The violation of the Bell inequality demonstrates that the classical theories based on local realism cannot completely explain the correlations predicted by quantum mechanics.
CHSH Game
Consider the following game.
Alice and Bob receive random binary variable $x$ and $y$ respectively, and must output binary variables $a$ and $b$.
If $a \oplus b = x \cdot y$, they win the game.
What is the maximum probability of winning?
CHSH Inequality
We can abstract this game using algebra:
Let $A_x$ denote Alice's output when she receives $x$. If Alice outputs $0$, $A_x = -1$; otherwise, $A_x = +1$.
$B_y$ is defined similarly to $A_x$.
Now we want to maximize:
$$ \begin{aligned} & \left\langle A_0B_0 \right\rangle + \left\langle A_0B_1 \right\rangle + \left\langle A_1B_0 \right\rangle - \left\langle A_1B_1 \right\rangle \\ =\;& \left\langle A_0\left(B_0 + B_1\right) \right\rangle + \left\langle A_1\left(B_0 - B_1\right) \right\rangle \end{aligned} $$
It is easy to see that either $B_0 + B_1$ or $B_0 - B_1$ equals zero. Therefore, the maximum of this formula is $2$.
This is the CHSH inequality:
$$\left\lvert \left\langle A_0B_0 \right\rangle + \left\langle A_0B_1 \right\rangle + \left\langle A_1B_0 \right\rangle - \left\langle A_1B_1 \right\rangle \right\rvert \leq 2$$
Quantum mechanics can violate the CHSH inequality.
Now assume we have a pair of qubits in the Bell state:
$$\left\lvert \psi \right\rangle = \frac{\left\lvert 0 \right\rangle \otimes \left\lvert 1 \right\rangle - \left\lvert 1 \right\rangle \otimes \left\lvert 0 \right\rangle}{\sqrt2}$$
where $\left\lvert 0 \right\rangle$ and $\left\lvert 1 \right\rangle$ are the eigenstates of one of the Pauli matrices:
$$\sigma_z = \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}$$
Now we define the observables as follows:
$$ \begin{aligned} & A_0 = \sigma_z,\ A_1 = \sigma_x = \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix} \\ & B_0 = -\frac{\sigma_z + \sigma_x}{\sqrt2},\ B_1 = \frac{\sigma_x - \sigma_z}{\sqrt2} \end{aligned} $$
According to the Born rule:
$$\left\langle A_0 \otimes B_0\right\rangle = \frac1{\sqrt2},\ \left\langle A_0 \otimes B_1\right\rangle = \frac1{\sqrt2},\ \left\langle A_1 \otimes B_0\right\rangle = \frac1{\sqrt2},\ \left\langle A_1 \otimes B_1\right\rangle = -\frac1{\sqrt2}$$
Hence, the sum is:
$$\left\langle A_0 \otimes B_0\right\rangle + \left\langle A_0 \otimes B_1\right\rangle + \left\langle A_1 \otimes B_0\right\rangle - \left\langle A_1 \otimes B_1\right\rangle = 2\sqrt2$$
Probability of Winning
$$P_{win} = \frac14\sum\limits_{x, y \in \left\{0, 1\right\}}\frac{1 + \left(-1\right)^{(x \cdot y)}\left\langle A_x \otimes B_y \right\rangle}2 = \frac12 + \frac18 \cdot 2\sqrt2 \approx 0.8536$$
In the classical case, the probability of winning is:
$$P_{win} = \frac12 + \frac18 \cdot 2 = 0.75$$