Average size of iterated sumset modulo $p-1$, Given a prime $p$, what is the average size of the iterated sumset, $|kA|$, modulo $p-1$, with $p$ a prime, and $k$ given, with $A$ chosen at random?
You can pick any type of prime you like for $p$, but we must be able to get arbitrarily large $p$'s of this type.  Note that the elements are chosen independently from the uniform distribution over $\mathbb{Z}_p$, and we assume that the size of $A$ is fixed.
 A: There is some ambiguity about how $A$ is chosen at random so I will define $\Omega$ to be the set of 
$$A\subseteq \mathbb{Z}/(p-1)\mathbb{Z} \ : \ |A|=K$$
where $K$ is some fixed integer and assume $A$ is chosen uniformly from $\Omega$. In this case the expected size of $kA$ is 
$$|kA|\sim \frac{|A|^k}{k!}.$$
We may be able to say a little more about the distribution.  I'll sketch the ideas. For each $1\le z \le p-1$ let $X_z$ denote the random variable 
$$X_z= 1 \quad \text{if} \quad  z\in kA,  \ \quad  
 0 \quad \text{otherwise}$$
so we want to calculate the expected value of 
$$X=X_1+\dots+X_{p-1}.$$
Using linearlity
$$\mathbb{E}(X)=\mathbb{E}(X_1)+\dots+\mathbb{E}(X_{p-1}).$$
  Fix some $1\le z \le p-1$ and consider $X_z$. We have 
$$X_z=1$$
if and only if there exists $a_1,\dots,a_k\in \mathbb{Z}/(p-1)\mathbb{Z}$ such that 
$$a_1+\dots+a_{k}=z,$$
Since the number of solutions to the above with $a_i=a_j$ for some $i\neq j$ is $O(p^{k-2})$, we may approximate 
$$\mathbb{E}(X_z)\sim \frac{1}{k!}\frac{1}{|\Omega|}\sum_{a_1,\dots,a_{k-1}\in \mathbb{Z}/(p-1)\mathbb{Z}}\sum_{\substack{A\in \Omega \newline a_1,\dots,a_{k-1},z-(a_1+\dots+a_{k-1})\in A}}1$$
where the $k!$ accounts for all possible rearrangements of $a_1,\dots,a_{k-1},z-(a_1+\dots+a_{k-})$. Simplifying 
\begin{align*}
\mathbb{E}(X_z)&\sim \frac{1}{k!}\frac{1}{|\Omega|}\sum_{a_1,\dots,a_{k-1}\in \mathbb{Z}/(p-1)\mathbb{Z}}\binom{p-1-k}{K-k} \newline
&\sim \frac{p^{k-1}}{k!}\frac{\binom{p-1-k}{K-k}}{\binom{p-1}{K}},
\end{align*}
then simplifying some more 
\begin{align*}
\mathbb{E}(X_z)\sim \frac{K^k}{k!(p-1)}.
\end{align*}
Summing over expectations 
$$\mathbb{E}(X)\sim  \frac{K^k}{k!}.$$
Combined with the central limit theorem, the above argument suggests the size of $kA$ may have normal distribution.
