Quote by Gordon Watson
Bill, you talkin' to me? (In that you cite gill1109.)
1. Not sure about your mystical being? Purpose =? (Is something more needed to clarify the OP?)
2. The CHSH inequality formulation will be the same, imho: since the experimental outcomes are ±1, no matter the settings a, b, etc.
3. To clarify the OP (if that's your issue): Having derived the expectation E(AB) for the classical setting  from your functions for A and B = ±1  what then is the related maximum value that that classical setting might yield for the CHSH inequality? That is: What a, b, c, d settings yield the maximum value in the CHSH formula, and what is that maximum?
4. Is it gill1109's +2?
5. Did you mean to say "the scenario is manifestly LOCAL"???
6. So  addressing your "What gives"  just give me your answers to the OP  or tell me why you can't. Especially as it seems that Bell might think you can; the given situation being wholly classical and involving no more than Bell's proposed (1964, etc.) analytical formulation.

Sorry for hijacking your thread Gordon, I was just responding to the portion by gill1109. To answer your questions, and more on topic.
 Without specifying the method by which the common pulse orientations are chosen, it is not possible to calculate an expectation value.Without ρ(x) we are hopeless to calculate a meaningful E(AB) even if A(a,x) and B(b,x) are clearly specified.
 The maxumum attainable is of course +2 as gill1109 calculated.
Aside:
However, note the following extremely important point
A(a1)B(b1)A(a1)B(b2)A(a2)B(b2)A(a2)B(b1)
→ A(a1)[B(b1)B(b2)]  A(a2)[B(b2)B(b1)]
→ A(a1)[B(b1)B(b2)] + A(a2)[B(b1)B(b2)]
→ [A(a1) + A(a2)]*[B(b1)B(b2)] > **!!!
if A(a1) = A(a2) = +1, and B(b1) = B(b2) = +1 Or,
A(a1) = A(a2) = 1 and B(b1) = B(b2) = 1, we obtain the maxium of 2.
If A(a1) = A(a2) = ±1, OR B(b1) = B(b2) = ±1, we get a value of zero.
And if A(a1) = A(a2) = 1 and B(b1) = B(b2) = +1 Or
A(a1) = A(a2) = +1 and B(b1) = B(b2) = 1, we obtain the minimum of 2.
This may seem like a pointless way to arrive at the same result as gill1109 except it is obvious from the emphasized expresion that the original 4 terms (A(a1)B(b1), A(a1)B(b2), A(a2)B(b2), A(a2)B(b1)) of products in the inequality originate from only 4 functions (A(a1), B(b1), A(a2), B(b2)) which must be factorizable. You can not use 4 different runs of an experiment (i, j, k, l) to obtain results from 8 functions (A(a1i), B(b1i), A(a1j), B(b2j), A(a2k), B(b2k), A(a2l), B(b1l)) and expect the inequality to work. It is a simple exercise to verify that for the case where 4 different runs of the experiment are performed, the maximum of the expression will be
A(a1i)B(b1i)  A(a1j)B(b2j) A(a2k)B(b2k) A(a2l)B(b1l) <=
4
NOT 2.
Some naively leave out the experiment identifyers (i,j,k,l) and fool themselves into thinking the result can be factorized.
In order for the results from 4 different experiments to be factorizable the following equalities must hold
A(a1i) = A(a1j)
A(a2k) = A(a2l)
B(b1i) = B(b1l)
B(b2j) = B(b2k)
Practically, this means if the experimental results consisted of a list of numbers (+1, 1) for each function and you obtained 8 columns for 4 different experiments, the data MUST be sortable such that 4 of the columns are duplicates, not only in the numbers of +1s and 1s but also in the switching pattern.
Therefore it is not sufficient that A*B for one experiment gives you a certain expection value for the paired product. For the inequality to have a maximum of 2, rather than 4, the value of one pair must constrain the value of a different pair in some manner.