Dear Professor Zeilberger,

An answer to the challenge in Emily Sergel's write-up (15 April 2021).

n = 6, with a = 10y, b = 62, c = 62, d = 10y, e = 1:

Gr := [[0,10*y,62,0],[0,0,1,62],[0,0,0,10*y]]:
Gl := [[0,10*y,62,0],[0,0,0,62],[0,0,0,10*y]]:
FindEqu(Gr,y,6)[4] / FindEqu(Gl,y,6)[4];

Left equilibrium [3,3], everybody 92 minutes; right equilibrium [6,0,0], everybody 121. Ratio 121/92 = 1.3152, against 92/83 = 1.1084. Both networks have a unique pure Nash equilibrium, so the answer does not depend on which equilibrium is selected.

More generally a = d = M*y, b = c = 6M+2, e = 1 gives ratio (12M+1)/(9M+2), increasing to 4/3; it beats 92/83 once M > 101/168.

And 4/3 is sharp. For n = 6 the ratio is always strictly below 4/3, at every pure Nash equilibrium of either network. The extremal case has an analytic proof (via t^2 - t + 1 >= 3/4, valid for all n); the other 195 pairs of equilibrium compositions are settled by exact rational LP with Farkas certificates. The same computation gives sup = 4/3 for even n and 4n^2/(3n^2+1) for odd n, verified for n <= 20.

I have a short note with the proofs, the 196 certificates for n = 6 as a JSON file, and a standard-library checker that re-verifies all of them without rerunning the search. My mail client will not attach it from here, so please just reply and I will send it straight away.

Disclosure: produced with the assistance of Claude (Anthropic) and Codex (OpenAI) models under my direction; every claim cross-checked by independent implementations and exact Farkas certificates.

Best regards,
Maximilian Thuemmler
maximilian.thuemmler@gmail.com
Independent researcher