= Solution
For deterministic classical responses, denote the two possible local answers by $x_0,x_1$, $y_0,y_1$ and $z_0,z_1$. Certainty in the <GHZ game> would require
$$
\begin{aligned}
x_0\oplus y_0\oplus z_0&=0,\\
x_0\oplus y_1\oplus z_1&=1,\\
x_1\oplus y_0\oplus z_1&=1,\\
x_1\oplus y_1\oplus z_0&=1.
\end{aligned}
$$
Taking the <exclusive or> of all four equations gives zero on the left, since every answer occurs twice, but one on the right. This contradiction rules out deterministic certainty. Pre-shared randomness cannot cure it: conditioning on the random seed fixes a deterministic response table, and none succeeds on every allowed input. Thus \b[noncommunicating classical players cannot win with certainty]. No uniform distribution on the referee's questions is needed for this worst-case claim.
For the quantum strategy, share the <GHZ state> and measure the <Pauli X gate> observable when the local question is 0, or the <Pauli Y gate> observable when it is 1. Return bit 0 for measurement <eigenvalue> $+1$ and bit 1 for <eigenvalue> $-1$. The four relevant operator identities are
$$
\begin{aligned}
(X\otimes X\otimes X)|GHZ\rangle&=|GHZ\rangle,\\
(X\otimes Y\otimes Y)|GHZ\rangle&=-|GHZ\rangle,\\
(Y\otimes X\otimes Y)|GHZ\rangle&=-|GHZ\rangle,\\
(Y\otimes Y\otimes X)|GHZ\rangle&=-|GHZ\rangle.
\end{aligned}
$$
For the first identity, the three bit flips exchange $|000\rangle$ and $|111\rangle$. For each remaining identity, the two Y actions contribute $i^2=-1$ on $|000\rangle$ and $(-i)^2=-1$ on $|111\rangle$, so they exchange the terms with a minus sign. Local measurements on separate systems commute, and the product of their observed <eigenvalues> is therefore fixed by the displayed joint <eigenvalue>. That product equals $(-1)^{x+y+z}$, giving exactly the required output parity for each promised input. Hence \b[the quantum strategy wins with certainty without communication]. This is the <GHZ theorem> in nonlocal-game form; it does not let a player control another player's individual outcome.
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