A Waiter-Client game[1] (also called:[2] Picker-Chooser game) is a kind of positional game. Like most positional games, it is described by its set of positions/points/elements (
X
l{F}
X
In a Waiter-Client game, Waiter wins if he manages to occupy all the elements of a winning-set, while Client wins if he manages to prevent this, i.e., hold at least one element in each winning-set. So Waiter and Client have, respectively, the same goals as Maker and Breaker in a Maker-Breaker game; only the rules for taking elements are different.
In a Client-Waiter game the winning conditions are reversed: Client wins if he manages to hold all the elements of a winning-set, while Waiter wins if he manages to hold at least one element in each winning-set.
In some cases, the Waiter is much more powerful than the player with the same goal in the Maker-Breaker variant. For example, consider a variant of tic-tac-toe in which Maker wins by taking k squares in a row and Breaker wins by blocking all rows.
Then, when the board is infinite, Waiter can win as Maker for any k >= 1.[3] Moreover, Waiter can win as Breaker for any k >= 2: in each round, Waiter picks a pair of squares that are not adjacent to the pairs picked so far (for example, in round i he picks the squares (2i,0) and (2i,1)). Moreover, even when the board is finite, Waiter always wins as Breaker when k >= 8. [4]
This leads to the following conjecture by József Beck: If Maker wins the Maker-Breaker game on
(X,l{F})
(X,l{F})
(X,l{F})
(X,l{F})
Suppose the winning-sets are all of size k (i.e., the game-hypergraph is k-uniform). In a Maker-Breaker game, the Erdos-Selfridge theorem implies that Breaker wins if the number of winning-sets is less than
2k-1
By the above conjecture, we would expect the same to hold in the corresponding Client-Waiter game - Waiter "should" win (as Breaker) whenever the number of winning-sets is less than
2k-1
{2k-1\over8(k+1)}
{2k-1\over3\sqrt{k+1/2}}