2023 AMC 10B Problem 21

All $3^{2023}$ assignments are equally likely. A sign filter counts the ones with every bin odd: $\frac{1}{8}\sum_{s \in \{\pm 1\}^3}(s_1 s_2 s_3)$ $\cdot (s_1 + s_2 + s_3)^{2023}.$ For $s = (1,1,1)$ and $s = (-1,-1,-1),$ both terms equal $3^{2023}.$ For each other sign choice, the sum in parentheses is $1$ or $-1,$ and the full term equals $-1,$ so these six terms total $-6.$ Thus the count is $\frac{2 \cdot 3^{2023} - 6}{8} = \frac{3^{2023} - 3}{4}.$ Dividing, the probability is $\frac{3^{2023} - 3}{4 \cdot 3^{2023}} = \frac{1}{4} - \frac{1}{4 \cdot 3^{2022}},$ a hair under $\frac{1}{4}.$ Thus, E is the correct answer.