For the same simple median-cut algorithm mentioned, suppose we decide to find medians in the order $R, G, B, R$ to make a 4-bit version of color. Thus for R, if we decide that a pixel's red value is less than the median, we proceed down the left branch of a binary tree and look at the green pixel values just for those left pixels, etc. So far, our best value for $\mathrm{R}$ is the median for all red values.
So in the end, for our 4-bit color we have 16 bottom $\mathrm{R}$ medians that are used, $8 \mathrm{~B}$ medians, and $4 \mathrm{G}$ medians. Hence a question: do we not actually have 4 bits for R, 3 bits for $\mathrm{B}$, and 2 bits for $\mathrm{G}$, making 9-bit accuracy?

Question

For the same simple median-cut algorithm mentioned, suppose we decide to find medians in the order $R, G, B, R$ to make a 4-bit version of color. Thus for R, if we decide that a pixel's red value is less than the median, we proceed down the left branch of a binary tree and look at the green pixel values just for those left pixels, etc. So far, our best value for $\mathrm{R}$ is the median for all red values. 
So in the end, for our 4-bit color we have 16 bottom $\mathrm{R}$ medians that are used, $8 \mathrm{~B}$ medians, and $4 \mathrm{G}$ medians. Hence a question: do we not actually have 4 bits for R, 3 bits for $\mathrm{B}$, and 2 bits for $\mathrm{G}$, making 9-bit accuracy?

Quizplus · Accepted Answer

The Answer of For the same simple median-cut algorithm mentioned, suppose we decide to find medians in the order $R, G, B, R$ to make a 4-bit version of color. Thus for R, if we decide that a pixel's red value is less than the median, we proceed down the left branch of a binary tree and look at the green pixel values just for those left pixels, etc. So far, our best value for $\mathrm{R}$ is the median for all red values. 
So in the end, for our 4-bit color we have 16 bottom $\mathrm{R}$ medians that are used, $8 \mathrm{~B}$ medians, and $4 \mathrm{G}$ medians. Hence a question: do we not actually have 4 bits for R, 3 bits for $\mathrm{B}$, and 2 bits for $\mathrm{G}$, making 9-bit accuracy?

For the Same Simple Median-Cut Algorithm Mentioned, Suppose We Decide R,G,B,RR, G, B, RR,G,B,R

For the Same Simple Median-Cut Algorithm Mentioned, Suppose We Decide $R, G, B, R$