Classical analog of Bell QM experiment correlation coefficient calculation

This question is motivated by recent experiments in QM entanglement. consider the following "simple/ simplified" classical analog of Bells experiment. it has a laser, a standard beamsplitter reflecting light in 2 directions "left/ right", and 2 polarizing beamsplitters, 1 at each left/ right arm. This creates 4 separate rays fed to 4 light detectors. 1 of the arms has a rotation angle relative to the other "θ".

Now consider a twist on the detection. The detectors measure light intensity, but convert it to binary on/off measurements ("pulses") based on a threshhold value T.

There is another twist. There is a motorized rotating polarizer screen in front of the laser prior to the 1st standard beamsplitter. It rotates at a constant speed.

The standard correlation coefficent for the Bell experiment is E(θ) = R++ + R-- - R-+ - R+- (where Rs are coincident pulses at opposite ends as in standard experiments).

In this experiment, E(θ) is calculated based on the detector threshhold measurements over short time Δt, counted/ integrated over long time scales/ many repeated measurements.

what is the formula for E(θ) (calculated using classical physics)?

note: can foresee there are some hidden subtlety(s) in calculating this formula.

 Quantum computer emulated by a classical system / La Cour, Ott, physorg

• What's the classical analog of entanglement? Apr 28 '18 at 19:35

To understand difficulty of violating Bell inequalities, it is better to look at this one (page 9 here) for binary variables $A,B,C$:

$$P(A=B)+P(A=C)+P(B=C) \geq 1$$

It can be easily checked by assigning any probability distribution among 8 possibilities, or just: "drawing 3 coins, 2 of them have to give the same value".

However, it is violated in QM.

To do it, instead of standard: "probability of alternative of disjoint events is sum of their probabilities" leading to above inequality,

we need to use Born rules: "probability of alternative of disjoint events is proportional to square of sum of their amplitudes".

Such Born rule can be easily understood if instead of ensemble of events in a given moment, like in Feynman path integrals we use ensemble of paths (discussion). This way amplitude corresponds to probability at the end of past or future half-paths, to draw some value in a given moment we need to draw it from both time directions, getting $P\propto \psi^2$ Born rule.