We have two red, two white and two green marbles in an urn

You can also just count the following way: $$ \frac{1}{5} + \frac{4}{5} \left[ \frac{1}{4} + \frac{2}{4}\frac{1}{3} + \frac{1}{4} \left(\frac{2}{3}\frac{1}{2} + \frac{1}{3}\right)\right] = \frac{2}{3}$$ In words: The first marble can be an arbitrary one (say red). Then the second one is either red (with $1/5$ probability) or it is not (with $4/5$ probability, say white). In this case either the third marble is white ($1/4$), or green ($2/4$ for the remaining not yet drawn one) in which case we must draw it again ($1/3$ probability), or we draw the first already drawn (red) one with $1/4$ probability. In the last case there remains the possibility to either now draw the green - not yet drawn - one with $2/3$ probability and we must draw it again to succeed - by then only 1 green and 1 white are left in the urn - which gives a factor $1/2$, or we draw the overall second (white) one with $1/3$ probability which just leaves 2 green marbles in the urn and we are done.

You may use the inclusion-exclusion principle. The number of ways when at least two consecutive balls have the same color is $$|R\cup W\cup G|=\underbrace{3\cdot 5 \cdot \frac{4!}{2!2!}}_{|R|+|W|+|G|}-\underbrace{3\cdot 2 \cdot \binom{4}{2}}_{|R\cap W|+|W\cap G|+|G\cap R|}+\underbrace{6}_{|R\cap W\cap G|}=60$$ where $R$ is the set of extractions where the two red balls are picked up back to back ($W$ and $G$ have similar definitions). Therefore the probability is $$p:=\frac{60}{\frac{6!}{2!2!2!}}=\frac{60}{90}=\frac{2}{3}.$$

Let $R$ denote the event that the red balls are picked out back to back.

Let $W$ denote the event that the white balls are picked out back to back.

Let $G$ denote the event that the green balls are picked out back to back.

Then applying inclusion/exclusion and symmmetry we find: $$P(R\cup W\cup G)=3P(R)-3P(R\cap W)+P(R\cap W\cap G)=$$$$3\frac5{\frac{6!}{2!4!}}-3\frac{12}{\frac{6!}{2!2!2!}}+\frac{6}{\frac{6!}{2!2!2!}}=\frac{15}{15}-\frac{36}{90}+\frac6{90}=\frac23$$