Theorem: Noiseless blend Bayes accuracy

Under exact observation,

In particular the reference accuracies at $\alpha =0.25,0.50,0.75,1$ are $1,9/16,1,1/4$. More distractor weight does not imply monotone information loss.

Proof. For one coordinate and $0<\alpha <1$, the four sign pairs give

For $\alpha \ne 1/2$, these values are distinct, so both signs can be recovered.

At $\alpha =1/2$, a coordinate is zero with probability $1/2$. A nonzero value identifies the visual sign; a zero leaves its two signs equiprobable. Optimal coordinate success is $1/2+(1/2)(1/2)=3/4$. The two coordinate experiments are independent, giving $9/16$.

At $\alpha =1$, the observation is independent of the four-class visual label. At $\alpha =0$, it is the visual center itself.