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The Greenberger–Horne–Zeilinger Game
The Greenberger–Horne–Zeilinger (GHZ) game is another interesting example of quantum pseudo-telepathy. Classically, the game has 75% winning probability. However, with a quantum strategy, the players will always win with winning probability equals to 1.

There are three players, Alice, Bob, and Carol playing against a referee. The referee poses a question $$\in \{0,1\}$$ to each of the players. The three players each respond with an answer $$\in \{0,1\}$$. The referee draws three questions x, y, z uniformly from the 4 options $$\{(0,0,0), (1,1,0),(1,0,1),(0,1,1)\}$$. As a clarification, if question triple $$(0,1,1)$$ is chosen, then Alice receives bit 0, Bob receives bit 1, and Carol receives bit 1 from the referee. Based on the question bit received, Alice, Bob, and Carol each respond with an answer a, b, c also in the form of 0 or 1. The players can formulate a strategy together prior to the start of the game. However, no communication is allowed during the game itself.

The players win if $$a \oplus b \oplus c = x \lor y \lor z$$, where $$\lor$$ indicates OR condition and $$\oplus$$ indicates summation of answers modulo 2. In other words, the sum of three answers has to be even if $$x=y=z=0$$. Otherwise, the sum of answers has to be odd.

Classical Strategy
Classically, Alice, Bob, and Carol can employ a deterministic strategy that always end up with odd sum (e.g. Alice always output 1. Bob and Carol always output 0). The players win 75% of the time and only lose if the questions are $$(0,0,0)$$.

In fact, this is the best winning strategy classically. We can only satisfy a maximum of 3 out of 4 winning conditions. Let $$a_0, a_1$$ be Alice's response to question 0 and 1 respectively, $$b_0, b_1$$ be Bob's response to question 0, 1, and $$c_0, c_1$$ be Carol's response to question 0, 1. We can write all constraints that satisfy winning conditions as

$$ a_0 + b_0 + c_0 = 0\mod 2$$

$$ a_1 + b_1 + c_0 = 1\mod 2$$

$$ a_1 + b_0 + c_1 = 1\mod 2$$

$$ a_0 + b_1 + c_1 = 1\mod 2$$

Suppose that there is a classical strategy that satisfies all four winning conditions, all four conditions hold true. Through observation, each term appears twice on the left hand side. Hence, the left side sum = 0 mod 2. However, the right side sum = 1 mod 2. The contradiction shows that all four winning conditions cannot be simultaneously satisfied.

Quantum Strategy
Now we have come to the interesting part where Alice, Bob, and Carol decided to adopt a quantum strategy. The three of them now share a tripartite entangled state $$ |{\psi}\rangle = \frac{1}{2} (|000\rangle - |011\rangle - |101\rangle - |110\rangle)$$, where they each pocesses a qubit to take measurement.

If question 0 is received, the player takes measurement in standard basis $$ \{|0\rangle, |1\rangle\}$$. If question 1 is obtained, the player takes measurement in diagonal basis $$ \{|+\rangle, |-\rangle\}$$. If measuring in standard basis, the player responds with answer 0 if the measurement result is 0 and answer 1 if the measurement result is 1. If measuring in diagonal basis, the player responds with answer 0 if $$|+\rangle$$ is measured, and answer 1 if $$|-\rangle$$ is measured. This quantum strategy in fact has winning probability = 1.

We will show one example where the referee chooses question triple $$(1,1,0)$$. Alice and Bob receive question 1 and measure in diagonal basis. Carol receives question 0 and measures in standard basis. To simulate measurement results, note that measurement in diagonal basis is equivalent to Hadamard transform followed by standard basis measurement.

$$(I \otimes H \otimes H) | \psi \rangle $$

$$= \frac{1}{2}(|++0\rangle - |+-1\rangle - |-+1\rangle - |--0\rangle)$$

$$=\frac{1}{4}\Big((|000\rangle + |010\rangle + |100\rangle + |110\rangle) - (|001\rangle + |011\rangle - |101\rangle - |111\rangle) - (|001\rangle - |011\rangle + |101\rangle - |111\rangle) - (|000\rangle - |010\rangle - |100\rangle + |110\rangle)\Big)$$

$$=\frac{1}{2}(|010\rangle + |100\rangle + |111\rangle - |001\rangle)$$

Due to quantum entanglement, the post measurement states collapse into one of the four possibilities. Regardless of the measurement outcomes, the sum of three measurement = 1 mod 2, which satisfies the winning condition.

The same procedure can be repeated for all four possible question triples $$\{(0,0,0), (1,0,1), (0,1,1), (1,1,0)\}$$. Regardless of the questions given by the referee, the players always win the GHZ game with the quantum magic. In the GHZ game, a quantum strategy (winning probability = 1) performs significantly better than a classical strategy (winning probability = 0.75).