Reconciling two different representations of Bell's inequality I have a troubling question. The original Bell inequality:
$$|P(\vec{a},\vec{b}) - P(\vec{a},\vec{c})| - P(\vec{b},\vec{c}) \leq 1$$
is maximally violated at angles of 60 and 120 degrees. This is clear to me, because:
$$P_{xy} = -cos(\theta_x - \theta_y)$$
And naturally:
$$-cos(60) = -0.5, -cos(120) = 0.5$$
hence the inequality becomes:
$$1.5 \nleq 1$$
However, some literature, including A simple proof of Bell’s inequality use the following variant of inequality:
$$P_{same}(A, B) + P_{same}(A, C) + P_{same}(B, C) \geq 1$$
Which they then proceed to prove as maximally violated at 60/120 degrees with:
$$\frac{1}{4} + \frac{1}{4} + \frac{1}{4} \ngeq 1$$
How do you arrive from the first to second conclusion? Rieffel and Polak even explicitly use the following approach, without, unfortunately explaining the reasoning (at least not well enough to me):
$$cos^2{\theta} + cos^2{\phi} + cos^2{(\theta + \phi)} ≥ 1$$
The math works because squaring 0.5 three times will indeed give us $\frac{3}{4}$, but why are they doing this?
 A: Where $P_{same}(A,B) + P_{same}(A,C) + P_{same}(B,C) \geq 1$ is coming from:
This is explained in one of the references of ''A simple proof of Bell's inequality'', lecture notes by Preskill, see Preskill 1-6, page 147. The argument goes as follows:
If you have a coin with three properties $A,B,C$ which can either be 0 or 1 (or just say you have three numbers $A,B,C$ which can each be either 0 or 1), two of them are always equal. For example (1,0,0) means $B=C$, (0,1,0) means $A=C$. They just cannot be all different because there are two choices (0 or 1) for three slots.
$P_{same}(A,B)= \bigg\{ \begin{matrix} 1 ,~(A=B) \\ 0,~ (A\neq B) \end{matrix}$ implies $P_{same}(A,B) + P_{same}(A,C) + P_{same}(B,C) \geq 1.$
(Maybe this was already clear. If yes, excuse the redundant explanation.)
How to compute the l.h.s. for a Bell state:
Coming back to the Bell experiment, the paper suggests three measurement directions $a$, $b$ and $c$ with $0^\circ, 60^\circ$ and $120^\circ$, respectively. They have the following eigenvectors $e_1$ and in direction of the measurement and $e_0$ perpendicular to it:
$$ |a_1\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix},~ |a_0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}~;~~ |b_1\rangle = \frac{1}{2} \begin{pmatrix}  \sqrt{3} \\ -1 \end{pmatrix},~ |b_0\rangle = \frac{1}{2} \begin{pmatrix} 1 \\ \sqrt{3} \end{pmatrix}~;~~ |c_1\rangle = \frac{1}{2} \begin{pmatrix}  \sqrt{3} \\ 1 \end{pmatrix},~ |c_0\rangle = \frac{1}{2} \begin{pmatrix}  1 \\ -\sqrt{3} \end{pmatrix}. $$
Each eigenvector is given by $e_1(\theta) = \begin{pmatrix}  \sin(2\theta) \\ \cos(2\theta) \end{pmatrix}$ (really just the direction of the measurement on the bloch sphere, i.e., with $2\theta$) for $\theta = 0, 60^\circ$ and $120^\circ$. The perpendicular one is hence $e_0(\theta) = \begin{pmatrix}  \cos(2\theta) \\ -\sin(2\theta) \end{pmatrix}$.
The considered bell state is:
$$ |\Phi^+\rangle = \frac{1}{\sqrt{2}} ( |00 \rangle + |11 \rangle )  = \frac{1}{\sqrt{2}} \Big[ \begin{pmatrix} 1 \\ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \\ 0\end{pmatrix}  + \begin{pmatrix} 0 \\ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \\ 1 \end{pmatrix} \Big]$$
In the "simple proof of Bell's inequality" it is shown in in eq. (3) that this entangled state corresponds to two entangled states with the same properties regarding $a, b$ and $c$. The probabilities $P_{same}(A,B)$ etc. are the chances that we measure $a_0$ in the first and $b_0$ in the second state, e.g.,
$$ | (\langle a_0 | \otimes \langle b_0| | \Phi^+ \rangle|^2 = 1/8,~ | (\langle a_1 | \otimes \langle b_1| | \Phi^+ \rangle|^2 = 1/8 \Rightarrow P_{same}(A,B) = 1/8 + 1/8 = 1/4.$$
These are also worked out in the paper below eq. (3).
I hope this clarifies where all this is coming from (and not making it more confusing).
