What's the probability that a particle jumps out of an interval? Suppose I have a small particle and put it on the center of an interval on a 1-D axis. If the particle undergoes a motion that satisfies: 

  
*
  
*It chooses its direction freely and randomly
  
*It chooses the span of each step of equal probability in range (0,r]
  
*The total number of the steps is unlimited
  

So I'd like to ask that what's the probability that the particle finally jump out of the interval specified as a function of r.
 A: The position at time $t$ is $x_t = \sum_{i=1}^t e_i$ where $e_i$ are iid, uniform in $[-r,r]$, $E(e_i)=0$, $Var(e_i)=r^2/3$. Hence $x_t$ tends to a gaussian with zero mean and variance $ t \, r^2/3$. Hence, when $t\to \infty$ the probability that it's inside some finite interval goes to zero.
A: This is a continuous-valued symmetrical random walk on the real line. At each step it moves (is 'displaced') by a number $D \sim Unif(-r, r).$ It is a Markov Chain because the movement of the particle
from step $i-1$ (now) to step $i$ (next) depends only on the distribution of $D$ and the position of the particle now..
[Condition 1 says displacement can be positive or negative
with equal probabilities, and Condition 2 says the absolute 
displacement is random in $Unif(0, r)$. That implies $D \sim
Unif(-r,r).$]
The resulting long-run behavior, with respect to
occupancy of the interval $(0,r]$, is not dependent on $r> 0$, which
is essentially a scale parameter. Over the long run this
process exhibits strange behaviors I won't discuss here.
The comment by @MichaelBurr gives the relevant answer.
The process will continue to go in and out of $(0, r]$ forever.
In particular, it will typically leave this interval for the
first time after only a few steps.
Note: In order to get a related process that settles to a long run
distribution, one can contrive the displacements to be 
biased towards the origin. (The farther from the origin,
the more likely movement is toward the origin.)
Below is a simulation (in R software) of one run of 5000 steps of the process
you specified, in which I chose $r = 1.$ From the graph you can
see it is outside $(0, 1)$ after only a few steps. (Index denotes
the step $i$.) Sometimes plots such as this one are called 'history plots'.
 r = 1
 m = 5000;  x = numeric(m);  x[1] = 0
for(i in 2:m) {
   d = runif(1, -r, r)
   x[i] = x[i-1] + d  }
plot(x, type = "l")


