I'm familiar with CHSH game and the strategy that allows Alice and Bob to succeed with a probability of $$\frac{1+\tfrac{1}{\sqrt 2}}{2}\approx 85\%$$ if they share a maximally entangled state such as $\lvert\Phi_+\rangle$ and use the measurements described by the observables $\sigma_x,\sigma_y$ on their respective qubits to determine their choice of move, depending on the bit they receive from Charlie. I've also seen the proof of the Tsirelson bound which effectively shows that $$\big\lVert A_1 \otimes B_1 + A_1\otimes B_2 + A_2\otimes B_1 - A_2\otimes B_2\big\rVert_2 \leq 2\sqrt{2}$$ if $A_1,A_2$ are Hermitian operators on a space $\mathcal H_A$ with spectrum $\pm 1$ and $B_1, B_2$ are Hermitian operators on $\mathcal H_B$ with spectrum $\pm 1$, and it happens that this maximum is attained when $A_1=B_1=\sigma_x$ and $A_2=B_2=\sigma_y$ with $\mathcal H_A=\mathcal H_B = \mathbb C^2$. (And the singular vector that gives the matrix norm its value happens to be maximally entangled.)
My question: Suppose we are only allowing Alice and Bob to start with some given quantum state that is entangled but not maximally entangled. For instance, say we give them the state $$\lvert\Psi\rangle = \tfrac{3}{5}\lvert 00\rangle + \tfrac{4}{5}\lvert 11\rangle$$ They can still outperform the classical maximum winning probability of $75\%$, and in fact by performing a $e^{\pi i/4}$ phase shift on the second component of their shared state and using the Pauli matrices just as is done in the optimal solution to the CHSH game, they can attain a winning probability of $$\frac{1+\tfrac{1}{\sqrt 2}\cdot\tfrac{24}{25}}{2}\approx 84\%$$ but I'd like to know if this solution is optimal for the given entangled state, or if some other slightly different choice of observables could give a better performance. More generally, does being given a different pure entangled state change their optimal strategy for the CHSH game, and how can we prove the answer?
