$${\left\langle\psi\right|}\hat{N}_{\text{O}}^{1\uparrow}{\left|\psi\right\rangle}=
{\left\langle\psi\right|}\hat{N}_{\text{O}}^{2\uparrow}{\left|\psi\right\rangle}=\frac{T}{2} ,
\label{27}$$ $${\left\langle\psi\right|}\hat{N}_{\text{O}}^{1\downarrow}{\left|\psi\right\rangle}=\frac{T}{2} ,
\label{28}$$ $${\left\langle\psi\right|}\hat{N}_{\text{O}}^{1\uparrow}\hat{N}_{\text{O}}^{2\uparrow} {\left|\psi\right\rangle}=
\frac{T^2}{2} \sin^2 \left( \frac{\theta_1\pm\theta_2}{2}\right)$$ and $${\left\langle\psi\right|}\hat{N}_{\text{O}}^{1\downarrow}\hat{N}_{\text{O}}^{2\uparrow} {\left|\psi\right\rangle}=
\frac{T^2}{2} \cos^2 \left( \frac{\theta_1\pm\theta_2}{2}\right) .$$

For the single terminal probability distributions in leads $i=1,2$ we get, in the presence and in the absence of an analyzer, respectively, $$\begin{aligned}
P^{\theta_i}(Q_i)&=&\left( \begin{array}{c}M\\Q_i \end{array}\right) 
\left (\frac{T}{2} \right )^{Q_i} \left (1-\frac{T}{2} \right )^{M-Q_i} \label{prob00}\\
P(Q_i)&=&\left( \begin{array}{c}M\\Q_i \end{array}\right)
\left (T \right )^{Q_i} \left (1-T \right )^{M-Q_i} ,
\label{prob0}\end{aligned}$$ so that the no-enhancement assumption reads: $$\left (1-\frac{T}{2} \right )^{(M-Q_i)} \left (\frac{1}{2} \right )^{Q_i} \le
(1-T)^{(M-Q_i)} ~~~~~~~i=1,2\; .
\label{noen1}$$ Note that the probabilities in Eqs. (\[prob00\]) and (\[prob0\]) do not depend on the angles $\theta_1$ and $\theta_2$ because the expectation values in Eqs. (\[27\]) and (\[28\]) are invariant under spin rotation. As a consequence, the effect of the analyzer is equivalent to a reduction of the transmission probability $T$ by a factor of 2, resulting in a shift of the maximum of the distribution. From Eq.(\[noen1\]) it follows that, for a given number $M=eVt/h$ of entangled pairs generated by the entangler, the no enhancement assumption can be verified only for certain values of $T$ and of $Q_i$. This makes clear that the CH inequality of Eq.(\[CHin\]) can be tested for violation 