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Suppose you have a pair of anti-correlated spin-1/2 particles. Then the probability of measuring one particle to have spin-up along an axis [itex]\vec{A}[/itex] is [itex]\frac{1}{2}[/itex]. If you then measure the spin of the second particle along axis [itex]\vec{B}[/itex], then the probability of getting spin-up will be eitherI agree that Bell derived certain inequalities. But I do not necessarily agree that the key assumptions required to obtain the inequalities are the ones you think they are. However, for the purpose of the discussions here, I do not care about the derivation, the inequalities are valid and we can start from there as I've told you previously, although I'll be happy to discuss in another thread why those inequalities are more general than you think.

What correlations? This is one of the issues.Please spell out how you have arrived at this conclusion.

[itex]sin^2(\frac{1}{2} \theta)[/itex]

if the first measurement had result spin-up, or

[itex]cos^2(\frac{1}{2} \theta)[/itex]

if the first measurement had result spin-down, where [itex]\theta[/itex] is the angle between [itex]\vec{A}[/itex] and [itex]\vec{B}[/itex].

Define [itex]S_A[/itex] to be either +1, if the result of the first measurement was spin-up, and -1, if the result was spin-down. Define [itex]S_B[/itex] to be either +1, if the result of the second measurement was spin-up, and -1, if the result was spin-down. Then we can define a "correlation function"

[itex]\langle S_A \cdot S_B \rangle[/itex]

to be the expectation value of the product of [itex]S_A[/itex] and [itex]S_B[/itex]. From the assumed probabilities, we conclude:

[itex]\langle S_A \cdot S_B \rangle = - cos^2(\frac{1}{2} \theta) + sin^2(\frac{1}{2} \theta) = cos(\theta)[/itex]

where I used a trigonometric identity about half-angles.

So that's the quantum prediction for correlation, in the spin-1/2 case.