Playing instead of colouring: exact thresholds of Maker-Breaker games for every regular three-variable equation

Maker and Breaker alternately claim numbers from {1,...,n}, Maker first; Maker wins if her numbers contain a solution of a(x−y)=bz. Every partition regular three-variable equation with non-trivial Rado numbers has this form. We determine the least n for which Maker wins exactly: for b ≥ 2 it is 2b+1 if a ≤ 2b, 2b+2 if a = 2b+1, and a if a ≥ 2b+2. For a = 1 this sharpens the bounds b+3 ≤ f ≤ 2b+2 that follow from Gaiser and Horn (Integers 24 (2024) #A42) to the exact value 2b+1. We also determine the threshold when x, y, z must be distinct, prove that Breaker taking q ≥ 2 numbers per turn forces n ≥ qa, prove a lower bound for q = 2 that our computations show to be sharp, and give first results on the Avoider–Enforcer version. The Rado number of the same equation is b²+b+1 for a < b: a Maker who plays needs only about 2b numbers, while a colouring needs about b². All values were also confirmed by exhaustive game-tree search. Companion paper: doi:10.5281/zenodo.24073906.