Occupation games on graphs in which the second player takes almost all vertices

Given a connected graph G = (V,E), two players take turns occupying vertices v ∈ V by putting black and white tokens so that the current vertex sets B,W ⊆ V are disjoint, B ∩W = ∅, and the corresponding induced subgraphs G[B] and G[W ] are connected any time. A player must pass whenever (s)he has no legal move. (Obviously, after this, the opponent will take all remaining vertices, since G is assumed connected.) The game is over when all vertices are taken, V = B∗∪W ∗. Then, Black and White get b = |B∗|/|V | and w = |W ∗|/|V |, respectively. Thus, the occupation game is one-sum, b+w = 1, and we could easily reduce it to a zero-sum game by simply shifting the payoffs, b′ = b− 1/2, w′ = w − 1/2. Let us also notice that b ≥ 0 and w ≥ 0; moreover, b > 0 and w > 0 whenever |V | > 1. [Let us remark that the so-called Chinese rules define similar payoffs for the classic game of GO, yet, the legal moves are defined in GO differently.] Like in GO, we assume that Black begins. It is easy to construct graphs in which Black can take almost all vertices, more precisely, for each ε > 0 there is a graph G for which b > 1 − ε. In this paper we show that, somewhat surprisingly, there are also graphs in which White can take almost all vertices.