Odds of Choosing Items out of a bag Let's say there is a bag filled with a myriad of balls, labeled 1 through 7. There are an infinite number of each ball (i.e. odds don't diminish when one is removed). Edit A better phrasing than infinite-balls would be balls being replaced. /Edit
I get to draw 12 balls out of the bag.
What are the chances I'd end up with at least 1 of each number (and obviously some repeat pulls)? What are the chances I'd end up with all-but-1 of the 7 numbers (i.e. $\frac{6}{7}$)? Or $\frac{5}{7}$, or any lower combination?
(I'm not particularly gifted with Combinatorics. I was playing with smaller values, like 3-draws of 2-different numbers having $2^3$ possible options and $\frac{6}{2^3}$ of them fulfilling the condition. I tried 3-draws of 3-different numbers having $\frac{6}{3^3}$ fulfilling the condition. I tried 4 draws of 4-draws of 4-different numbers and got $\frac{24}{4^4}$ fulfilling the condition. Needless to say, I couldn't see a readily repeatable pattern to use on higher values :-( )
 A: You do need Stirling numbers (second kind) for this. The Stirling number $S(n,k)$ denotes the number of ways to partition an $n$ element set into $k$ non-empty subsets. To make discussion easier, let me replace the bad by a $7$-sided dice, and drawing from the bag by throwing the dice. These are equivalent for our purposes.
Let us first consider the chances of getting each number at least once. The total space will consist of all possible $12$-element sequences of $1,2,\dots,7$ - there are $7^{12}$ of these. How many of these correspond to the event we are interested in? Well, each such event gives a partition of $1,2,\dots,12$ into $7$ non-empty sets (namely, the set of throws when different numbers came out). Each such partition corresponds to precisely $7!$ events (given a partition, there are that many ways to assign numbers $1,2,\dots 7$ to different subsets in this partition). Hence, there are the total of $7! \cdot S(12,7)$ sequences of throws with each result occurring at least once. The probability you are looking for is:
$$ \frac{7! \cdot S(12,7)}{7^{12}}$$
Mathematica tells me this is: $\frac{9218880}{40353607} \simeq 0.22... $
For the second question of getting $6$ out of $7$ possible results, note that any such combination of throws corresponds to a partition of $1,2,\dots,12$ into $6$ subsets, and for  each partition there are $7!$ combinations of throws. Hence the probability is
$$ \frac{7! \cdot S(12,6)}{7^{12}}$$
Note that for $5$ out of $7$, the same goes through, except now each partition corresponds to ${7 \choose 2} \cdot 5! = \frac{7!}{2!}$ combinations of throws. I hope you should be able to generalise it without much problem.
P.S. The reason why you couldn't see a pattern is probably that there are Stirling numbers involved. The reason why they deserve a separate name is that they aren't easily expressible in terms of simpler expressions, so a pattern would not be obvious. 
A: The disadvantage with Stirling #s is that you need recourse to a table.
However, the question can easily be solved using inclusion-exclusion:
$Pr =\Large{{7^{12} - 7\cdot6^{12} + 21\cdot5^{12}-35\cdot4^{12} +35\cdot3^{12} - 21\cdot2^{12} + 7\cdot 1^{12}}\over 7^{12}}$
