Particle moves in square, what is the expected distance before first return to edge? There is a unit square with a particle moving in it. After the particle collides with an edge, the angle of reflection is random and is drawn from the uniform distribution on $[-\frac{\pi}{2}, \frac{\pi}{2}].$ 
The question is to find the average distance the particle covers before it returns to the same edge next time (I guess that after a large number of collusions the starting point is not important). 
To me, this sounds like a question about the stationary distribution of a Markov chain with a continuum of states. However, the problem actually is taken from a physics olympiad for high school students. It is claimed that the answer is $2\sqrt{2}$. 
If it is not a mistake, there probably is an intuitive non-rigorous argument why the answer is $2\sqrt{2}$.
Update 1: I was asked to post the original text of the problem here. It is somewhat different from what I wrote above, but I believe that this is basically the same question:

Problem: In a computer model, movement of a particle inside of a square is
  simulated. Square has sides of length L, the speed of the point is V.
  After a collusion with an edge point bounces at a random angle
  (equiprobable from -90 to 90 degrees) with the same speed. Estimate the
  number of collusions with one of the sides after a large period of
  time T.
Answer: $\frac{TV}{2\sqrt{2}L}$.

Update 2: There were attempts to do a simulation (see comments below), and the results tend to be somewhat smaller than $2\sqrt{2}$.
Also, in my simulation the distribution of collision points is not uniform (points close to angles are more frequent) and distribution of distance from bounce to bounce is asymmetric and bimodal.
 A: One possible heuristic reasoning behind the answer (for the "official" problem) is the following:
1.The events of hitting each side are equiprobable (good estimate if T is big) ,so
the number of hits for a fixed side is estimated as : $\frac{(total-no-of-hits)}{4}$
2.Total traveled distance in time $T$ is $TV$
3.The longest traveled distance between two hits (between any two sides) is $\sqrt{2}L$(on the diagonal). The shortest is 0 . A rough estimate for the average distance  between two hits will be $\frac{\sqrt{2}L}{2}$ (this is the most problematic part).
4.Now we can estimate the total number of hits of any side as $\frac{2TV}{\sqrt{2}L}$
5.For a fixed side the total number of hits will be $\frac{2TV}{4\sqrt{2}L}=\frac{TV}{2\sqrt{2}L}$
Update: With these estimates the answer to the initial question is indeed $2\sqrt{2}L$ . If we introduce two random variables , X- the length of the path between two collisions  and Y - the number of sides hit until the particle returns to the side where it started will be $Z=X + YX$ . Since X and Y are independent , using the estimates from above and observing that Y follows a geometric distribution with $p=\frac{1}{3}$ we conclude that the expectation of Z is : $E[Z] = \frac{\sqrt{2}L}{2} + 3\frac{\sqrt{2}L}{2}$ = $2\sqrt{2}L$
