Expected number of people having X tickets, given Y tickets distributed among Z people I have had this problem when trying to describe this probability, and I do not know what it's called or how to solve it.
Example: 100 (Y) tickets are randomly distributed to 100 (Z) people, each is independent, people can get multiple tickets. The expected number of tickets for any one person is 1, of course, but it will not be distributed that way. It is more likely that some people will get 0 tickets, some will get 2, 3, 4, and so on. So what is the expected number of people who have exactly a particular number (X) of tickets? I know that number will likely be a decimal.
On the same subject, it would be nice to know if the same technique can be used to calculate a range of tickets i.e. "expected number of people who have 3 or more tickets".
 A: We can use an indicator variable trick for this. Fix $x \geq 0$ and let $A_{k, x} = \{\text{person $k$ has exactly $x$ tickets}\}$. The expected number of people with exactly $x$ tickets is 
$$\mathbb E \left[\sum_{k=1}^{100} 1_{A_{k, x}} \right] = \sum_{k=1}^{100} \mathbb E[1_{A_{k, x}}] = \sum_{k=1}^{100} \mathbb P(A_{k, x}) = 100 \cdot \mathbb P (A_{k, x})$$
by linearity of expectation, so we really only need to look at the probability of a single person doing this. (Note: this feels like cheating, because the events $A_{k, x}$ are not independent of one another! But fortunately, it doesn't matter when using linearity of expectation.)
That probability can be computed with binomial considerations, i.e.
$$\mathbb P(A_{k, x}) = \binom{100}{x} \left( \frac{1}{100} \right)^{x} \left( \frac{99}{100} \right)^{100 - x}$$
since that person either succeeds (with probability $1/100$) or fails (with probability $99/100$) independently on each trial, and there are $\binom{100}{x}$ ways to select which  $x$ of the $100$ trials are a success. Hence, the answer you want is $\fbox{$100 \binom{100}{x} \left( \frac 1 {100} \right)^x \left( \frac{99}{100} \right)^{100-x}$}$.
For the "at least" version, the idea is the same, but we would need a new expression to replace $\mathbb P(A_{k, x})$. The "easy" way to do this is to replace $\mathbb P(A_{k, x})$ with $\sum_{y = x}^{100} \mathbb P(A_{k, y})$, which is conceptually straightforward but somewhat computationally obnoxious. You could replace the binomial sum with some approximation (such as a normal approximation to the binomial) to lower the burden a bit, if you are OK with a slight error.
