CHSH and absolute value, plus impossibility of complete definiteness

Along the proof of CHSH inequality, the following point is reached :

$$C=int|A(a,x)B(b,x)-A(a,x)B(b’,x)|+|A(a’,x)B(b,x)+A(a’,x)B(b’,x)|dx$$

Then factorizes the A in each absolute value and deduces $Cleq 2$.

Now if one codes on a computer the function A(a,x) becomes a routine. Now suppose that due to impossibility of complete definiteness (analytically x could contain time and space parameters too, but thrn would not be averaged over) the arguments a,x are not sufficient to compute A, but A contains a call to the random generator :

int A(real a, real x, real p) { If(rand()<(1+pows(cos(a-x),p)/2) Return(1) Else Return(-1) }

Where

pows(x,p)=sign(x)exp(p*log(fabs(x)))

Then one cannot factorize the result of the A’s above and single values of C are in {0,2,4}.

Looking for the C as function of p gives a maximum at p=1/2, giving $Cleqapprox 2.33$. Taking away the absolute values with the same function A gives but 2 as maximum.

So my question is : is CHSH derivation correct since putting or not the absolute values gives always 2 whereas simulation with a call to random gives another average.