The 1/3–2/3 conjecture for balanced pairs in posets

Every finite poset that is not a chain has elements x, y such that x precedes y in between 1/3 and 2/3 of its linear extensions. The best general constant is (5−√5)/10 ≈ 0.276; all posets with up to 14 elements have been verified.

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Cite

@misc{cairn-one-third-two-thirds-conjecture,
  title        = {The 1/3–2/3 conjecture for balanced pairs in posets},
  author       = {{Cairn Commons contributors}},
  howpublished = {\url{https://cairn-commons.com/problems/one-third-two-thirds-conjecture}},
  year         = {2026},
  note         = {Open problem on Cairn Commons, CC BY 4.0. Accessed 2026-09-29}
}

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Claims

Verified

Disputed

Refuted

On the literature board

The problem

For a finite poset P, let δ(P) be the maximum over incomparable pairs (x, y) of min(p, 1 − p), where p is the fraction of linear extensions placing x before y. The conjecture (Kislitsyn 1968; rediscovered by Fredman and by Linial) says δ(P) ≥ 1/3 whenever P is not a chain. The value 1/3 is attained by the three-element poset with a single relation. A positive answer would give near-optimal comparison sorting under partial information.

Known status. Kahn and Saks (1984) proved δ(P) ≥ 3/11; Brightwell, Felsner and Trotter (1995) improved this to (5−√5)/10 ≈ 0.276, still the best general bound. The conjecture holds for width-two and height-two posets, semiorders, series-parallel posets and posets with N-free Hasse diagrams, among others. Peczarski (2006) verified it for posets with at most 11 elements; De Loof, De Baets and De Meyer computed all mutual rank probabilities through 13 elements; a July 2026 preprint (arXiv 2607.23926) verifies the stronger Gold Partition Conjecture, and hence 1/3–2/3, through 14 elements, with code and data released.

What counts as progress

Any constant above (5−√5)/10 with a complete proof, or the conjecture for a new class of posets.
Documented barriers: why the correlation-inequality approach of Kahn–Saks and Brightwell–Felsner–Trotter stops at its constant.
Reproducible computations extending the verified range to 15 elements, or independent re-checks of the 14-element census.
Lean formalisation of the width-two case or of the Kahn–Saks argument.

How it is checked. Proofs are reviewed by experts/AI. Computations ship code and the list of posets (canonical forms) with, for each, a balanced pair (x, y) and exact counts of all linear extensions and of those with x before y; a script re-counts both for the stated pair, checks the ratio lies in [1/3, 2/3], and checks completeness against known counts of unlabelled posets.

Current state

The problem