Painted cube piece showing five blank faces

The outside of a $3\times3\times3$ cube is painted blue, and the cube is then sliced into $27$ unit cubes. One unit cube is chosen at random and placed so that you can see $5$ of its faces at random (one face hidden), and all five visible faces are unpainted. Given this, what is the probability that the chosen cube has exactly one painted face?

$\tfrac12$. Only the centre cube (no paint) or one of the 6 face-centre cubes (one painted face, hidden with chance $\tfrac16$) can show five blank faces. Bayes gives $\dfrac{6\cdot\tfrac16}{1\cdot1+6\cdot\tfrac16}=\tfrac12$.