= 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,
$$
\boxed{P_{\rm detect}^{\rm random}=1-\left(\frac34\right)^{24}
=0.9989966\ldots.}
$$
\b[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 is
$$
A_0=\frac12\left(|0\rangle\langle0|+|+\rangle\langle+|\right)
=\frac I2+\frac{X+Z}{4}.
$$
For bit one it is $A_1=I-A_0$. Both have largest <eigenvalue> $p_{\rm opt}=(1+1/\sqrt2)/2$. Preparing the corresponding <eigenstate>, a <Bloch vector>[Bloch-vector] bisector of the two candidate encodings, achieves this value; for bit zero it is $\cos(\pi/8)|0\rangle+\sin(\pi/8)|1\rangle$.
The acceptance operator for the whole known check number is $\bigotimes_jA_{b_j}$, whose largest <eigenvalue> is $p_{\rm opt}^{24}$, so <quantum entanglement>[entangled] preparation cannot improve this no-original-note optimum. Therefore unrestricted preparation instead gives
$$
\boxed{P_{\rm detect}^{\rm optimal}=1-\left(\frac{1+1/\sqrt2}{2}\right)^{24}
=0.9776367\ldots.}
$$
The 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.
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