A fair die is rolled n times. What is the probability that at least 1 of the 6 values never appears? A fair die is rolled $n$ times. What is the probability that at least $1$ of the $6$ values never appears? I went about calculating the complement of this, because it seemed to be easier. However, I am having trouble with it.
I was able to calculate the complement for $n=6$ and $n=7$ using a formula for putting $n$ items into $6$ boxes and requiring that each box had one item. For $n=7$ one box had to have two items and there are six ways to do that, so I accounted for this in the formula: $$\frac{6\times(7!/2!)}{6^7}$$ It seems that it will be quite complicated to apply this method for $n>7$, but I can't seem to figure out any other way. I thought to use a combination to choose the 6 from n which have to be the values $1$ through $6$: $${n\choose 6}6!/6^n$$ but this seems to undercount by quite a bit. Would the answer be something like this? This is not homework, just self-study. Thanks!
 A: This is a practical way to allow calculating the probability with an Excel or similar. For $n=20$ I have the probability at $0.14665$. For $n=50$ I have the probability at $0.00066$. 
I am assuming $n\ge6$, otherwise the probability is $1$.
The probability of not getting a $1$ is $({5\over 6})^n$. The probability of getting a $1$ and not getting a $2$ is $(1-({5\over 6})^n)({5\over 6})^n)$.
Let's define $$p = ({5\over 6})^n$$
The probability of not getting a $1$ is $p$. The probability of getting a $1$ and not getting a $2$ is $(1-p)p = p-p^2$. The probability of not getting a $1$ and/or not getting a $2$ is $p+p-p^2 =2p - p^2$. The probability of getting $1$ and $2$ and not getting $3$ is $(1-(2p-p^2))p = p-2p^2+p^3$. The probability of not getting at least one of $1$ or $2$ or $3$ is therefore $2p-p^2+p-2p^2+p^3=3p-3p^2 +p^3$.
Let's define $$q = 3p-3p^2 +p^3$$
 $q$ works for $4$, $5$ and $6$ just as it had for $1$, $2$ and $3$. So your probability for $n \ge 6$ should be $$P=2q - q^2$$
