Rolling a k-sided fair die n times and not see all k numbers fall If we throw a k-sided fair die n times, what is the probability that we will never get one of its k numbers? Further, what is the probability that two, three, or more of its numbers will never occur in the results? (presume the faces are unique - numbered 1 to k)
More formally, if the occurrence count of the number $i$ after $n$ throws with a fair, uniquely numbered die is $X_i$ and $\sum_{i}{X_i}=n$, let's define a "zero occurrence" set $S = \{ x_i \mid x_i = 0 \}$. S contains all the numbers that didn't occur during the $n$ throws. If we consider $Y$ to be the size of the set S, $Y = \left\vert S\right\vert$, the question is asking for a closed form expression of the pmf of $Y$ that would allow computing the probabilities.
I have very little so far:
\begin{equation}P(Y=1)= P(X_1=0, X_2 > 0, \ldots, X_k>0) + P(X_1>0, X_2 = 0, \ldots, X_k>0) + \ldots + P(X_1>0, X_2 > 0, \ldots, X_k=0) = k P(X_1=0, X_2 > 0, \ldots, X_k>0)=\sum\limits_{\{x_1,x_2,\ldots, x_k\}\in P} \frac{n!}{x_1!\ldots x_k!}{p_1^{x_1}\ldots p_k^{x_k}} \end{equation}
I struggle to get anything useful from the multinomial pmf, since to calculate its cdf, you have to sum over different partitions of n elements into k-1 bins. Finding the set of possible partitions $P$, where $P = \{\{x_1,x_2,\ldots,x_k\}\mid x_1 = 0 \land \sum_i{x_i} = n \}\}$ is np hard and there are ${n-1}\choose{k-2}$ probabilities we would have to sum up (we have k-1 bins to redistribute n throws). I wonder whether there is a way to derive a result which doesn't involve partitioning and summing up the probabilities.
 A: 
If we throw a k-sided fair die n times, what is the probability that
  we will never get one of its k numbers? Further, what is the
  probability that two, three, or more of its numbers will never occur
  in the results? (presume the faces are unique - numbered 1 to k)

Maybe I have misunderstood the problem but it looks much easier to me.
The probability we will never get one of its k numbers in n rolls is just
$\left(\frac{k-1}{k}\right)^n$
For instance, the probability that number 3 does not occur if a 6 sided (fair) dice is rolled 7 times is
$\left(\frac{6-1}{6}\right)^7\approx 27.91\%$
And the probabilty that two, three, or more of its numbers will never occur in the results can be expressed in a  similar way.
A: There are $\binom ky$ ways to choose $y$ sides that occur, $\def\stir#1#2{\left\{{#1\atop#2}\right\}}\stir ny$ ways to partition the $n$ throws into $y$ non-empty subsets and $y!$ ways to assign the $y$ subsets to the $y$ sides, so the probability that exactly $y$ sides occur in $n$ throws is
\begin{align}
\frac{y!\binom ky\stir ny}{k^n}=k^{-n}\binom ky\sum_{j=0}^y(-1)^{y-j}\binom yjj^n\;,
\end{align}
where $\stir ny$ is a Stirling number of the second kind.
