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Let $p_{ab}$ denote the conditional winning probability in the CHSH game. With both settings zero, computational-basis measurements of the Bell state always agree, so $p_{00}=1$. When just one setting is one, Bob's conditional pure state from part (i), or the corresponding calculation with Alice and Bob exchanged, gives agreement probability ${\cos }^2\theta$. Hence $p_{01}=p_{10}={\cos }^2\theta$.

For $a=b=1$, winning requires different output bits. Alice obtains each of her two outcomes with probability $1/2$. Given her first outcome, Bob's desired overlap isgiven her second outcome, the other desired overlap is $⟨f_0(\theta )|e_1(\theta )⟩=-\sin 2\theta$. Thus $p_{11}={\sin }^2(2\theta )$. Averaging over the four equally likely inputs givesThe maximum winning probability for noncommunicating classical players is $3/4$, the game form of the CHSH inequality. Setting $t=\cos 2\theta$, strict improvement requires $t-t^2>0$, or $0<t<1$. On the stipulated interval this givesThe endpoints give equality with the classical bound. Within this reflected real-basis CHSH strategy, the maximum is $13/16$ at $t=1/2$, namely $\theta =\pi /6$; this follows by completing the square in $t$.

For the Bell state $|{\varphi }_{+}⟩$, let both zero-input measurements use the computational basis, and let the one-input bases be rotated by $\theta$ for Alice and $-\theta$ for Bob in the real qubit plane. In the CHSH game, the four conditional winning probabilities are $1,{\cos }^2\theta ,{\cos }^2\theta ,{\sin }^2(2\theta )$. They follow by contraction of the Bell state with vectors from the real orthonormal bases used for the basic measurements. Their average is the displayed expression. On $0\le \theta \le \pi /2$, this beats $3/4$ exactly when $0<\theta <\pi /4$ and reaches $13/16$ at $\theta =\pi /6$, since $t-t^2=1/4-(t-1/2{)}^2$.