5. Large strategy sets Proposition 2.3 establishes a trivial existence result where the bound on the Lipschitz constant is independent of the number of strategies in the game. Proposition 5.1 below shows that for games with unbounded strategy sets this bound is the best possible (up to a constant). 5.1. Proposition. For every even n there is a game in L ( n, 2 n/ $^{2}$, 1 /n ) with no pure 1 / 8 – equilibrium. To prove the proposition we construct another ‘mass version’ of matching pennies. Again we divide the players into groups of females and males, but this time every player has a coin for each of her/his opponents. The strategy of every player encodes all their coins. Proof. Let the number of players be even n = 2 k , and let the strategy set of each player be { +1 , − 1 } $^{k}$. We divide the players into two groups, females and males, and denote their strategy profiles by ¯ x = ( x$_{1}$, . . . , x$_{k}$ ) and ¯ y = ( y$_{1}$, . . . , y$_{k}$ ) respectively. The strategy of female i is given by the vector ( x$_{i}$ [ j ]) k j $_{=1}$, and similarly the strategy of male j is given by the vector ( y$_{j}$ [ i ]) k i $_{=1}$. The payoff to female i is u$_{i}$ (¯ x, ¯ y ) = 1 4 k ∑ j x$_{i}$ [ j ] · y$_{j}$ [ i ] , and the payoff to male j is v$_{j}$ (¯ x, ¯ y ) = − 1 4 k ∑ i x$_{i}$ [ j ] · y$_{j}$ [ i ] . The game has Lipschitz constant 2 / 4 k = 1 /n but no pure 1 / 8–equilibrium: For every strategy profile of the opponents, every player can guarantee 1 / 4. Therefore, in every 1 / 8–equilibrium every player should get at least 1 / 8. But this is impossible since the sum of all the players’ payoffs is 0 in every profile. □ 11