Source
Balduzzi, Tuyls, Pérolat & Graepel, LIT-1508, read in NOTE-1155 from arXiv:1806.02643v2: Proposition 4 (p. 6, proof p. 13), Theorem 1 (pp. 6–7, proofs pp. 14–15), Lemma 1 (p. 14), Example 1 (p. 6).
What was actually shown
- Uniqueness (Proposition 4). The Nash equilibria of a two-player zero-sum game form a convex polytope. For an antisymmetric matrix the set is symmetric, and entropy is strictly concave, so one symmetric equilibrium has maximum entropy.
- The key lemma. The game’s value is zero, so p is a Nash equilibrium exactly when Ap ≤ 0. Every agent then scores at most zero against p. Copy agent c. Each split (…, αp_c, (1 − α)p_c) of a Nash equilibrium is one of the padded game, and merging the copies of a padded equilibrium gives one of the original.
- From the lemma to invariance. This step is the reader’s. If the original equilibrium is unique, every padded equilibrium has total mass p_c on the copies. Entropy is then maximized by the even split. The Nash average A′p′ equals the original on every original agent, and the copies get the copied agent’s value. The paper’s Example 1 (rock, paper, scissors with C doubled: (⅓, ⅓, ⅓) becomes (⅓, ⅓, ⅙, ⅙), and every Nash average stays 0, while uniform averaging gives B the edge) is this case.
- Interpretability (Theorem 1, P3). p* is uniform exactly when div A = 0, and when A = grad(r) the game decouples, so p* is uniform on argmax r.
What could have come out otherwise: in the paper’s Example 1, uniform averaging moves (div A′ = (−1.15, 1.15, 0, 0)) and Nash averaging does not.
What this does not say
- That Nash averaging is invariant to copies in general. The paper says so (Theorem 1, P1) and does not prove it. With a segment of equilibria, the maximum-entropy selection on the padded game weighs the copied agent’s mass by the extra log 2 of its split and moves. Take rock-paper-scissors, a null agent D tying all three, and E, which loses to all three and beats D. Copying D moves the selection from (¼, ¼, ¼, ¼, 0) to (⅕, ⅕, ⅕, ⅕, 0, ⅕) and E’s Nash average from −0.5 to −0.2 (NOTE-1155, checked by script).
- That it resists near-copies. Continuity (P2) is of the payoff. The equilibrium itself can jump (the paper’s Example 2).
- That it resists a non-redundant addition. A new agent beating the leader changes the equilibrium completely. That is the method working.
- That the equilibrium measures skill. It is the unexploitable team for this data set; the authors call it adversarial and “a harsh master”.
- Anything ordinal. Payoffs are log-odds and the results use their magnitudes.
Connections
The cleanest formal answer the record holds to one capture route: a scoring layer that can be moved by padding the evaluated population. It narrows CASE-124’s μ route and is distinct from CASE-155’s share-dependent c (NOTE-1155). The value-based invariant of Balduzzi et al. (2019), THEORY-944, needs no uniqueness and holds for behavioural mixtures as well as copies.
Note of 2026-10-30: the second check
The second check at the status review (ADR-065 §5) followed each step.
Uniqueness. The value is 0, so p is an equilibrium strategy of either player exactly when Ap ≤ 0. The equilibrium set {p in the simplex : Ap ≤ 0} is a convex polytope, the same for both players, and entropy, being strictly concave, has one maximiser on it.
The lemma. Copy agent c as c′. In the padded matrix A′, A′p′ is A m(p′) on the original agents and (A m(p′))_c on c′, where m(p′) merges the masses of c and c′. So p′ is an equilibrium of the padded game exactly when m(p′) is one of the original.
Invariance. If the original equilibrium p is unique, every padded equilibrium is a split of it, with αp_c on c and (1 − α)p_c on c′. Its entropy is H(p) + p_c h(α), h being the binary entropy, which is largest at α = ½. The Nash averages A′p′ on the original agents are Ap, and the copy gets (Ap)_c.
P3. If div A = 0, the uniform u has Au = 0, so it is an equilibrium, and it has the most entropy on the whole simplex. Conversely Σ_i (Au)_i = 0 for antisymmetric A, so Au ≤ 0 forces Au = 0, which is div A = 0. If A_ij = r_i − r_j, then (Ap)_i = r_i − r·p ≤ 0 for every i forces r·p = max r, so p lies on argmax r, and the maximum-entropy one is uniform there.
The counterexample. E’s mass in an equilibrium would raise the sum of the rock, paper and scissors rows by 3p_E > 0, so E has none, and Ap ≤ 0 then makes the three equal. With mass λ on them and 1 − λ on D, E’s row needs λ ≥ ½. Maximising −λ ln(λ/3) − (1 − λ) ln(1 − λ) gives ln(3(1 − λ)/λ) = 0, so λ = ¾, and E’s Nash average is −¾ + ¼ = −½. With D copied the second term becomes −(1 − λ) ln((1 − λ)/2), which gives λ = 3/5, p* = (⅕, ⅕, ⅕, ⅕, 0, ⅕), and E’s average −3/5 + 2/5 = −⅕.
A script (numpy, with an exact linear-programming solver, in the session’s scratchpad) checked these. In 400 random antisymmetric matrices of 3 to 7 agents with a unique equilibrium, one agent copied, every split tried was an equilibrium of the padded game, the solver’s padded equilibrium merged back to the original, and the even split left every original Nash average unchanged. P3 held on 100 matrices each way. The counterexample is the control: there a grid search over λ agreed with the closed forms above, and the “invariant” split (¼, ¼, ¼, ⅛, 0, ⅛) is an equilibrium of lower entropy than the selection.