Skip to content

The Dark Horse — how the Borda count can elect a nobody with zero honest support

A strategic pathology of the Borda count (and any strictly-ranked point method): a candidate every voter honestly ranks last can win, because each faction is tempted to bury its rivals behind the harmless nobody — and if they all do, the nobody backs into victory. It's a prisoner's dilemma in ballot form. Neither STAR nor Ranked Robin rewards this trap — a score (or a pairwise comparison) lets you oppose a rival without being forced to prop anyone up. (Condorcet methods like RR have a separate, sincere dark-horse seam that STAR's support-reading also avoids — see Keep it fair.)

→ Part of the strategic-pathology set: The strategic pathologies — five Molochs, and where STAR stands. Framing from Jameson Quinn (Center for Election Science). Related: strategic voting · what makes a good winner?.

The scenario (Quinn's "Dark Horse 3")

Four candidates. D is a nobody — the honest last choice of every single voter. Three factions:

Voters Honest preference
34 A > B > C > D
33 B > A > C > D
33 C > A > B > D

A is the honest winner every sensible way you count: the utilitarian favorite and the Condorcet winner (A beats B, C, and D head-to-head).

Honest Borda → A. All-strategic Borda → D. (verified with pref_voting)

Under Borda, each voter must hand out 3, 2, 1, 0 points. Honestly, that elects A:

HONEST Borda:   A 234,  B 200,  C 166,  D 0     → A wins

But the only way to hurt your rival under Borda is to rank someone above them — and the safe someone is the harmless D. So each faction is tempted to rank D second, burying its rivals at the bottom. If all three factions do it:

ALL-STRATEGIC Borda:  D 200,  A 168,  B 133,  C 99   → D wins

D — the honest last choice of 100% of voters — wins. Nobody wanted D; everyone only used D as a weapon against their rivals, and the weapon won the election.

Why it's a prisoner's dilemma

  • If one faction buries its rivals behind D, that faction profits.
  • If all factions do it, they all lose to D — the worst outcome for everyone.
  • Honesty is the socially best play, but each faction's individual incentive is to defect. Classic prisoner's dilemma — which is exactly why Quinn calls Dark Horse a "Moloch."

The root cause: Borda forces you to award points in rank order, so opposing a rival requires elevating someone else. There's no way to say "I oppose both my rival and the nobody equally."

STAR and Ranked Robin can't be dark-horsed

A score ballot breaks the trap: you give your rival a 0 and the nobody D a 0. Opposing a rival never props up anyone. Scored honestly (favorite 5, the compromise ~1, D a 0):

STAR Scoring Round
   A -- 236 -- First place
   B -- 232 -- Second place
   C -- 198
   D --   0            ← the nobody stays a nobody
Automatic Runoff
   A -- 34   vs   B -- 33     → A wins

[Divergence from STAR]  STAR = A ;  Ranked Robin (RCV-RR) = A
  • STAR → A (dark_horse_star) — D can never rise, because no voter has any reason to give D a point.
  • Ranked Robin → A — burying a rival below D doesn't help you win a single pairwise matchup, so the incentive to do it never exists; A beats everyone head-to-head.

That's Quinn's own prescription, satisfied: "don't force people to dishonestly support D merely in order to oppose some other candidate." A score ballot doesn't force it.

Keep it fair

  • This is a Borda problem, not a "ranked ballot" problem in general — Ranked Robin uses the same ranked ballots and is immune to this dark horse, because it reads pairwise wins, not positional points. It's positional point methods (Borda and its kin) that are dark-horse-prone in the strategic sense above.
  • But Condorcet has its own, milder dark horse — and this is where STAR pulls ahead of Ranked Robin. As Quinn notes, a Condorcet method can crown "somebody 90% of everybody's never heard of" — a bland unknown each side ranks second only because they rank them above the other side. That's not the strategic Borda trap; it's a sincere one, and it's a genuine Ranked Robin limit: a ranked ballot sees order, not strength, so a thinly-supported compromise everyone tolerates can beat a candidate a majority loves. STAR reads that strength — the unknown scored a 1 by everyone never survives the scoring round — so STAR resists both the strategic Borda dark horse and the sincere Condorcet one, exactly the preference-vs-support distinction.
  • It's the mildest Moloch. Dark Horse is "outstandingly evil but not particularly powerful" (Quinn): it needs a strictly-ranked point method almost nobody proposes for public office, and iterated "social glue" tends to hold honesty in place. The reason to name it is mechanism design — pick a method that doesn't create the trap at all.
  • STAR isn't strategy-proof in general (Gibbard forbids that) — see its own chicken/Burr seam and honest limits. The claim here is narrow and true: STAR has no Dark Horse.

Reproduce it

# STAR / Ranked Robin side (LH engine):
python3 STARVote_LH_tabulation_engine/starvote_larry_hastings.py method_comparisons/dark_horse_borda/cases/dark_horse_star.yaml

# Borda side (pref_voting — the LH engine doesn't tabulate Borda):
uv run python -c "
from pref_voting.profiles import Profile
from pref_voting.scoring_methods import borda
n={'A':0,'B':1,'C':2,'D':3}
def P(r): return Profile([tuple(n[c] for c in o) for _,o in r], rcounts=[c for c,_ in r])
honest=[(34,'ABCD'),(33,'BACD'),(33,'CABD')]
strat =[(34,'ADBC'),(33,'BDAC'),(33,'CDAB')]
print('honest ->',[list(n)[w] for w in borda(P(honest))])
print('strategic ->',[list(n)[w] for w in borda(P(strat))])"

Source of the framing: Jameson Quinn, "The Six Voting Molochs" (advocacy-adjacent — Quinn favors Approval/3-2-1, not STAR specifically, which makes the "STAR has no Dark Horse" point an outside verdict rather than our own cheerleading). The Borda math is standard and reproduced here with pref_voting.