Expected number of rolls to get all colors on 6-sided die colored in with 3 colors

Label the possible states of the game according to which colors have previously been seen. Thus we consider $8$ possible states $\{r,b,g\}$ where each symbol can be either $0$ or $1$ according to whether the associated color has been seen. Similarly, we denote by $E[r,b,g]$ the expected number of throws it will take to finish from the state $\{r,b,g\}$. The answer you want is $E=E[0,0,0]$. We'll proceed by backwards induction, starting with the observation that $E[1,1,1]=0$.

I. Missing one color. Say we are in state $\{1,1,0\}$, so we are only missing Green. We throw the die. We finish if we get a Green, probability $\frac 16$. With probability $\frac 56$ we stay in the state $\{1,1,0\}$. Thus $$E[1,1,0]=\frac 16\times 1+\frac 56\times (E[1,1,0]+1)\implies E[1,1,0]=6$$

Similarly,$$E[1,0,1]=\frac 26\times 1+\frac 46\times (E[1,0,1]+1)\implies E[1,0,1]=3$$ And $$E[0,1,1]=\frac 12\times 1+\frac 12\times (E[0,1,1]+1)\implies E[0,1,1]=2$$

II. Missing two colors. Say we are in state $\{1,0,0\}$. As before we roll the die and see that $$E[1,0,0]=\frac 26\times (E[1,1,0]+1)+\frac 16\times (E[1,0,1]+1)+\frac 36\times(E[1,0,0]+1)$$ $$=\frac {14}6+\frac {4}6+\frac 36\times(E[1,0,0]+1)\implies E[1,0,0]=7$$

Similarly, we get: $$E[0,1,0]=\frac {26}4\;\;\&\;\;E[0,0,1]=\frac {19}5$$

Finally, $$E=E[0,0,0]=\frac 36\times (E[1,0,0]+1)+\frac 26\times (E[0,1,0]+1)+\frac 16\times (E[0,0,1]+1)=7.3$$

Define $X_r$ to be the number of times to throw to see just a red side. Then $E(X_r) = 2$ (the reciprocal of the chance of a red in a throw) by a standard result.

Similarly we define $X_b$ for blue, with $E(X_b) = 3$ and $X_g$ with $E(X_g) = 6$.

Now define for any two colours $X_{i,j}$ the number of times to throw to see both colours $i$ and $j$ at least once.

Then by splitting on the options of the first throw we have

$$X_{r,b} = \frac{1}{2}(1+X_b) + \frac{1}{3}(1+X_r) + \frac{1}{6}(X_{r,b}+1)$$

We can take the expectation and substiture the known ones:

$$E(X_{r,b}) = \frac{1}{2}\cdot 4 + \frac{1}{3}\cdot 3 + \frac{1}{6}(E(X_{r,b}) + 1)$$

from which we can solve $E(X_{r,b})$. Do the same for the two other combinations of 2 colours.

Then finally $$E_{x,r,b} = \frac{1}{2}(1+E(X_{b,g})) + \frac{1}{3}(1+E(X_{r,g})) + \frac{1}{6}(1+E(X_{r,b}))$$

where we now know all the values on the right hand side.