 Solution

The first three angle differences have cosine $1/\sqrt{2}$, while the fourth has cosine $-1/\sqrt{2}$. The first three winning conditions require agreement and the fourth requires disagreement. Each consequently succeeds with the same probabilityUniform independent selection of the four setting pairs gives this same average. Equivalently the CHSH inequality expression is $E_{11}+E_{12}+E_{21}-E_{22}=2\sqrt{2}$, attaining the CHSH Tsirelson bound.