 Solution

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 $|0⟩⟨1{|}_C$ or its adjoint. Their partial traces vanish. Hence both preparations have the same reduced density matrixFor any POVM effect $E$ acting jointly on $AB$, the outcome probability is $\mathrm{Tr}(E{\sigma }_{AB})$ 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 $B$.^[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.