If you have a Bell state, e.g., the spin-0 state,
$$|\Psi \rangle=\frac{1}{\sqrt{2}} (|HV \rangle-|VH \rangle),$$
then the single photons are precisely unpolarized, i.e.,
$$\hat{\rho}_A =\mathrm{Tr}_B |\Psi \rangle \langle \Psi|=\frac{1}{2} \hat{1}, \quad \hat{\rho}_B=\mathrm{Tr}_A |\Psi \rangle \langle \Psi|=\frac{1}{2} \hat{1},$$
and thus if Alice measures 48% H and 52% V, her polarization experiment is somehow inaccurate. As has been correctly stressed over and over again that cannot be used to transmit a signal, because all that B has are also precisely unpolarized photons. With a correctly working polarization measurement he'll simply get a random sequence of H and V polarized photons with the 50%:50% probabilities, no matter what A does with her photons. Only if both detectors work precisely you can figure out the correlations predicted by QED given the two-photon Bell state, and this can be done only by comparing A's and B's measurement protocols, which is possible only by exchange of this information, which is possible only with real-world signals, which propagate at most at the speed of light.