Jilang said:
If one of the variables is pre-existing and the other is random until measured how can they they be lumped together?
The whole idea of hidden variables is to explain EPR in terms of purely local interactions. So if it's purely local, then Alice's result depends on \lambda, which is state information that she shares with Bob, \alpha, which is Alice's choice of settings, and possibly \lambda_A, which is some other variable local to Alice (maybe it describes the details of Alice's detector). Similarly, Bob's result depends on \lambda, \beta, which is Bob's choice of settings, and possibly \lambda_B, some details about Bob's local situation.
So what I think you're suggesting is that \lambda_A and \lambda_B might be random, determined at the moment that Alice and Bob, respectively, perform their measurements. I'm pretty sure that can't possibly make any difference, unless you somehow say that the settings of \lambda_A and \lambda_B are correlated, which would just push the problem back to how their correlations are enforced.
In any case, the perfect anti-correlations of EPR imply that \lambda_A and \lambda_B can have no effect in a local model.
Let P_A(A | \alpha, \lambda, \lambda_A) be the probability that Alice gets measurement result A (\pm 1) given shared hidden variable \lambda, setting \alpha and local random variable \lambda_A. Similarly, let P_B(B | \beta, \lambda, \lambda_B) be the probability that Bob gets B given his setting \beta, the value of the shared hidden variable, \lambda and his local random variable \lambda_B.
The anti-correlated EPR probabilities tells us that if \alpha = \beta, then there is no possibility of Alice and Bob getting the same result. What that means is that for all possible values of \lambda, the product
P_A(A | \alpha, \lambda, \lambda_A) P_B(A | \alpha, \lambda, \lambda_B) = 0
This means that if P_A(A | \alpha, \lambda, \lambda_A) \neq 0, then P_B(A |\alpha, \lambda, \lambda_B) = 0. Since there are only two possible results for Bob, if one of the results has probability 0, then the other has probability 1. So we conclude:
This means that if P_A(A | \alpha, \lambda, \lambda_A) \neq 0, then P_B(A |\alpha, \lambda, \lambda_B) = 0. Since there are only two possible results for Bob, if one of the results has probability 0, then the other (-A) has probability 1. So we conclude:
If P_A(A | \alpha, \lambda, \lambda_A) \neq 0, then P_B(-A |\alpha, \lambda, \lambda_B) = 1.
But it's also true that P_A(-A | \alpha, \lambda, \lambda_A) P_B(-A | \alpha, \lambda, \lambda_B) = 0, so if P_B(-A | \alpha, \lambda, \lambda_B) = 1, then P_A(-A | \alpha, \lambda, \lambda_A) = 0 and so P_A(+A | \alpha, \lambda, \lambda_A) = 1. So we have:
If P_A(A |\alpha, \lambda, \lambda_A) \neq 0 then P_A(A | \alpha, \lambda, \lambda_A) = 1
Similarly,
If P_B(B |\alpha, \lambda, \lambda_B) \neq 0 then P_B(B | \alpha, \lambda, \lambda_B) = 1
That means that the probabilities for Alice's possible results are either 0 or 1, and similarly for Bob. That means that Alice's result is actually a deterministic function of the parameters \alpha, \lambda, \lambda_A, and similarly, Bob's result is a a deterministic function of \beta, \lambda, \lambda_B. So there are two functions, F_A(\alpha, \lambda, \lambda_A) which returns +1 or -1, giving the result of Alice's measurement, and a second function, F_B(\alpha, \lambda, \lambda_B) giving the result of Bob's measurement.
Now, again, perfect anti-correlation means that if \alpha = \beta, then F_A(\alpha, \lambda, \lambda_A) = - F_B(\alpha, \lambda, \lambda_B). That has to be true for all values of \lambda_A. That means that F_A(\alpha, \lambda, \lambda_A) doesn't actually depend on \lambda_A, and similarly F_B(\beta, \lambda, \lambda_B) doesn't actually depend on \lambda_B. So extra hidden variables, if they are local and uncorrelated, have to be irrelevant.