= Solution
Use the <optimal three-to-one qubit random access code>. For Alice's bits $(b_1,b_2,b_3)$, put $s_j=(-1)^{b_j}$ and send the <pure state> with <Bloch vector>
$$
\boxed{r_b=(s_1,s_2,s_3)/\sqrt3.}
$$
These are the eight vertices of a cube inscribed in the <Bloch sphere>. If an explicit ket is wanted, choose $\cos\theta_b=s_3/\sqrt3$ and $e^{i\varphi_b}=(s_1+is_2)/\sqrt2$, and prepare $\cos(\theta_b/2)|0\rangle+e^{i\varphi_b}\sin(\theta_b/2)|1\rangle$.
Bob wanting bit $j$ performs a <Pauli measurement> of $\sigma_x,\sigma_y,\sigma_z$ respectively, reporting zero for <eigenvalue> $+1$ and one for $-1$. Its success for every input and every requested bit is
$$
\boxed{p_*=\frac12\left(1+\frac1{\sqrt3}\right)\simeq0.788675.}
$$
Only one chosen bit is extracted; the protocol does not recover all three simultaneously and uses no shared <entanglement>.
To prove that no one-qubit protocol gives a greater guaranteed probability, allow Alice arbitrary mixed encoding vectors $|r_b|\le1$ and Bob any binary <positive operator-valued measure>. Write his decision difference as $D_j=E_{j,0}-E_{j,1}=t_jI+v_j\cdot\sigma$. Its <eigenvalues> lie in $[-1,1]$, implying $|v_j|\le1-|t_j|\le1$. On input $b$, success for bit $j$ is $[1+s_j(t_j+v_j\cdot r_b)]/2$. Average over all eight strings and three choices. The terms in $t_j$ cancel, leaving
$$
p_{\mathrm{avg}}=\frac12+\frac1{48}\sum_b r_b\cdot V_b
\le\frac12+\frac1{48}\sum_b|V_b|,\qquad V_b=\sum_{j=1}^3s_jv_j.
$$
The <Cauchy-Schwarz inequality> and cancellation of cross terms over all sign choices give
$$
\frac18\sum_b|V_b|\le\sqrt{\frac18\sum_b|V_b|^2}
=\sqrt{\sum_j|v_j|^2}\le\sqrt3.
$$
Hence $p_{\mathrm{avg}}\le p_*$. Any worst-case guarantee is no greater than this average; the cube protocol attains it uniformly, proving optimality both for guaranteed success and for the uniform-input average. Biased prior information about the bits would define a different optimization problem.
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