Maxmin Strategy in Battle of the Sexes

Kevin Leyton-Brown and Yoav Shoham Essentials of Game Theory, page 15:

Given the following payoff table for a typical Battle of the Sexes game, calculate the maxmin strategies and values for each player.

	L	W
L	2 1	0 0
W	0 0	1 2

Proof: Player 2 must minimize $$u_1(s)=2s_1(L)s_2(L)+s_1(W)s_2(W)=$$ $$2s_1(L)s_2(L)+(1-s_1(L))(1-s_2(L))=$$ $$s_2(L)(3s_1(L)-1)+1-s_1(L)$$ When $s_1(L) > 1/3$ and $3s_1(L)-1$ is positive, player 2 minimizes by playing $s_2(L)=0$ and now the expression collapses to $1-s_1(L)$ so that $u_1(s) < 2/3$. If $s_1(L) < 1/3$ and $3s_1(L)-1$ is negative, player 2 minimizes with $s_2(L)=1$ and the expression collapses to $2s_1(L)$ showing $u_1(s) < 2/3$. When $s_1(L)=1/3$ then the expression invariably maximizes as $u_1(s)=2/3$. This shows $$s_1(L)=1/3$$ $$s_1(W)=2/3$$ is player 1's maxmin strategy with maxmin value $2/3$. Symmetrically, similar holds for player 2.$~\square$