Probability of two points being split into different partitions. Suppose $x < y \in [0, 1]$ and $U_1, U_2, ...$ are random variables distributed as uniform on $(0, w)$ with $w < 1$. Define $S_n = \sum_{i = 1} ^ n U_i$. We'll say that $(S_i)$ splits $x$ and $y$ if, for some $n$, $x < S_n < y$. I'm interested in what sorts of things could be said about the probability that $x$ and $y$ are split. If it helps, take $w$ to be small and $x, y$ close to $1$ and $|x - y| < w$.
I feel that this is quite related to whatever subject studies things like Poisson processes. Obviously, I know that if $|x - y| > w$ then $x$ and $y$ will be split, but beyond that I'm not sure what theory I would look to, and I don't know whether this sort of problem is tractable or not (it might be trivial for all I know). Only guess that I have is that near $|x - y| = w$ the probability should go to $0$ like $1 - \frac{|x - y|}{w}$, and this should be a lower bound for any $|x - y| < w$. I may also be interested in results or pointers for $U_i$ not uniform, but I need it to be the case that for points sufficiently far away the probability of being split is exactly $1$. 
EDIT: In light of the solution posted by PinkElephant below, I'll reword this as follows. Set $w = 1$ and remove the restriction on $x, y \in [0, 1]$ so that just $x < y \in \mathbb R$. What is the probability that $x$ and $y$ are split for large values of $x, y$ but for $|x - y|$ fixed, i.e. $x, y \to \infty$ but $y - x$ constant. It seems to me that asymptotically things should only depend on $y - x$; for $x, y$ near $0$ it seems as though there is dependence on the fact that we started the sum from $0$ that I think we escape from for large $x, y$. 
Update: I have a heuristic argument that gives the asymptotic probability of $x$ and $y$ being split as $1 - (1 - |x - y|)^2_+$ where $(a)_+$ denotes $\max\{0, a\}$. The argument is as follows: it should not matter asymptotically where $x$ and $y$ lie, so we can assume $x= k$ for some positive integer $k$. $x$ and $y$ will be split if, for the first value of $S_i$ in the interval $[x, x + 1]$, we have $x \le S_i \le y$. Consider the Markov chain $T_j$ consisting of the fractional parts of those $S_i$ such that $(S_{i - 1}) > (S_i)$, where $(a)$ denotes the fractional part of $a$. The probability that $x$ and $y$ are split should be $P(T \le (y - x))$ where $T$ is drawn from the stationary distribution of the $T_j$. I took a blind guess that the stationary distribution was given by the density $f(t) = 2(1 - t)$, drew a bunch of samples from the Markov chain I described, and verified empirically that $f(t) = 2 (1 - t)$ is the answer I'm supposed to get. 
If anyone wants to take a stab at verifying that $1 - (1 - |x - y|)_+^2$ is indeed the correct answer asymptotically, it would be much appreciated.
 A: If we scale $x$, $y$, and $w$ by the same positive constant, the answer is unchanged, so for simplicity I will take $w=1$ and remove the restriction that $x,y\in[0,1]$. Define $f(x,y)$ to be the probability that $x$ and $y$ are split. For arbitrary $w$, the probability that $x$ and $y$ are split is $f\left(\frac{x}{w},\frac{y}{w}\right)$.
For $0<x\leq y$, $f(x,y)$ satisfies the equation
$$
f(x,y)=\int_0^1f(x-t,y-t)\,dt
$$
(We extend $f(x,y)$ to the region $\{y\geq x\}\subset\mathbb{R}^2$ by setting $f(x,y)=1$ for $x<0$, $y\geq0$, and $f(x,y)=0$ for $x<y<0$)
This equation can be differentiated to give the differential-delay equation
$$
f_x(x,y)+f_y(x,y)=f(x,y)-f(x-1,y-1)
$$
The solution is piecewise real-analytic, with boundaries between the pieces lying on the lines $y=x$, $y=x+1$, the line segments $y\in [x,x+1],x\in\mathbb{Z}_{\geq0}$, and the line segments $y\in[x,x+1],y\in\mathbb{Z}_{\geq0}$.
It's not too hard to recursively compute values of $f(x,y)$ on the various pieces of the domain. I computed some values with Mathematica:
For $0<x\leq y<1$, $f(x,y)=(y-x)e^x$.
For $0<x\leq 1\leq y\leq x+1$, $f(x,y)=(y-x)e^x+1-e^{y-1}$.
For $1\leq x\leq y\leq 2$, $f(x,y)=e^{x-1}-e^{y-1}+e^x\left(x^2-2x+2y-xy\right)$.
If you want to compute a specific value of $f(x,y)$ exactly, I don't know if there's a faster way than to continue this recursive computation. If you're only interested in the limiting value as $x,y\to\infty$ with $y-x$ fixed, this might be possible to compute.
A: Given $x$, let $S_x$ be the random variable $S_{\min \{n, S_n \ge x\}}$. It has values in $[x;x+1]$, and let $g_x(y)$ be its density :
$\forall x \le y_1 \le y_2 \le x+1, \int_{y_1}^{y_2} g_x(y)dy = P(S_x \in [y_0;y_1])$.
As $x$ grows, the weight put on $g_x(y)$ for $y$ near $x$ is displaced and redistributed to all the other $y \in [x;x+1]$. In particular, if $x_1 \le x_2 \le y_1 \le y_2 \le x_1+1$, then $y_1$ and $y_2$ receive the same amount of weight when going from $x_1$ to $x_2$, so : $g_{x_2}(y_1) - g_{x_1}(y_1) = g_{x_2}(y_2) - g_{x_1}(y_2)$, and this implies that the function $g_x(y_2)-g_x(y_1)$ is constant for the range of $x$ where it makes sense.
From this we can deduce that there are two functions $F(x),G(y)$ such that $g_x(y) = G(y)+F(x)$ whenever $g_x(y)$ makes sense. 
Now, we must have that forall $x>1$, $g_x(x+1)=0$, thus $F(x) = -G(x+1)$, and so $g_x(y) = G(y)-G(x+1)$.
Moreover, forall $x$, $1 = \int_x^{x+1} g_x(y)dy = -G(x+1)+ \int_x^{x+1} G(y)dy$.
Differentiating this you get $0 = -G'(x+1)+G(x+1)-G(x)$, so the problem reduces to studying the delay differential equation 
$$f'(x) = f(x) - f(x-1)$$ 
Our initial condition is $G(x) = 1$ for $0 \le x < 1$, and $G(1)=0$. The goal is to show that for $a \in [0;1], G(x) - G(x+a) = 2a + o(1)$.
The solution we are interested in can be defined piece by piece, it turns out that
$G(x+1) = 1 - e^x\sum_{0 \le k \le x} (\frac{k-x}e)^k/k!$

Suppose $f$ is a solution to the differential equation.
Put $g(x) = \int_{x-1}^x f(t) dt - f(x)$. Then $g'(x) = f(x) - f(x-1) -f'(x) = 0$.
So $g(x) = C$ and by substracting $2Cx$ from $f(x)$, we can assume without loss of generality that $g(x) = 0$.
Define $h(x) = \int_{x-1}^x (f(t)-f(x))^2 dt$. Then 
$$h'(x) = (f(x)^2 - f(x-1)^2) -2(f(x)^2-f(x-1)f(x)+f'(x)\int_{x-1}^x f(t)dt) + 2f(x)f'(x) \\ = (f(x)^2 - f(x-1)^2) -2(f(x)^2-f(x-1)f(x)) = -(f(x-1)-f(x))^2 = -f'(x)^2$$
$h$ is obviously positive, and decreasing, so it has a limit, and in particular, for any $a\in [0;1], h(x)-h(x+a) \to 0$, uniformly in $a$.
Finally, $(f(x+a)-f(x))^2 = \left(\int_x^{x+a} f'(t)dt \right)^2 \le a \int_x^{x+a} f'(t)^2 dt = a(h(x+a) - h(x)) \to 0$. Hence in the original context, $g_x(y) \to 0$ (uniformly) as $x \to \infty$
