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Same matrix, different plurality — three electorates the pairwise table cannot tell apart

Three 12-ballot electorates. Identical pairwise results — every head-to-head count, every margin, the same Condorcet winner, the same Borda scores. Ranked Robin, Minimax, Ranked Pairs and Kemeny cannot tell them apart, and must all return the same winner on all three. Choose-One returns a different winner on each — one per candidate. That is the whole of what a method reads, demonstrated in ballots you can count by hand.

Level: 301 · deep dive Concept: what a method reads · summability · the C1/C2/C3 tiers. LH-only — the lesson is the contrast across three electorates, which a single BetterVoting election cannot show.

The three electorates

Same cast throughout (Ada, Ben, Cal) — it's one election with the ballots rearranged, so a new cast would imply a different election.

Ballots (12 each) First choices Choose-One winner
P1 4× Ada>Ben>Cal · 4× Cal>Ben>Ada · 2× Ben>Cal>Ada · 1× Ben>Ada>Cal · 1× Ada>Cal>Ben Ada 5, Cal 4, Ben 3 Ada
P2 2× Ada>Ben>Cal · 2× Cal>Ben>Ada · 3× Ben>Ada>Cal · 2× Ben>Cal>Ada · 1× Ada>Cal>Ben · 2× Cal>Ada>Ben Ben 5, Cal 4, Ada 3 Ben
P3 3× Ada>Ben>Cal · 3× Cal>Ben>Ada · 3× Ben>Ada>Cal · 1× Ben>Cal>Ada · 2× Cal>Ada>Ben Cal 5, Ben 4, Ada 3 Cal

The pairwise table — the same one, three times

All three electorates produce this, exactly:

Round-Robin — every pair, head-to-head (For – Against):
   Ben  beats Ada   7 – 5
   Ada  ties  Cal   6 – 6
   Ben  beats Cal   7 – 5

Verified: the three reports are byte-identical on those lines. So Ranked Robin elects Ben in all three — and so would Minimax, Ranked Pairs, Schulze and Kemeny, because there is nothing in their input that differs.

Borda too. A Borda score is just a row of that table added up: with margins Ada −2, Ben +4, Cal −2, the scores are Ada 11, Ben 14, Cal 11 in every one of the three electorates (why that identity holds).

Why the ballots differ but the table doesn't

The trick is that a ballot and its exact mirror image cancel out pairwise. Ada>Ben>Cal and Cal>Ben>Ada together put one vote on each side of every head-to-head — every count and every margin is untouched. But they do not cancel in the first-choice tally: one hands a first preference to Ada, the other to Cal. Swap mirror pairs in and out and you can move the plurality winner anywhere you like while the pairwise table sits perfectly still.

That is not a quirk of these particular ballots. It's the structural fact: first-choice counts are not recoverable from the pairwise matrix.

What this does and does not show

It shows that Choose-One's winner is not a function of the pairwise data — the sense in which plurality sits outside the matrix (Fishburn C3). It makes vote splitting precise rather than rhetorical: in P1 and P3 the plurality winner isn't even the candidate a majority prefers head-to-head.

It does not show that plurality is "more complex," or that reading less is a defect in itself. Choose-One publishes the smallest possible precinct summary — one number per candidate — and is the cheapest method here to audit. The tiers classify which statistic a rule reads, not how big it is or how hard it is to count; the two ideas come apart at exactly this method (summability).

Note P2, where plurality happens to agree with everyone else and elects Ben. Nothing forces the disagreement — the point is that plurality is free to differ, not that it always does.

Run it

.venv/bin/python STARVote_LH_tabulation_engine/starvote_larry_hastings.py method_comparisons/same_matrix_different_plurality/cases/same_matrix_p1_ranked_robin.yaml
Electorate Ranked Robin Choose-One
P1 page · …p1_ranked_robin.yaml → Ben page · …p1_plurality.yaml → Ada
P2 page · …p2_ranked_robin.yaml → Ben page · …p2_plurality.yaml → Ben
P3 page · …p3_ranked_robin.yaml → Ben page · …p3_plurality.yaml → Cal

The Choose-One files encode each ballot as a single 1 — which is exactly what a Choose-One ballot is, and makes visible what it discards.