Maker-breaker percolation games II: Escaping to infinity

Let Λ be an infinite connected graph, and let v 0 be a vertex of Λ. We consider the following positional game. Two players, Maker and Breaker, play in alternating turns. Initially all edges of Λ are marked as unsafe. On each of her turns, Maker marks p unsafe edges as safe, while on each of his turns Breaker takes q unsafe edges and deletes them from the graph. Breaker wins if at any time in the game the component containing v 0 becomes finite. Otherwise if Maker is able to ensure that v 0 remains in an infinite component indefinitely, then we say she has a winning strategy. This game can be thought of as a variant of the celebrated Shannon switching game. Given ( p , q ) and ( Λ , v 0 ) , we would like to know: which of the two players has a winning strategy? Our main result in this paper establishes that when Λ = Z 2 and v 0 is any vertex, Maker has a winning strategy whenever p ≥ 2 q , while Breaker has a winning strategy whenever 2 p ≤ q . In addition, we completely determine which of the two players has a winning strategy for every pair ( p , q ) when Λ is an infinite d-regular tree. Finally, we give some results for general graphs and lattices and pose some open problems.