On the probability and expectation of rolling 3 dice and get 4, 5, 6 I am looking at the following questions on rolling 3 dice at the same time:

*

*Probability of getting 4, 5, 6 without particular order;

*If get a result different from 4, 5, 6, then roll again. What is the expectation of number of rolls;

*If get a result different from 4, 5, 6, then roll the dice which is not in the set {4, 5, 6} again, e.g. first roll gives 4, 5, 1, then roll the third dice only until get a 6. What is the expectation of number of rolls.

I am stuck with the third part and do not know how to approach to such question. For the first one, I get the probability $= \left( \frac{1}{6} \right)^3 \times 3\,! = \frac{1}{36}$. And the second question, I use the mean of a geometric distribution and get the expectation $=36$.
 A: For the 3rd question:
(This solution assumes each dice rolled will count as one roll, e.g. throwing 3 dice at the same time will count as 3 rolls)
The expected number of rolls before getting 4, 5 and 6 on three separate dice with re-rolls for the dice not having 4, 5 or 6 can be viewed as throwing one dice several times until each number has been accounted ones (you just take away the dice that gets a suitable number then move to the next dice). A solution is:
\begin{equation}
E_3 = 1 +\frac{3}{6}*E_2+\frac{3}{6}*E_3\\
E_2 = 1+\frac{2}{6}*E_1+\frac{4}{6}*E_2\\
E_1 = 1+\frac{1}{6}*E_0+\frac{5}{6}*E_1\\
E_0 = 0
\end{equation}
where $E_3$ is the expected throws to get all three numbers, $E_2$ is the expected throws to get two of the numbers, $E_1$ is the expected throws to get the last missing number and $E_0$ is the expected number if you already got them all ($=0$, if you already got  4, 5 and 6 then you don't need to throw more). The fraction before each expected value indicates the probability of moving forward to that step. The 1 in the beginning is the throw that you just did. This equation can now be solved from the bottom up
\begin{equation}
E_1 = 6\\
E_2 = 9\\
E_3 = 11
\end{equation}
So the expected number of rolls if you're missing all {4, 5 and 6} is 11 throws, the expected number if you're missing just 2 values is 9 throws, and the expected number if you're missing just one value is 6 throws.
