Finding the Probability of a Sequence of Numbers in Materials Testing There are $n$ numbers on a wheel. For this example, let's say $n = 20$. You are going to spin $x$ times. For this example, let's say $x = 7$. The chances of spinning the same number all 7 times is $\left(\frac{1}{n}\right)^x$ or 1 in 1,280,000,000 for this example. For materials testing, the smaller the probability, the more likely someone cheated. What formula can I use to find the probability of having the same number occur 4 times and a different number occur 3 times? Or the odds that 7 different numbers will occur? Or the odds that 3 numbers are the same, 2 numbers are the same, 2 more numbers are the same? etc. The numbers do not have to be in order or consecutive and all 20 numbers can occur each spin.
My thought is this is a complex equation that can simply be solved. 
Also, what are the odds that all 7 numbers are in the bottom half, bottom quarter, middle quarter, etc. of the numbers?
Please note: This sounds similar to roulette, but I promise this isn't roulette. I just know roulette terminology more than probability. Thanks for the help!
 A: To answer the first question, first consider the probability of this event happening:
$$1,1,1,1,1,2,2$$
Where $1$ and $2$ can be replaced with any integer $[0,n]$ so long as they are different. The first choice is free, however every $1$ after it has a $\frac{1}{n}$ chance of agreeing. The odds of the first $2$ being a number that is not $1$ is simply $\frac{n-1}{n}$, and the odds of every number after it also being $2$ is again $\frac{1}{n}$ for every number. However, it is important to note that permutations such as
$$2,1,2,1,1,1,1\ \ \text{and}\ \ 1,1,1,2,1,2,1$$
increase the chances of the desired event happening. In order to account for those we have to consider every permutation of $1,1,1,1,1,2,2$ without repeats, which would simply be $\frac{7!}{5!2!}$. Therefore, if we take our set of $n$ items, the odds of getting one element of the set $k_1$ times, another element of the set $k_2$ times, all the way to $k_j$ would be shown by:
(The sum of $k_1, k_2, \cdots, k_j$ = $x$)
$$ P = \Big[x!\prod_{i=1}^j \frac{1}{k_i!}\Big] \cdot \Big[\Big(\frac{1}{n}\Big)^{x-j}\Big] \cdot \Big[\prod_{i=0}^j \frac{n-i}{n}\Big]$$
The first product takes care of the fact that order does not matter, the second product checks the odds of the same number being hit again, and the third product takes care of the fact that $1,1,1,1,1$ does not count as getting 2 1's and 3 1's, but simply 5 1's.
As for the second question:
$$P_2 = P \cdot \Big( \frac{1}{a} \Big)^j$$
where $a$ is how you are partitioning the set. (One half $\Rightarrow a=2$, One third $\Rightarrow a=3$, etc)
Note: If you care for the derivation of $\frac{7!}{5!2!}$, it is simply a generalization of the familiar $C_n^k$ to multiple subsets instead of just 2 . The first product in my solution answers the question "How many ways can you partition a set into $j$ subsets of sizes $k_1,k_2,\cdots,k_j$". The proof if found in Theorem 1.4 of http://www.amazon.com/Probability-Theory-Concise-Course-Mathematics-ebook/dp/B00DP7ULZY/ref=tmm_kin_title_0?_encoding=UTF8&sr=&qid= (Click the picture to see the first chapter of the book).
