BV2161 — Borda's SCC paradox: the winner flips when a loser exits

▶ Live on BetterVoting: vote · results ↗ (election q3h4fk).

7 voters, three candidates. Every live count picks C — STAR and Choose-One agree, and so does Borda's paper count. The paradox is what happens when B, a losing candidate, drops out: Borda flips to A.

Source

Dan S. Felsenthal, "Review of Paradoxes Afflicting Various Voting Procedures Where One Out of m Candidates (m ≥ 2) Must Be Elected", University of Haifa / LSE, revised 26 May 2010; Appendix §A5, Example 15.

Profile note (arithmetic correction): the paper prints the third bloc as c > a > b, but its own Borda totals — 6, 7, 8, summing to 21 — are only consistent with c > b > a. This case uses the arithmetic-consistent profile:

No. of voters    Preference ordering
      2          A > C > B
      2          B > A > C
      3          C > B > A

Borda, worked (the paper paradox)

Borda points (2/1/0): A 6, B 7, C 8 → C elected. Now let B — who lost — withdraw, ceteris paribus. Borda on the remaining pair: A 4, C 3 → A elected. A loser's exit flipped the winner: SCC (the subset choice condition — the formal spoiler) violated by Borda's count.

Honest note: with B gone, A beats C head-to-head 4–3, so any method elects A in the two-candidate contest — on a cyclic profile (B>A 5–2, A>C 4–3, C>B 5–2) every method's winner is exit-sensitive. Felsenthal's specific charge is that Borda's point arithmetic is what did the flipping. The cycle is also why no Ranked Robin or IRV race exists here (both would hit random ties on BV).

The live races

Race	BetterVoting	LH engine	Agree?
STAR (5/3/1 map: 19/21/23)	C (beats B 5–2 in the runoff)	C	✓
Choose-One (Plurality)	C (3 of 7)	C	✓
Borda (paper only)	—	C 8 → B exits → A 4–3	n/a

Files: star yaml · plurality yaml · frozen export · mirrors: star, plurality.