Maker and Breaker alternatively select 1 and q edges of K_n (the complete graph on n vertices) until all edges have been claimed. Maker wins if his graph has a triangle. What is the smallest q=q(n) such that Breaker has a winning strategy?

The best known strategy for Maker is to claim edges incident to some vertex x (and never miss a one-step win). If Breaker managed to block x completely, say after m rounds, then Breaker made n-1-m + {m\choose 2}\le m q(n) moves. This implies that q(n) is approximately at least \sqrt{2n}.

On the other hand, Breaker can win if q> 2\sqrt{n}: for each edge xy claimed by Maker, Breaker selects \sqrt{n} edges at x and \sqrt{n} edges at y. Thus Maker’s graph has maximum degree at most (n-1)/\sqrt{n}+1 and Breaker can always block all immediate threats.

Reference: V.Chvatal and P.Erdos, Biased positional games, Ann. Discrete Math. 2 (1978), 221-229.

I would bet on the \sqrt{2n} bound

Oleg