Quantum correlations
Consider two entangled qubits in the singlet state
|\psi\rangle
= \frac{1}{\sqrt{2}} \left( |01\rangle-|10\rangle \right)
and note that the projector |\psi\rangle\langle\psi| can be written as
|\psi\rangle\langle\psi|
= \frac{1}{4} \left(
\mathbf{1}\otimes\mathbf{1}- \sigma_x\otimes\sigma_x - \sigma_y\otimes\sigma_y - \sigma_z\otimes \sigma_z
\right)
where \sigma_x, \sigma_y, and \sigma_z are our old friends the Pauli matrices.
Also recall that any single-qubit observable with values \pm1 can be represented by the operator
\vec{a}\cdot\vec\sigma
= a_x\sigma_x + a_y\sigma_y + a_z\sigma_z,
where \vec{a} is a unit vector in the three-dimensional Euclidean space.
So if Alice and Bob both choose observables, then we can characterise their choice by vectors \vec{a} and \vec{b}, respectively.
If Alice measures the first qubit in our singlet state |\psi\rangle, and Bob the second, then the corresponding observable is described by the tensor product
A\otimes B
= (\vec{a}\cdot\vec\sigma)\otimes(\vec{b}\cdot\vec\sigma).
The eigenvalues of A\otimes B are the products of eigenvalues of A and B.
Thus A\otimes B has two eigenvalues: +1, corresponding to the instances when Alice and Bob registered identical outcomes, i.e. (+1,+1) or (-1,-1); and -1, corresponding to the instances when Alice and Bob registered different outcomes, i.e. (+1,-1) or (-1,+1).
This means that the expected value of A\otimes B, in any state, has a simple interpretation:
\langle A\otimes B\rangle = \Pr (\text{outcomes are the same}) - \Pr (\text{outcomes are different}).
This expression can take any real value in the interval [-1,1], where -1 means we have perfect anti-correlations, 0 means no correlations, and +1 means perfect correlations.
We can evaluate the expectation value in the singlet state:
\begin{aligned}
\langle\psi|A\otimes B|\psi\rangle
& = \operatorname{tr}\Big[
(\vec{a}\cdot\vec\sigma)\otimes(\vec{b}\cdot\vec\sigma) |\psi\rangle\langle\psi|
\Big]
\\& = -\frac{1}{4} \operatorname{tr}\Big[
(\vec{a}\cdot\vec\sigma)\sigma_x \otimes(\vec{b}\cdot\vec\sigma)\sigma_x
+ (\vec{a}\cdot\vec\sigma)\sigma_y \otimes(\vec{b}\cdot\vec\sigma)\sigma_y
+ (\vec{a}\cdot\vec\sigma)\sigma_z \otimes(\vec{b}\cdot\vec\sigma)\sigma_z
\Big]
\\& = -\frac{1}{4} \operatorname{tr}\Big[
4(a_x b_x + a_y b_y + a_z b_z) \mathbf{1}\otimes\mathbf{1}
\Big]
\\& = -\vec{a}\cdot\vec{b}
\end{aligned}
where we have used the fact that \operatorname{tr}(\vec{a}\cdot\vec\sigma)\sigma_k = 2a_k (for k=x,y,z).
So if Alice and Bob choose the same observable \vec{a} = \vec{b}, then the expected value \langle A\otimes B\rangle will be equal to -1, and their outcomes will always be opposite: whenever Alice registers +1 (resp. -1) Bob is bound to register -1 (resp. +1).