Given $x$ and $y$ in $\mathbb{R}$, and let $\mathcal{H} = \{ h \mid \mathbb{R} \to \mathbb{N} \}$ be a family of hash functions where $ h(x) = \left\lfloor x + \sum^C\_{i=1} U\_i \right\rfloor$ for some constant $C$. Morever, $U\_i$ is independently and uniformly choosen at random from [0,1]. 


What is the probability that $h\_1(x) = h\_2(y)$ occurs, where $h\_1$ and $h\_2$ are choosen independently at random from $\mathcal{H}$? I know that it looks similar to a *hat kernel* for $C=5$, see [Plot](https://i.stack.imgur.com/DYfpq.png) (scaled to $[0,1]$). Moreover the distribution of the sum is an Irwin-Hall distribution. Any suggestions how to proceed?


I posted the question on math.stackexchange a week ago, but did not get an answer.



Probability of collision of some family of hash functions
Let $n:=C$. 
For any integer $i$ and real $x$, let 
\begin{equation}
 p\_{i,x}:=P(h(x)=i)=P(i-x\le S<i+1-x)=F(i+1-x)-F(i-x), 
\end{equation}
where $S:=\sum^n\_{i=1} U\_i$, $F(s):=\frac1{n!}\,\sum\_{j=0}^n(-1)^j\binom nj (s-j)\_+^n$, $u\_+:=0\vee u$. Of course, $F$ is the cdf of the Irwin--Hall distribution. 
Then the probability in question is 
\begin{equation}
 \sum\_i p\_{i,x}p\_{i,y}, 
\end{equation}
where $\sum\_i$ denotes the summation over all integers $i$ or, equivalently in this case, the summation over all integers $i$ such that $x\vee y-1\le i\le x\wedge y+n$. 
In particular, it follows (as pointed out by Max Alekseyev) that this probability will be $0$ in the case when $|x-y|>n+1$. 


