TITLE: Expected number of rolls to get all colors on 6-sided die colored in with 3 colors
QUESTION [6 upvotes]: If I have a die that has 3 red sides, 2 blue sides, and 1 green side, how many rolls do I expect until every color has appeared at least once?
I have run some tests and I’m getting numbers around 7.31, but clearly I’m looking for a mathematical solution.
Thanks in advance.

REPLY [6 votes]: 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, $$E=E[0,0,0]=\frac 36\times (E[1,0,0]+1)+\frac 26\times (E[0,1,0]+1)+\frac 16\times (E[0,0,1]+1)=7.3$$