Probability distribution of number of times required to roll all of a certain set of k numbers? Consider a 12-sided fair die.  What is the distribution of the number T of rolls required to roll a 1, a 2, a 3, and a 4?
Taking inspiration from the Coupon Collector's Problem, I believe that the expected number of rolls to achieve the goal would be
$$E[T] = \sum\limits_{i=0}^3 \frac{12}{4-i} = 25$$
Similarly, the variance would be
$$Var[T] = \sum\limits_{i=0}^3 \frac{1-\frac{4-i}{12}}{(\frac{4-i}{12})^2} = 180$$
But applying Chebyshev here does not yield very useful bounds.  My question is therefore, how would you compute, for example, $P(T=16)$ or $P(T<30)$?
Ideally this could be generalized to a set of k required numbers, not just 4 as in the example.
 A: Personally I would call $P(n,b,k,j)$ the probability that after rolling $n$ dice with $b$ sides, $j$ of the target $k$ sides had been found, with the formula
$$P(n+1,b,k,j)= \frac{b-k+j}{b}  P(n,b,k,j) +  \frac{k+1-j}{b}  P(n,b,k,j-1)$$
starting from $P(0,b,k,j)=0$ for $j \not = 0$ and  $P(0,b,k,0)=1$.
Then $$\Pr(T \le t) = P(t,b,k,k)$$ and $$\Pr(T = t) = P(t,b,k,k)-P(t-1,b,k,k).$$
So in your example it is not difficult to calculate $\Pr(T = 16) \approx 0.0380722$ and $\Pr(T \le 30) \approx 0.7305294$.  Your expected value of $T$ and variance appear to be correct.
A: I think it's the sum of four different geometric distributions.
For example, you start off and you haven't rolled any of the numbers. The probability of rolling one of them is 4/12 so you start a geometric distribution with that parameter. Then you suddenly roll one of them. Now you need to roll one of the other three, each roll has a probability of 3/12 of getting one of them, so it's another geometric distribution.
By this logic I think T ~ Geo(4/12) + Geo(3/12) + Geo(2/12) + Geo(1/12) which of course is probably some hideous ditribution define in terms of convolutions (or you know, could miraculously be nice and of closed-form, I don't know what.)
