 Solution

Alice verifies each submitted qubit in her recorded secret basis and checks its bit against the revealed number. For a forger who chooses X or Z at random and prepares the corresponding known-bit eigenstate, each bit passes with probability $3/4$. With independent basis choices,The detection probability depends on Daisy's preparation strategy. The question does not restrict her to legitimate X/Z eigenstates. Once the bit is known, for bit zero the averaged acceptance operator isFor bit one it is $A_1=I-A_0$. Both have largest eigenvalue $p_{\mathrm{opt}}=(1+1/\sqrt{2})/2$. Preparing the corresponding eigenstate, a Bloch-vector bisector of the two candidate encodings, achieves this value; for bit zero it is $\cos (\pi /8)|0⟩+\sin (\pi /8)|1⟩$.

The acceptance operator for the whole known check number is ${⨂}_j{A}_{b_j}$, whose largest eigenvalue is $p_{\mathrm{opt}}^{24}$, so entangled preparation cannot improve this no-original-note optimum. Therefore unrestricted preparation instead givesThe random-basis value is the elementary protocol calculation; the second value states the best forging strategy allowed by arbitrary state preparation and no access to the original quantum note.