I'm thinking about a modified version of the CHSH game and I'm trying to convince myself that in this case, entanglement cannot be used to gain an advantage or else it would imply signalling is possible through entanglement (i.e. contradicting the no-signalling theorem).
The rules are: the referee generates $a$ and $b$ uniformly from $\{0, 1\}$ and computes $a\oplus b = w$. The referee sends $a$ to Alice and $b$ to Bob. The game is won if both Alice and Bob can each send a bit $w_a, w_b$ back to the referee such that $w_a=w$ and $w_b = w$. Alice and Bob cannot classically communicate but they can distribute entanglement before the game starts.
What I have tried is assuming Alice and Bob share an entangled state $\rho \in \mathcal{H}_A\otimes\mathcal{H}_B$ and have arbitrary measurements $\{ M_a \}, \{M_b\}$. Assume Alice and Bob coordinate with a distribution $c(x_a, x_b|a,b)$, where $x_a$ and $x_b$ are variables to help them decide for $w_a$ and $w_b$. Then
$$ \begin{align} c(x_a, x_b|a,b) &= tr((M_a\otimes M_b) \rho) \\ &= tr_A(tr_B((M_a\otimes M_b) \rho))\\ &= \dots\\ &= tr(\rho)? \end{align} $$
This isn't correct as it is, but I'm looking for something that would imply that $a$ and $b$ wouldn't affect $x_a, x_b$, or that they are uniformly random which would then imply Alice and Bob couldn't strategize any better than guessing. I'm not sure if this argument makes sense at all since I think it would also contradict that there is an advantage to the normal CHSH game.
Any tips to showing if there is an advantage or not would be helpful. Thanks.