Greenberger, Horne, and Zeilinger (1989)

Let there be Alice, Bob and Charlie. Each get a single-bit prompt

x,y,z\in \{0,1\}

They are allowed a single bit

a,b,c\in\{\pm 1\}

No communication between them is allowed once the game starts.

Before the game starts, they can agree on a strategy and share resources. But after the prompt comes in, they are isolated.

The referee picks one of four prompts uniformly at randomly (25% chance of picking one)

$x$	$y$	$z$	required $abc$
0	0	0	$-1$
1	1	0	$+1$
0	1	1	$+1$
1	0	1	$+1$
$x$ is given to Alice, $y$ is given to Bob, $z$ is given to Charlie
These four are valid as the parity is 0, i.e.,

x\oplus y \oplus z = 0

So basically

abc = \begin{cases} -1 & \text{if } x+y+z = 0 \\ +1 & \text{if } x+y+z = 2 \end{cases}

Local realism caps this at 75% as classically if they were to guess, there would be 2^2=4 possible Function per player and 4^3=64 total strategies. Can any of them win all four scenarios? There are no strategies that win all of the time. The best that one can do is 75%.

Quantum strategy allows Alice, Bob and Charlie to share the GHZ state

\ket{\text{GHZ}} = \tfrac{1}{\sqrt{2}}\big(\ket{000} - \ket{111}\big)

If you receive a 0 bit, then measure $\sigma^X$ otherwise if you receive a 1 bit, then measure $\sigma^Y$ This turns the GHZ state of all four operator products $\sigma^X\sigma^X \sigma^X$ and $\sigma^X \sigma^Y\sigma^Y$ and all its cyclic permutations.

These all have the right eigenvalues to win every scenario with probability one.

The strategy

Promote $a_i, b_i, c_i$ to be Observable $\hat{A_i},\hat{B_i},\hat{C_i}$ Let

\hat{A}_0=\hat{B}_0=\hat{C}_0=\sigma^X

\hat{A}_1=\hat{B}_1=\hat{C}_1=\sigma^Y

These all mutually commutes

\sigma^X\otimes \sigma^X \otimes \sigma^X=-1

\sigma^Y\otimes \sigma^Y\otimes \sigma^X = +1

and all cyclic permutations...

Because

\sigma^X\sigma^X\sigma^X\ket{\psi}=-\ket{\psi}\quad \sigma^Y\sigma^Y\sigma^X\ket{\psi}=\ket{\psi}

hence the win percentage is 100%.