While in the standard game each player picks one element per turn, in the biased game each player takes a different number of elements.
More formally, for every two positive integers p and q, a (p:q)-positional game is a game in which the first player picks p elements per turn and the second player picks q elements per turn.
The main question of interest regarding biased positional games is what is their threshold bias - what is the bias in which the winning-power switches from one player to the other player.
In this game, the elements are all edges of a complete graph on n vertices, and the winning-sets are all triangles (=cliques on 3 vertices).
Suppose we play it as a Maker-Breaker game, i.e., the goal of Maker (the first player) is to take a triangle and the goal of Breaker (the second player) is to prevent Maker from taking a triangle.
Using a simple case-analysis, it can be proved that Maker has a winning strategy whenever n is at least 6.
Therefore, it is interesting to ask whether this advantaged can be biased by letting Breaker pick more than 1 element per turn.
Indeed, it is possible to prove that:[1] In an unbiased Maker-Breaker game, the Erdos-Selfridge theorem gives a winning condition for Breaker.
This condition can be generalized to biased games as follows:[3] [2]: 30–32 The strategy uses a potential function which generalized the function of Erdos-Selfridge.
The potential of a (non-broken) winning-set E with |E| untaken elements is defined as
If Maker wins the game then there exists a set E with |E|=0, so its potential is 1; therefore, to prove that Breaker wins, it is sufficient to prove that the final potential-sum is less than 1.
Indeed, by assumption, the potential-sum at Breaker's first turn is less than 1; and if Breaker always picks an element that maximizes the potential-drop, it is possible to show that the potential-sum always weakly decreases.
elements, for some fixed k, Breaker's winning condition simplifies to:
This condition is tight: there are k-uniform set-families with
[4] In an unbiased Maker-Breaker game, a theorem by Beck gives a winning condition for Maker.
This condition can be generalized to biased games as follows:[3] If
, then Maker has a winning-strategy in the (p:q) game when playing first.
In a biased Avoider-Enforcer game, the following conditions guarantee that Avoider has a winning strategy:[2]: 47–49