Source
Balduzzi, Tuyls, Pérolat & Graepel, LIT-1508, read in NOTE-1155 from arXiv:1806.02643v2: Proposition 1 (p. 3), Proposition 3 (p. 4) and appendix A (p. 12), with the proofs on p. 13.
What was actually shown
- Proposition 1. Batch Elo updates are stationary exactly when, for every agent, the predicted win probabilities against all others sum to the observed ones. Only the row sums are fitted.
- Proposition 3. If p_ij = σ(r_i − r_j) with Σr = 0, the row mean of logit P is r. An Elo rating generating P exists if and only if log(p_ij/p_ji) + log(p_jk/p_kj) + log(p_ki/p_ik) = 0 for all i, j, k. On the complete graph that is the Hodge statement that a curl-free flow is a gradient (THEORY-894).
- Copies (appendix A). Rock-paper-scissors at win rate 0.9 gives every agent rating 0. Copying C moves A and B apart. The paper prints ±63; the batch fixed point at the stated α, recomputed by the reader, is ±71.9. Either way A falls because it loses to the duplicated agent, and B rises because it beats it. The row means of the log-odds here are ±95.4: with curl, the batch fixed point is not the row mean.
The reader re-derived both propositions. The steps are written out in the note of 2026-10-30 below.
What this does not say
- That Elo fails when wins cycle and succeeds when they do not. The condition is on log-odds, cardinal. A record whose wins are acyclic can have curl and be misfit by Elo; the condition is not TERM-077’s ordinal one, and CLAIM-434 holds the reason they come apart.
- That a curl-free fit means the agents differ in one skill. It means the log-odds are additive; nothing about mechanism.
- That only copies distort Elo. Any change to the evaluated population moves the row means. Copies are the clean case.
Connections
Bears on CLAIM-451 only negatively: in a fully cyclic record every option gets the same Elo, so a scalar mean cannot tell cyclic tension from indifference. The copy effect is the simplest form of a scoring layer that can be moved without changing what is scored. That is the kind of route that CASE-124’s μ arrow and CASE-155’s share-dependent scoring formalize.
Note of 2026-10-30: the re-derivation, written out
The second check at the status review (ADR-065 §5) followed each step.
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Proposition 1. Batch Elo updates r_i ← r_i + η Σ_j (p_ij − σ(r_i − r_j)), with p_ij the observed win rate. It is stationary exactly when Σ_j σ(r_i − r_j) = Σ_j p_ij for every i, so only the row sums are fitted.
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Proposition 3, forward. If p_ij = σ(r_i − r_j), then logit p_ij = r_i − r_j, and (1/n) Σ_j logit p_ij = r_i − (1/n) Σ_j r_j = r_i when Σ r = 0. The term j = i is logit ½ = 0 = r_i − r_i.
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Proposition 3, converse. Write A_ij = log(p_ij / p_ji). It is antisymmetric, and it is logit p_ij, since p_ij + p_ji = 1. If A_ij + A_jk + A_ki = 0 for all i, j, k, then A_ij = A_ik + A_kj. Put r_i = (1/n) Σ_k A_ik. Then r_i − r_j = (1/n) Σ_k (A_ik − A_jk) = (1/n) Σ_k (A_ik + A_kj) = A_ij, so p_ij = σ(r_i − r_j). If some triangle sum is nonzero, no ratings r have A_ij = r_i − r_j, since any such A sums to zero round every triangle.
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When the log-odds are curl-free, batch Elo reaches the row means. Those ratings reproduce P, so they match its row sums, and the row-sum equations are the Bradley–Terry likelihood equations, whose solution is unique up to a shift.
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Copies. Take rock-paper-scissors at 0.9, with A beating B, B beating C and C beating A: every rating is 0. Add C′, a copy of C. By symmetry r_C = r_C′ = 0 and r_A = −r_B = −x. B’s observed row sum is 0.1 + 0.9 + 0.9 = 1.9, so the fixed point needs σ(2x) + 2σ(x) = 1.9, which gives x = 0.4140 in natural units, or 71.91 Elo points at 400/ln 10 points per unit. A falls because it loses to the copied agent, and B rises because it beats it. The row means of the log-odds are instead ±(ln 9)/4 = ±0.549, or ±95.42 Elo points.
A script (numpy, in the session’s scratchpad) checked these steps. In five random curl-free matrices of 3 to 7 agents, the row means of the log-odds equalled the generating ratings and batch Elo converged to them. In five random matrices with curl, batch Elo matched the row sums to 10⁻¹³, but neither it nor the row means reproduced P (largest misfits 0.02 to 0.46), and the two differed by up to 0.64. As a control, removing the cyclic part of the same matrices made the row means fit exactly. The copy example’s fixed point came out at ±71.91 both by iterating the update and by bisection on the equation in step 5.
Note of 2026-10-30: the clause the narrowing drops
The title said, without condition, that Elo ratings are the row means of a win-probability matrix’s log-odds. That is the paper’s gloss, “Elo is, essentially, a uniform average in logit space” (p. 4), and Proposition 3 proves it only when the matrix is generated by ratings. With curl, batch Elo’s fixed point is not the row mean: in the copy example the row means are ±95.4 Elo points and the fixed point ±71.9. Read unconditionally, the clause is shown false, so it is recorded here only (ADR-047’s kind 3), and the title now states the condition.