= Solution
\b[No test on Alice and Bob's systems gives any information about which global state was prepared, even with arbitrarily many independent copies.] In the pure GHZ <density matrix>, the off-diagonal terms contain $|0\rangle\langle1|_C$ or its adjoint. Their <partial traces> vanish. Hence both preparations have the same <reduced density matrix>
$$
\sigma_{AB}=\operatorname{Tr}_C(|\mathrm{GHZ}\rangle\langle\mathrm{GHZ}|)
=\operatorname{Tr}_C\rho
=\tfrac12\bigl(|00\rangle\langle00|+|11\rangle\langle11|\bigr).
$$
For any <POVM> effect $E$ acting jointly on $AB$, the outcome <probability> is $\operatorname{Tr}(E\sigma_{AB})$ in either case. <Equal reduced states give identical local statistics>, where local here means local to the entire accessible laboratory $AB$, not separate operations on $A$ and $B$.<Foundations I: States and Ensembles by John Preskill>[§2.3.1, printed pp. 14–17, reduced density operators and accessible expectation values]{ref}
For $n$ independent copies, taking the <partial trace> of each inaccessible $C$ system yields $\sigma_{AB}^{\otimes n}$ under either hypothesis. Even an arbitrary collective <quantum measurement> on all $2n$ accessible <qubits> therefore has identical outcome <probabilities>. Adaptive local processing cannot change equality of the accessible <density matrices>. With equal prior <probabilities> the optimal guessing <probability> remains $1/2$; with other priors, no observation updates their ratio.
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