Three red balls and three green balls on a circle

Suppose all three red balls are randomly placed on the circle. They divide the circle into three disjoint arcs.The probability of having 0 red balls such that both neighbors are green is

\[{3 \choose 1}(\frac{1}{3})^3=\frac{1}{9}.\]

The probability of having 1 red balls such that both neighbors are green is
\[{3 \choose 1} {3 \choose 2} (\frac{1}{3})^2 \frac{2}{3}=\frac{6}{9}.\]

Indeed we choose one of the three arcs to have 2 green balls, and 2 out of the three red balls to land in that arc. The last green ball should land on the remaining 2 arcs which has the probability $\frac{2}{3}.$

The probability of having 2 red balls such that both neighbors are green is 0 (impossible). 

The probability of having 3 red balls such that both neighbors are green is

\[\frac{2}{3} \times \frac{1}{3}=\frac{2}{9}.\]

Indeed, the first green ball can land on any arc, the second ball should land on a different arc than the first one which has the probability $\frac{2}{3}$, the thirs ball should land on the only remaining arc which has the probability $\frac{1}{3}$.

So expectation of the number of red balls such that both their neighbours are green is
\[0\times \frac{1}{9}+1 \times \frac{6}{9}+3 \times \frac{2}{9}=\frac{12}{9}=\frac{4}{3}.\]