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Worst-case Strategy-proofness

T0 review · 1 major / 0 minor · reviewed 2026-06-26 · grok-4.3

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Pith's one-line read The anti-plurality rule with fixed-order tie-breaking satisfies worst-case strategy-proofness exactly when the numbers of agents and alternatives meet a specific numerical condition.

desk verdict The paper defines WCSP as an axiom strictly between strategy-proofness and NOM-worst, then gives a numerical necessary-and-sufficient condition for anti-plurality with fixed tie-breaking. read the letter →

arxiv 2606.22021 v1 pith:SMM34J37 submitted 2026-06-20 econ.TH

Kazumasa Ikeda This is my paper · ORCID
classification econ.TH
keywords worst-casestrategy-proofnessanti-pluralityrulevotingrulesnon-obviousmanipulabilitytie-breakingsocialchoice
verification ladder T0 review T1 audit T2 compute T3 formal

The pith

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The reading

The paper defines worst-case strategy-proofness, an axiom requiring that no voter can gain in the worst-case outcome by misreporting preferences. This axiom lies strictly between full strategy-proofness and the weaker non-obvious manipulability-worst. Several common rules, including plurality, Borda, and Dowdall, satisfy the weaker notion yet fail WCSP. For the anti-plurality rule with fixed-order tie-breaking the paper supplies a necessary and sufficient condition stated purely in terms of the counts of agents and alternatives.

What carries the argument

Worst-case strategy-proofness (WCSP), which demands that truthful reporting remains optimal for every agent even when all ties are resolved against the deviator.

What would settle it

For any specific pair of agent count n and alternative count m that the derived condition claims satisfies WCSP, exhibit one preference profile and one misreport under which the deviator obtains a strictly better outcome in every possible tie resolution.

Watch

Extended reading notes

Core claim

The central claim is that the anti-plurality rule with fixed-order tie-breaking meets worst-case strategy-proofness if and only if the numbers of agents and alternatives obey a stated numerical relation; the relation is derived directly from the requirement that no profitable worst-case deviation exists under the rule.

Load-bearing premise

The model assumes agents hold complete transitive preferences over a finite set of alternatives and that the tie-breaking rule is fixed and known in advance.

Editorial extensions

If this is right

  • Plurality, Borda, and Dowdall rules all violate WCSP.
  • Anti-plurality satisfies WCSP precisely under the identified numerical condition on agents and alternatives.
  • WCSP is strictly stronger than NOM-worst yet weaker than full strategy-proofness.

Reading between the lines

Editorial extensions of the paper, not claims the author makes directly.

  • The numerical condition may indicate for which small electorates anti-plurality can be used without fear of worst-case manipulation.
  • Similar conditions could be derived for other scoring rules or for randomized tie-breaking.
  • The gap between WCSP and NOM-worst suggests that many rules remain vulnerable only when agents consider the most adverse tie outcomes.
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Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, simulated authors' rebuttal, and a circularity audit.

Referee Report

1 major / 0 minor

Summary. The paper introduces worst-case strategy-proofness (WCSP) as a non-manipulability axiom that is weaker than strategy-proofness but stronger than NOM-worst. In a standard voting model with complete transitive preferences, it shows that plurality, Borda, and Dowdall rules violate WCSP, and states a necessary and sufficient numerical condition (in terms of the numbers of agents |N| and alternatives |A|) under which the anti-plurality rule with fixed-order tie-breaking satisfies WCSP.

Significance. If the stated characterization holds and is correctly derived, the result would be a clean, parameter-free delineation of the boundary for WCSP compliance in one specific rule. This could usefully extend the literature on intermediate manipulability axioms by identifying exact size thresholds separating compliance from violation.

major comments (1)
  1. [Abstract] Abstract: the manuscript asserts the existence of a necessary and sufficient condition for anti-plurality with fixed-order tie-breaking to satisfy WCSP, but provides neither the derivation, supporting lemmas, nor any verification that the condition indeed captures WCSP. Without these elements the central claim cannot be evaluated.

Simulated Author's Rebuttal

1 responses · 0 unresolved

We thank the referee for their careful reading and for highlighting this issue with the presentation of the central result. We address the comment below.

read point-by-point responses

  1. Referee: [Abstract] Abstract: the manuscript asserts the existence of a necessary and sufficient condition for anti-plurality with fixed-order tie-breaking to satisfy WCSP, but provides neither the derivation, supporting lemmas, nor any verification that the condition indeed captures WCSP. Without these elements the central claim cannot be evaluated.

    Authors: The manuscript states the necessary and sufficient numerical condition on |N| and |A| under which anti-plurality with fixed-order tie-breaking satisfies WCSP. We agree, however, that the derivation, supporting lemmas, and explicit verification that the condition is indeed necessary and sufficient are not adequately developed or presented. We will revise the manuscript to supply these elements in full. revision: yes

Circularity Check

0 steps flagged · score 0.0 of 10

No significant circularity; characterization is independent of inputs

full rationale

The paper defines the new axiom WCSP independently (weaker than strategy-proofness, stronger than NOM-worst from an external 2020 citation) and states a necessary-and-sufficient numerical condition on |N| and |A| for anti-plurality with fixed-order tie-breaking. No equations, fitted parameters, self-citations, or ansatzes appear in the abstract or described derivation chain that would reduce the result to its own inputs by construction. The modeling assumptions are standard and explicitly separated from the characterization result. This is a normal non-circular theoretical characterization.

Assumptions & free parameters 0 free parameters · 1 assumptions · 0 invented entities

The paper's central contribution rests on the introduction of a new axiom whose definition is internal to the work; no free parameters, external axioms, or invented entities are mentioned.

assumptions (1)
  • ad hoc to paper WCSP is a well-defined non-manipulability axiom that is weaker than strategy-proofness and stronger than NOM-worst
    The axiom is newly introduced by the paper.

how reviews work

0 comments
Cite this review

Pith. "Pith review of Worst-case Strategy-proofness." pith.science (2026). https://pith.science/paper/SMM34J37

@misc{pith2026260622021,
  author       = {Pith},
  title        = {Pith review of: Worst-case Strategy-proofness},
  year         = {2026},
  howpublished = {\url{https://pith.science/paper/SMM34J37}},
  note         = {Machine review of arXiv:2606.22021}
}
read the original abstract

We introduce a new non-manipulability axiom called worst-case strategy-proofness (WCSP). This axiom is weaker than strategy-proofness and stronger than non-obvious manipulability-worst (NOM-worst) by Troyan and Morrill (2020). WCSP focuses on non-manipulability in a worst-case scenario. We examine the implications of WCSP in a voting model. Although many voting rules, such as the plurality rule, the Borda rule, and the Dowdall rule, satisfy NOM-worst, they violate WCSP. We obtain a necessary and sufficient condition for the anti-plurality rule with fixed-order tie-breaking to satisfy WCSP in terms of the numbers of agents and alternatives.

Figures

Figures reproduced from arXiv: 2606.22021 by the authors.

Figure 1
Figure 1. Conceptual illustration of WCSP [PITH_FULL_IMAGE:figures/full_fig_p007_1.png] view at source ↗
Figure 2
Figure 2. Conceptual illustration of the three axioms [PITH_FULL_IMAGE:figures/full_fig_p008_2.png] view at source ↗

Discussion (0). Continue with ORCID to comment.

Reference graph

Works this paper leans on

60 extracted references · 1 canonical work pages

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    Troyan, P., Morrill, T. (2020). Obvious manipulations,Journal of Economic Theory, 185, 104970. A Appendix A.1 Proof of Lemma 9 We shall show that ifn≥3, then no plurality rule is WCSP. Assume thatn≥3. Let f:P n →Xbe a plurality rule. Take any distincta, b, c∈X. There are three...

    2020

  18. [28]

    Therefore,fis not WCSP

    Iff(P) =c, thenP −1 is worst for (P 1, f(P 1,·)) andf(P ′ 1, P−1) =b P 1 c=f(P). Therefore,fis not WCSP. Case 2.n= 3k+ 4 withk∈Z ≥0: LetP∈ P 3k+4 andP ′ 1, P ′ 3 ∈ Pbe given by the following table: 6 P1 P2 P3 P4 P5 · · ·P k+4 Pk+5 · · ·P 2k+4 P2k+5 · · ·P 3k+4 P ′ 1 P ′ 3 b b ...

  19. [29]

    Then, for eachx∈X\ {c},S(P ′ 1, P−1, c)> S(P ′ 1, P−1, x) and f(P ′ 1, P−1) =c

    Assume thatf(P) =b. Then, for eachx∈X\ {c},S(P ′ 1, P−1, c)> S(P ′ 1, P−1, x) and f(P ′ 1, P−1) =c. Thus,P −1 is worst for (P ′ 1, f(P ′ 1,·)) andf(P) =b P ′ 1 c=f(P ′ 1, P−1)

  20. [30]

    Then, for eachx∈X\ {b},S(P ′ 3, P−3, b)> S(P ′ 3, P−3, x) and f(P ′ 3, P−3) =b

    Assume thatf(P) =c. Then, for eachx∈X\ {b},S(P ′ 3, P−3, b)> S(P ′ 3, P−3, x) and f(P ′ 3, P−3) =b. Thus,P −3 is worst for (P ′ 3, f(P ′ 3,·)) andf(P) =c P ′ 3 b=f(P ′ 3, P−3). Then,fis not WCSP. 20 Case 3.n= 3k+ 5 withk∈Z ≥0: LetP∈ P 3k+5 andP ′ 1, P ′ 2, P ′ 5 ∈ Pbe given by...

  21. [31]

    Then,P −5 is worst for (P 5, f(P 5,·))

    Assume thatf(P) =b. Then,P −5 is worst for (P 5, f(P 5,·)). For eachx∈X\ {a}, S(P ′ 5, P−5, a)> S(P ′ 5, P−5, x) andf(P ′ 5, P−5) =a P 5 b=f(P)

  22. [32]

    Then,P −2 is worst for (P ′ 2, f(P ′ 2,·))

    Assume thatf(P) =aandf(P ′ 2, P−2) =b. Then,P −2 is worst for (P ′ 2, f(P ′ 2,·)). By assumption,f(P) =a P ′ 2 b=f(P ′ 2, P−2)

  23. [33]

    Then, (P ′ 2, P−(1,2)) is worst for (P1, f(P 1,·))

    Assume thatf(P) =aandf(P ′ 2, P−2) =c. Then, (P ′ 2, P−(1,2)) is worst for (P1, f(P 1,·)). For eachx∈X\ {b},S(P ′ 1, P ′ 2, P−(1,2), b)> S(P ′ 1, P ′ 2, P−(1,2), x) andf(P ′ 1, P ′ 2, P−(1,2)) = b P 1 c=f(P ′ 2, P−2). Therefore,fis not WCSP. Considering the three cases, we con...

  24. [34]

    Then,S(P ′ 1, P2, b)> S(P ′ 1, P2, c)> S(P ′ 1, P2, a) andf(P ′ 1, P2) =b P 1 c=f(P)

    Iff(P) =c, thenP 2 is worst for (P 1, f(P 1,·)). Then,S(P ′ 1, P2, b)> S(P ′ 1, P2, c)> S(P ′ 1, P2, a) andf(P ′ 1, P2) =b P 1 c=f(P)

  25. [35]

    Iff(P) =a, thenP 1 is worst for (P 2, f(P 2,·)) andf(P 1, P ′

  26. [36]

    Iff(P ⋆) =b, then,P ⋆ 2 is worst for (P ⋆ 1 , f(P ⋆ 1 ,·)) andf(P ′′ 1 , P ⋆ 2 ) =a P ⋆ 1 b=f(P ⋆)

  27. [37]

    Iff(P ⋆) =c, then,P ⋆ 1 is worst for (P ⋆ 2 , f(P ⋆ 2 ,·)) andf(P ⋆ 1 , P ′′ 2 ) =a P ⋆ 2 c=f(P ⋆)

  28. [38]

    Also, f(P ′′ 1 , P ⋆ 2 ) =a P 1 b=f(P 1, P ⋆ 2 )

    Iff(P) =b,f(P ⋆) =aandf(P 1, P ⋆ 2 ) =b, thenP ⋆ 2 is worst for (P 1, f(P 1,·)). Also, f(P ′′ 1 , P ⋆ 2 ) =a P 1 b=f(P 1, P ⋆ 2 )

  29. [39]

    Also, f(P 1, P ′

    Iff(P) =b,f(P ⋆) =aandf(P 1, P ⋆ 2 ) =a, thenP 1 is worst for (P ⋆ 2 , f(P ⋆ 2 ,·)). Also, f(P 1, P ′

  30. [40]

    Therefore,fis not WCSP

    =b P ⋆ 2 a=f(P 1, P ⋆ 2 ). Therefore,fis not WCSP. Case 2.m≥4 andn= 2: Take anyd∈Xwithd̸=a, b, c. LetP, P ′ ∈ P 2 andP ′ 1 ∈ Pbe given by the following table: 6 P1 P2 P ′ 1 P ′ 2 P ′′ 1 a d b c b b c a d a ... ... ... ... ... c b c b d d a d a c Then, for eachx, y∈X,S(P, x) =S...

  31. [41]

    Then,P 2 is worst for (P 1, f(P 1,·)) andf(P ′ 1, P2) =b P 1 d= f(P)

    Assume thatf(P) =d. Then,P 2 is worst for (P 1, f(P 1,·)) andf(P ′ 1, P2) =b P 1 d= f(P). 22

  32. [42]

    Sincef(P 1, P ′

    Assume thatf(P)̸=d. Sincef(P 1, P ′

  33. [43]

    Then,f(P ′′ 1 , P ′

    =c=t m−1(P1),P ′ 2 is worst for (P 1, f(P 1,·)). Then,f(P ′′ 1 , P ′

  34. [44]

    Therefore,fis not WCSP

    =b P 1 c=f(P 1, P ′ 2). Therefore,fis not WCSP. Case 3.m= 3 andn= 2k+ 3 withk∈Z ≥0: LetP∈ P 2k+3 andP ′ 1, P ′ 2, P ′ 3 ∈ Pbe given by the following table: 6 P1 P2 P3 P4 · · ·P k+3 Pk+4 · · ·P 2k+3 P ′ 1 P ′ 2 P ′ 3 a b c c· · ·c a· · ·a b c a b c a b· · ·b b· · ·b a b c c a b...

  35. [45]

    Iff(P) =a, thenP −2 is worst for (P 2, f(P 2,·)) andf(P ′ 2, P−2) =c P 2 a=f(P)

  36. [46]

    Iff(P) =b, thenP −3 is worst for (P 3, f(P 3,·)) andf(P ′ 3, P−3) =a P 3 b=f(P)

  37. [47]

    Therefore,fis not WCSP

    Iff(P) =c, thenP −1 is worst for (P 1, f(P 1,·)) andf(P ′ 1, P−1) =b P 1 c=f(P). Therefore,fis not WCSP. Case 4.m= 3 andn= 2k+ 4 withk∈Z ≥0: LetP∈ P 2k+4 andP ′ 1, P ′ 2, P ′ 3 ∈ Pbe given by the following table: 6 P1 P2 P3 P4 P5 · · ·P k+4 Pk+5 · · ·P 2k+4 P ′ 1 P ′ 2 P ′ 3 a...

  38. [48]

    Iff(P) =a, thenP −3 is worst for (P 3, f(P 3,·)) andf(P ′ 3, P−3) =c P 3 a=f(P)

  39. [49]

    Iff(P) =b, thenP −2 is worst for (P 2, f(P 2,·)) andf(P ′ 2, P−2) =c P 2 b=f(P)

  40. [50]

    23 Therefore,fis not WCSP

    Iff(P) =c, thenP −1 is worst for (P 1, f(P 1,·)) andf(P ′ 1, P−1) =b P 1 c=f(P). 23 Therefore,fis not WCSP. Case 5.m≥4 andn= 2k+ 3 withk∈Z ≥0: LetP∈ P 2k+3 be given by the following table: 6 P1 · · ·P k+1 Pk+2 · · ·P 2k+3 a· · ·a c· · ·c ...· · · ... b· · ·b b· · ·b ...· · · ....

  41. [51]

    Then, sinces(1)−s(m−1)> s(1)−s(2),S(P ′ 1, P−1, b) =S(P, b) +s(1)−s(m− 1)> S(P, b) +s(1)−s(2) =S(P, c) =S(P ′ 1, P−1, c)

    =c. Then, sinces(1)−s(m−1)> s(1)−s(2),S(P ′ 1, P−1, b) =S(P, b) +s(1)−s(m− 1)> S(P, b) +s(1)−s(2) =S(P, c) =S(P ′ 1, P−1, c). Then,f(P ′ 1, P−1)̸=c. Hence, f(P ′ 1, P−1)P 1 f(P) =c. Therefore,fis not WCSP. Case 6.m≥4 andn= 2k+ 4 withk∈Z ≥0: Take anyd∈Xwithd̸=a, b, c. LetP∈ P 2...

  42. [52]

    Then, sinces(1)−s(m−1)> s(1)−s(2),S(P ′ 1, P−1, c) =S(P, c) +s(1)−s(m− 24 1)> S(P, c) +s(1)−s(2) =S(P, d) =S(P ′ 1, P−1, d)

    =d. Then, sinces(1)−s(m−1)> s(1)−s(2),S(P ′ 1, P−1, c) =S(P, c) +s(1)−s(m− 24 1)> S(P, c) +s(1)−s(2) =S(P, d) =S(P ′ 1, P−1, d). Then,f(P ′ 1, P−1)̸=d. Hence, f(P ′ 1, P−1)P 1 f(P) =d. Therefore,fis not WCSP. Considering all six cases, we have: •Ifn= 2, thenfis not WCSP (Cases...

  43. [53]

    Then,f(P ′ 1, P2)̸=candf(P ′ 1, P2)P 1 f(P) =c

    Iff(P) =c, thenP 2 is worst for (P 1, f(P 1,·)) andS(P ′ 1, P2, b) = 1 + 1 2 >1 + 1 m = S(P ′ 1, P2, c). Then,f(P ′ 1, P2)̸=candf(P ′ 1, P2)P 1 f(P) =c

  44. [54]

    Then,f(P 1, P ′ 2)̸=aandf(P 1, P ′ 2)P 2 f(P) =a

    Iff(P) =a, thenP 1 is worst for (P 2, f(P 2,·)) andS(P 1, P ′ 2, b) = 1 + 1 2 >1 + 1 m = S(P1, P ′ 2, a). Then,f(P 1, P ′ 2)̸=aandf(P 1, P ′ 2)P 2 f(P) =a. 25 Therefore,fis not WCSP. Case 2.m= 3 andn= 3: LetP, P ′ ∈ P 3 be given by the following table: 6 P1 P2 P3 P ′ 1 P ′ 2 P...

  45. [55]

    Iff(P) =c, thenP −1 is worst for (P 1, f(P 1,·)) andf(P ′ 1, P−1) =b P 1 c=f(P)

  46. [56]

    Iff(P) =b, thenP −2 is worst for (P 2, f(P 2,·)) andf(P ′ 2, P−2) =a P 2 b=f(P)

  47. [57]

    Therefore,fis not WCSP Case 3.m≥4 andn= 3: Take anyd∈Xwithd̸=a, b, c

    Iff(P) =a, thenP −3 is worst for (P 3, f(P 3,·)) andf(P ′ 3, P−3) =c P 3 a=f(P). Therefore,fis not WCSP Case 3.m≥4 andn= 3: Take anyd∈Xwithd̸=a, b, c. LetP∈ P 3 be given by the following table: 6 P1 P2 P3 a d c b b d c a b ... ... ... d c a Then, for eachx∈X\ {a, b, c, d}, S(P...

  48. [58]

    Then, S(P ′ 1, P−1, d) =S(P, d) = 1 + 1 2 + 1 m and S(P ′ 1, P−1, c) =S(P, c) +s(1)−s(3) = 2 + 1 m

    =d. Then, S(P ′ 1, P−1, d) =S(P, d) = 1 + 1 2 + 1 m and S(P ′ 1, P−1, c) =S(P, c) +s(1)−s(3) = 2 + 1 m . Thus,S(P ′ 1, P−1, c)> S(P ′ 1, P−1, d) andf(P ′ 1, P−1)̸=d. Hence, f(P ′ 1, P−1)P 1 f(P) =d. Therefore,fis not WCSP. Case 4.n= 2k+ 4 withk∈Z ≥0: LetP∈ P 2k+4 be given by t...

  49. [59]

    Then, S(P ′ 1, P−1, c) =S(P, c) = 2 + 1 2 + 1 m + 3k 2 and S(P ′ 1, P−1, b) =S(P, b) +s(1)−s(2) = 3 + 3k 2

    =c. Then, S(P ′ 1, P−1, c) =S(P, c) = 2 + 1 2 + 1 m + 3k 2 and S(P ′ 1, P−1, b) =S(P, b) +s(1)−s(2) = 3 + 3k 2 . 27 Thus,S(P ′ 1, P−1, b)> S(P ′ 1, P−1, c). Then,f(P ′ 1, P−1)̸=c. Hence, f(P ′ 1, P−1)P 1 f(P) =c. Therefore,fis not WCSP. Case 5.n= 2k+ 5 withk∈Z ≥0: LetP∈ P 2k+5...

  50. [60]

    Then, S(P ′ 1, P−1, c) =S(P, c) = 3 + 1 m + 3k 2 and S(P ′ 1, P−1, b) =S(P, b) +s(1)−s(2) = 3 + 2 m + 3k 2

    =c. Then, S(P ′ 1, P−1, c) =S(P, c) = 3 + 1 m + 3k 2 and S(P ′ 1, P−1, b) =S(P, b) +s(1)−s(2) = 3 + 2 m + 3k 2 . Thus,S(P ′ 1, P−1, b)> S(P ′ 1, P−1, c). Then,f(P ′ 1, P−1)̸=c. Hence, f(P ′ 1, P−1)P 1 f(P) =c. Therefore,fis not WCSP. 28 Considering all five cases, we have: •If...

  51. [61]

    Sinces(1)> s(m−1),S(P ′ 1, P−1, x2) =S(P, x 2)+s(1)−s(m−1)> S(P, x 1) =S(P ′ 1, P−1, x1)

    =x 1. Sinces(1)> s(m−1),S(P ′ 1, P−1, x2) =S(P, x 2)+s(1)−s(m−1)> S(P, x 1) =S(P ′ 1, P−1, x1). Then,f(P ′ 1, P−1)̸=x 1. Thus, f(P ′ 1, P−1)P 1 f(P) =x 1. Therefore,fis not WCSP. Case 2.s(1)> s(2) andn=m+ 2kwithk∈Z ≥0: LetP∈ P m+2k be given by the following table: 6 29 P1 P2 ·...

  52. [62]

    Sinces(1)−s(m−1)≥s(1)−s(2),S(P ′ 1, P−1, x1) =S(P, x 1)+s(1)−s(m−1)≥ S(P, x1) +s(1)−s(2) =S(P, x 2) =s(P ′ 1, P−1, x2)

    =x 2. Sinces(1)−s(m−1)≥s(1)−s(2),S(P ′ 1, P−1, x1) =S(P, x 1)+s(1)−s(m−1)≥ S(P, x1) +s(1)−s(2) =S(P, x 2) =s(P ′ 1, P−1, x2). By the definition of≻,f(P ′ 1, P−1)̸=x 2. Thus, f(P ′ 1, P−1)P 1 f(P) =x 2. Therefore,fis not WCSP. Considering all three cases, we have: •Ifs(1) =s(2)...

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Reviewed June 26, 2026 · model on record in the stance chip above.