While the original Bell inequality might leave some hope for violation, here is one which seems completely impossible to violate - for three binary variables A,B,C:

Pr(A=B) + Pr(A=C) + Pr(B=C) >= 1

It has obvious intuitive proof: drawing three coins, at least two of them need to give the same value.

Alternatively, choosing any probability distribution pABC among these 2^3=8 possibilities, we have:

Pr(A=B) = p000 + p001 + p110 + p111 ...

Pr(A=B) + Pr(A=C) + Pr(B=C) = 1 + 2 p000 + 2 p111

... however, it is violated in QM, see e.g. page 9 here: http://www.theory.caltech.edu/people/pr ... /chap4.pdf

If we want to understand why our physics violates Bell inequalities, the above one seems the best to work on as the simplest and having absolutely obvious proof.

QM uses Born rules for this violation:

1) Intuitively: probability of union of disjoint events is sum of their probabilities: pAB? = pAB0 + pAB1, leading to above inequality.

2) Born rule: probability of union of disjoint events is proportional to square of sum of their amplitudes: pAB? ~ (psiAB0 + psiAB1)^2

Such Born rule allows to violate this inequality to 3/5 < 1 by using psi000=psi111=0, psi001=psi010=psi011=psi100=psi101=psi110 > 0.

I have just refreshed https://arxiv.org/pdf/0910.2724 (previous thread) adding section III about violation of this inequality using ensemble of trajectories: that proper statistical physics shouldn't see particles as just points, but rather as their trajectories to consider e.g. Boltzmann ensemble - it is in Feynman's Euclidean path integrals or its thermodynamical analogue: MERW (Maximal Entropy Random Walk: https://en.wikipedia.org/wiki/Maximal_e ... andom_walk ).

For example looking at [0,1] infinite potential well, standard random walk predicts rho=1 uniform probability density, while QM and uniform ensemble of trajectories predict different rho~sin^2 with localization, and the square like in Born rules has clear interpretation:

Do you know a different way to understand violation of this inequality?