Expected number of rolls to get all colors on 6-sided die colored in with 3 colors 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.
 A: 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, $$E=E[0,0,0]=\frac 36\times (E[1,0,0]+1)+\frac 26\times (E[0,1,0]+1)+\frac 16\times (E[0,0,1]+1)=7.3$$
A: 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.
A: Okay, the idea is simple. 
Let the first color you hit be $x$ with probability $pr(x)$. On your next throw you will get the a different color with probability $(1-pr(x))$. The probability that you will hit the next color will be geometrically distributed, so that your mean number of throws will be $ \frac{1}{1-pr(x)} $. 
Then, let the second color be $y$ with probability $pr(y)$. The probability that will hit the third color will be $(1-pr(x) -pr(y))$. Again the number of throws you will need is geometrically distributed. Such that the mean number of times that you will have to throw will be $ \frac{1}{1-pr(x)-pr(y)} $.
Now you just need to use some combination theory to solve the problem, as the colors you hit the first, second, and third time can have a different order.
Br,
Carl
