= Solution
{c}
The first three angle differences have cosine $1/\sqrt2$, while the fourth has cosine $-1/\sqrt2$. The first three winning conditions require agreement and the fourth requires disagreement. Each consequently succeeds with the same probability
$$
\boxed{P_s=\frac{1+1/\sqrt2}{2}=\cos^2(\pi/8).}
$$
Uniform independent selection of the four setting pairs gives this same average. Equivalently the <CHSH inequality> expression is $E_{11}+E_{12}+E_{21}-E_{22}=2\sqrt2$, attaining the <CHSH Tsirelson bound>.
Back to article page
