Indeed, they are equivalent - true if we choose any probability distribution among 2^3=8 possibilities, but can be violated by physics, QM ... and other models with Born rule: pAB? ~ (psiAB0 + psiAB1)^2, like realistic and "4D local": ensemble of trajectories.

Jarek wrote:Mikko, substituting Pr(A≠B) = 1 - Pr(A=B) we get:
Pr(A≠B) + Pr(A≠B) + Pr(A≠B) <= 2
It is a bit different than what you have written - I am not certain about its status (?)

Apparently you meant to write Pr(A≠B) + Pr(A≠C) + Pr(B≠C) <= 2. There are two ways to compare these inequalities. One may regard them as two different inequalities about the same (or different) variables. They can be inferred from the same general assumptions by largely similar reasoning and used for similar purposes. Another way is to regard them as the same inequality but using different variables:

Start with Pr(A=B) + Pr(A=C) + Pr(B=C) >= 1.
Define X, Y, Z as X = A, Y = B, Z ≠ C where the last definition is unambiguous because C is a binary variable.
Now we have Pr(X=Y) + Pr(X≠Z) + Pr(Y≠Z) >= 1.
Move the first term to other side: Pr(X≠Z) + Pr(Y≠Z) >= 1 - Pr(X=Y).
Rewrite the right side: Pr(X≠Z) + Pr(Y≠Z) >= Pr(X≠Y).

Mikko, substituting Pr(A≠B) = 1 - Pr(A=B) we get:
Pr(A≠B) + Pr(A≠B) + Pr(A≠B) <= 2
It is a bit different than what you have written - I am not certain about its status (?)

Anyway, the general problem is that violation of such looking obvious inequalities (by QM and physics) is believed to require giving up local realism.
Using ensemble (uniform, Boltzmann) of paths also allows to violate it in similar way (through Born rules) - this is realistic model, and in fact required if we e.g. think of general relativity: where we need to consider entire spcatime, particles are their paths.
It is not local in "evolving 3D" picture, but it is local in 4D spacetime/Einstein's block universe view - where particles are their trajectories, ensembles of such objects we should consider.

Ok, I should specify my question: what other realistic models (with some objective situation) allow to violate this looking obvious inequality?
Do you agree that considering ensembles of paths allows for that?

Jarek wrote: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

This could be generalized a little by writing it like a triangle inequality:

Pr(A≠B) + Pr(A≠C) ≤ Pr(B≠C)

This way it is not necessary to restrict to binary variables. In addition, the analogy with the triangle inequality of geometry helps to understand why this inequality makes sense. These considerations are of course unimportant to current discussion.

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

Of course there is the standard quantum mechanics and e.g. the Copenhagen interpretation.