What are the probabilities of $x = 0, x=1, \dots, x=n$ for a given individual, where $n$ is the number of different objects distributed among $k < n$ different individuals?
My understanding is that the total number of possibilities (length of the "sample space") is given by $k^n$. If so, the remaining (harder) question is the number of cases in which $x=0, x = 1, \dots, x=n$ out of the $k^n$ total of possibilities. I browsed through the archives but couldn't find much help. I found a formula here (p. 20):
http://www.pitt.edu/~bonidie/cs441/lecture21.pdf
that, adapted to my notation and needs, became:
$$\Pr(x=n_1) = \frac{n!/(n_1! n_2! \dots n_k!)}{k^n},$$
subject to $\sum_{i=1}^k n_i = n$, where $n_i = 0, 1, \dots, n$ is the number of different objects that may be assigned to individual $i$ -- and I'm using $i=1$ for my given individual. Does this formula seem right to you?
And, what if $n < k$?
I'll appreciate all your replies. Thanks in advance!
We could make use of that formula (we would want to sum over valid $n_2, \dots, n_k$, which you don't seem to be doing) but there are easier ways to solve the problem.
If we assume that each object is independently allocated to an individual, and is equally likely to go to anyone, there are $k^n$ equally likely outcomes. The probability that a specific individual receives $x$ objects is $\binom nx (\frac1k)^x (\frac{k-1}{k})^{n-x}$, or $\binom nx \frac{(k-1)^{n-x}}{k^n}$. This is just a binomial distribution: each object has a $\frac1k$ chance of going to this individual, and we want to know the probability that some $x$ objects do this and the rest do not.
The formula $\frac{n!}{n_1!\,n_2!\dotsb n_k!}$ is the number of ways we can split up $n$ objects into $n_i$ objects given to the $i^{\text{th}}$ individual. So if $X_i$ is the number of objects the $i^{\text{th}}$ individual gets, then $$ \Pr[X_1 = n_1,\,X_2 = n_2,\dots, X_k = n_k] = \frac{\frac{n!}{n_1!\,n_2!\dotsb n_k!}}{k^n}. $$ If we just care about $X_1$ and not any of the others, we would want to take a sum $$ \Pr[X_1 = n_1] = \sum_{n_2 + \dots + n_k = n - n_1} \frac{\frac{n!}{n_1!\,n_2!\dotsb n_k!}}{k^n} $$ where the sum ranges over all $(n_2, n_3, \dots, n_k)$ that add up to $n-n_1$. This can be simplified: if we factor out a $\binom n{n_1} \frac1{k^n} = \frac{n!}{n_1! (n-n_1)!} \frac1{k^n}$, we get $$ \Pr[X_1 = n_1] = \binom n{n_1} \frac1{k^n} \left(\sum_{n_2 + \dots + n_k = n - n_1} \frac{(n-n_1)!}{n_2!\,n_3!\dotsb n_k!}\right) $$ and this simplifies to $\binom n{n_1} \frac1{k^n} (k-1)^{n-n_1}$ because the sum remaining in the parentheses is what we get if we apply the multinomial theorem to the expansion of $(\underbrace{1 + 1 + \dots + 1}_{k-1})^{n-n_1}$. So we get the same formula we got earlier, after a lot more effort.
Last thing: all of this is valid no matter which of $k$ or $n$ is larger, so there are not two distinct cases here.