full report → cases/reinf_south_c3_b3_rr.yaml

   Ada  beats Cara   2 – 1
   Ada  beats Ben    3 – 0
   Cara beats Ben    3 – 0
Winner — Ranked Robin: Ada (beats every opponent head-to-head — the Condorcet winner)

North (6 voters) — a perfect rock-paper-scissors tie. Ada, Ben, and Cara each win one matchup and lose one; every method ties them (Ranked Robin's Copeland count is 1–1 all round). Ada is among the winners. full report → cases/reinf_north_c3_b6_rr.yaml

   Ada  beats Ben    4 – 2
   Ben  beats Cara   4 – 2
   Cara beats Ada    4 – 2      → 3-way tie (a Condorcet cycle)

Combined (9 voters) — a new Condorcet winner appears: Cara. full report → cases/reinf_combined_c3_b9_rr.yaml

   Ada  beats Ben    7 – 2
   Cara beats Ada    5 – 4
   Cara beats Ben    5 – 4
Winner — Ranked Robin: Cara (beats every opponent head-to-head — the Condorcet winner)

Cara now beats everyone head-to-head, so every Condorcet method elects her — and Ada, the only candidate who won both halves, loses. That is the reinforcement paradox.

Why additive methods keep the promise — and Condorcet methods can't¶

The split is not an accident of these numbers; it's structural.

Score, Approval, and Plurality just add points across ballots. A candidate's combined score is her South score plus her North score. Ada leads South and ties North, so she leads the sum — always. This is Young's theorem (1975): additive scoring rules satisfy reinforcement by construction. No electorate can ever paradox them.
Condorcet methods count head-to-head majorities, which don't add up. Merging two electorates can create a pairwise majority that existed in neither half — here, Cara's 5–4 edges over both rivals emerge only in the union. Brandt, Dong & Peters prove this is unavoidable: at three candidates, every Condorcet extension shows the reinforcement paradox once there are ≥ 8 voters (found via a SAT solver; the bound is tight).

Two older theorems that sharpen both halves¶

The Brandt–Dong–Peters result above is a tight three-candidate statement. The classical literature bounds the same two claims more broadly, and the pair is worth having side by side.

The Condorcet half generalizes past three candidates. Zwicker's Proposition 2.5 states it without any ceiling: all Condorcet extension SCFs for three or more alternatives violate reinforcement. The proof is elementary — no SAT solver — and it runs on exactly the profiles in this folder: the symmetric 6-voter cycle as one electorate, a 3-voter district with a clear Condorcet winner as the other. (The convergence is worth noting: two papers three decades apart reach for the same construction. It is the canonical witness, which is why the same ballots keep reappearing.) So what BDP add is tightness at m = 3; what Zwicker adds is no upper limit on m — with Pareto assumed, extending the proof to four or more candidates is straightforward.

The additive half is an exact characterization, not just "scoring rules pass." The version usually quoted — Young's — runs one direction: additive scoring rules are reinforcing. The full result is an iff, and it names the whole class:

Theorem (Smith 1973; Young 1975). The anonymous, neutral, and reinforcing SCFs are exactly the compound scoring rules.

A compound scoring rule allows a cascade of score vectors: ties under w₁ are broken by score differences under a second vector w₂ (say, plurality score to separate tied Borda winners), a third if any remain, and so on for any finite number. Add one further axiom — continuity, aka the Archimedean property (for any s and any t with a unique winner x, enough copies of t eventually carry the merged election: f(s + j·t) = {x} for all large j) — and the class narrows to the simple, one-vector scoring rules.

That converse is what makes this page's verdict sharp rather than anecdotal. STAR fails reinforcement, so STAR is provably not a compound scoring rule — no cascade of score vectors, however elaborate, reproduces it. Its scoring round is a scoring rule; the automatic runoff is not, and the theorem says that is exactly the step where the promise had to break. Same for Ranked Robin, RCV-IRV, and every other method here with an elimination or runoff stage. Conversely it explains why Score, Approval and Plurality can never be paradoxed this way: they are simple scoring rules, and the theorem covers them by construction.

Whichever way the cycle falls — the other two branches¶

One detail the table above glosses, and it matters for anyone reproducing this.

North is a perfect three-way tie, so an anonymous, neutral rule can only return all three candidates as co-winners — and then the single South district above is enough, because Ada is inside that set. But the LH engine is resolute: it must print one name, and it gets there by spending neutrality on a published lot order (the lot_numbers field). With [Ada, Ben, Cara] it prints Ada — which is why the original trio works.

Change the lot and the original South no longer springs the trap: if North resolved to Ben, then North and South would share no winner at all, reinforcement's hypothesis would never fire, and nothing would look wrong. That is not a hole in the theorem — Zwicker's proof handles it by permuting the second district — but it is a hole in a demonstration built from one pair. So the other two branches are now built too. North is unchanged and invariant under the rotation Ada→Ben→Cara→Ada, so rotating South gives each branch:

If the cycle resolves to…	South district	South elects	Merged 9 voters elect
Ada	page · `reinf_south_c3_b3_rr`	Ada	Cara ⚠️
Ben	page · `reinf_south_ben_c3_b3_rr`	Ben	Ada ⚠️
Cara	page · `reinf_south_cara_c3_b3_rr`	Cara	Ben ⚠️

Every row is engine-verified, and every merged winner is a strict Condorcet winner with the same 5–4 / 5–4 / 7–2 signature, rotated:

Ben branch  (cases/reinf_combined_ben_c3_b9_rr.yaml)
   Ada   beats Ben    5 – 4
   Ada   beats Cara   5 – 4
   Ben   beats Cara   7 – 2
Winner — Ranked Robin: Ada   (both halves had said Ben)

Cara branch (cases/reinf_combined_cara_c3_b9_rr.yaml)
   Ben   beats Ada    5 – 4
   Ben   beats Cara   5 – 4
   Cara  beats Ada    7 – 2
Winner — Ranked Robin: Ben   (both halves had said Cara)

So the failure does not depend on a lucky lot draw. Whichever candidate the cycle is resolved to, there is a South district that agrees with that choice and a merged electorate that overturns it — which is the case analysis in Zwicker's proof, made runnable.

Where STAR lands — and the honest catch¶

STAR is a score ballot plus an automatic runoff, so it's half-and-half — and this case pins down exactly which half wins. On the combined 9 ballots:

Abridged for the lesson — not verbatim engine output

Scoring Round
   Ada   -- 29 -- First place     ← the pure Score result: Ada. Consistent. No paradox here.
   Cara  -- 27 -- Second place
   Ben   -- 16
Automatic Runoff Round
   Cara  -- 5 -- First place       ← the runoff is a pairwise step, and it
   Ada   -- 4                         catches Cara's head-to-head win — flips to Cara.
 Cara wins.

[Runoff Reversal]  Score Round Winner = Ada, Runoff Winner = Cara
[Condorcet Winner] Cara — matches the STAR winner
[Divergence from STAR] Choose-One = Ada, Approval = Ada (both differ from STAR)

STAR's scoring round elects Ada — the consistent, additive answer. But STAR's runoff re-imports the very pairwise flip that trips the Condorcet methods, and elects Cara. So STAR does not escape this paradox: it inherits it through the runoff. Saying so plainly is the point — STAR buys a majority-honest final round, and the price is that the final round can side with a merge-created majority against the additive winner. (Pure Score would have kept Ada; STAR's runoff is where consistency gives way.)

Want the whole count? Full LH report → cases/cases_tabulated/reinf_combined_c3_b9_star_tabulated.txt.

What to take away¶

"Both halves agreed, so the whole must agree" is a promise only some methods keep. Additive point methods (Score, Approval, Plurality) keep it; Condorcet methods provably cannot at ≥ 8 voters; STAR keeps it in the scoring round but can lose it in the runoff.
It's not a bug in Ranked Robin — it's a theorem about all of Condorcet. The three-candidate maximin result is the rigorous frame; this is its smallest concrete instance.
Fairness cuts both ways. The same page that credits Condorcet methods for guaranteeing the head-to-head winner has to concede they break consistency — and that STAR, via its runoff, can too. That candor is the repo's whole method.

Cases (all engine-verified) — the original trio: reinf_north_c3_b6_rr.yaml · reinf_south_c3_b3_rr.yaml · reinf_combined_c3_b9_rr.yaml · reinf_combined_c3_b9_star.yaml; the two rotated branches: reinf_south_ben_c3_b3_rr.yaml · reinf_combined_ben_c3_b9_rr.yaml · reinf_south_cara_c3_b3_rr.yaml · reinf_combined_cara_c3_b9_rr.yaml.

Sources: Felix Brandt, Chris Dong & Dominik Peters, "Condorcet-Consistent Choice Among Three Candidates" (arXiv:2411.19857, 2024) — profiles P1/P2 from the Theorem 2 proof, cast as Ada/Ben/Cara across North/South towns; the tight ≥ 8-voter bound at three candidates. Lean: neutral. · William S. Zwicker, "Introduction to the Theory of Voting," in Handbook of Computational Social Choice (CUP 2016), §2.6 — Proposition 2.5 (no ceiling on the candidate count) and Theorem 2.4 (Smith 1973; Young 1975), the compound-scoring-rule characterization and the continuity/Archimedean axiom. Lean: neutral.