Because this line of argument leads inevitably to Bell inequalities, as I and others have been trying to explain to you since you started posting here. Suppose we have some large number of particle pairs, from the above you should agree that in each pair, the two particles should have some definite set of predetermined results like [+ on a, - on b, + on c] or [- on a, +on b, + on c] etc.? And for any collection of things (like particle pairs) where each member of the collection either does or doesn't have each of three possible properties A, B, and C (say A=+ on angle a, B=+ on angle b, C=+ on angle c, so "not A"=- on angle a, "not B"=- on angle b, and "not C"=- on angle c), simple arithmetic shows the whole collection must satisfy this inequality:
Number(A, not B) + Number(B, not C) ≥ Number(A, not C)
There's a proof on
this page, but I think their proof is not as simple as it could be, the simplest way of seeing it is this:
Number(A, not B) = Number(A, not B, C) + Number(A, not B, not C) [since any member of the group satisfying A, not B must either have or not have property C]
Number(B, not C) = Number(A, B, not C) + Number(not A, B, not C)
Number(A, not C) = Number(A, B, not C) + Number(A, not B, not C)
And plugging this into the above inequality and cancelling like terms from both sides gives:
Number(A, not B, C) + Number(not A, B, not C) ≥ 0
Which obviously must be true since the number with any given set of properties must be ≥ 0!
Anyway, whether you like my proof or the one on the page I linked to better, hopefully you agree that if we knew the complete set of three predetermined properties for a collection of particle pairs, the inequality Number(A, not B) + Number(B, not C) ≥ Number(A, not C) would be satisfied? If so, it's a short step from there to the statement that if you
measure two properties for a large number of particle pairs, P(A, not B|measured a and b) + P(B, not C|measured b and c) ≥ P(A, not C|measured a and c) (basically the only extra assumption needed is that the probability the experimenters will pick a given pair of axes to measure is uncorrelated with the triplet of predetermined results prior to measurement). I discuss this more on post #11
here, but we can also discuss it here if you agree with the inequality Number(A, not B) + Number(B, not C) ≥ Number(A, not C) for
all particle pairs but don't agree that for measurements this implies P(A, not B|measured a and b) + P(B, not C|measured b and c) ≥ P(A, not C|measured a and c).