Optimal three-to-one qubit random access code ($p_{\ast }=\frac{1}{2}(1+1/\sqrt{3})$)

Encode signs $s_j=(-1{)}^{b_j}$ by the unit Bloch vector $(s_1,s_2,s_3)/\sqrt{3}$. Perform the requested Pauli measurement and interpret a positive outcome as zero. Every input and index has success $p_{\ast }$. To prove optimality, write each binary-measurement difference as $D_j=t_{j}I+v_j\cdot \sigma$, where positivity gives $|v_j|\le 1-|t_j|$. Averaging over all eight inputs cancels the $t_j$ terms. For any encoding vectors of norm at most one, the average success is at most $1/2+(1/48){\sum }_s|{\sum }_j{s}_j{v}_j|$. The Cauchy-Schwarz inequality bounds the sum by $8\sqrt{{\sum }_j|v_j{|}^2}\le 8\sqrt{3}$, proving the optimum even for mixed encodings and general binary measurements. The constant-success construction also maximizes worst-case success.