A discrete time random walk in $\mathbb{R}^n$. Consider a random walk where at $t=0$ you start at the origin of $\mathbb{R}^n$, and at each time $t\in \mathbb{N}$ you randomly move to a point that's within a ball of radius $1$ from your current location (each point being chosen uniformly at random).
There's a lot of questions one could ask about this, but my main question is: at time $t=2$, what is the probability that you're once again within distance $1$ from the origin?  
 A: We can assume WLOG, due to the circular symmetry of the issue, that if $O$ is the center of the disk, the first line segment  $OM_1$, is along the real positive $x$ axis with length $OM_1=R_1$. 
Let us call $M_2$ the second point, $R_2$ the distance $M_1M_2$. Let $\theta$ be the polar angle of $M_1M_2$ with respect to the $x$ axis, i.e. such that $\vec{M_1M_2}=\binom{R_2 \cos(\theta)}{R_2 \sin(\theta)}$ . 
Cosine law applied to triangle $OM_1M_2$ gives:
$$\text{squared distance} \ (OM_2)^2=R_1^2+R_2^2+2 R_1R_2 \cos(\theta)$$
(with a plus sign in front of the last term because we have to consider angle $\pi-\theta$.)
The issue is now to compute the probability of event :
$$\tag{1}R_1^2+R_2^2+2 R_1R_2 \cos(\theta) < 1.$$
with random variables $R_1, R_2$ uniform on [0,1] and $\theta$ uniform on $[0,2 \pi]$ (in fact $[0,\pi]$ would be enough).
I thought at first that an equivalent problem is as follows : find the pdf $f$ of the random variable "distance of two random points in the unit disk". Then the result is  $\int_0^1 f(\rho)d\rho$ with
$$\tag{2}f(\rho)=\tfrac{4}{\pi}\rho\left(cos^{-1}(\tfrac{\rho}{2})-\tfrac{\rho}{2}\sqrt{1-(\tfrac{\rho}{2})^2}\right) \ \ \ $$
(see Appendix 2 for explanations).
But this is not the case : (2) corresponds to the case of a uniform distribution on the disk, i.e., with a pdf having a top hat shape.
But here, the pdf is :
$$\tag{3}\varphi(r)=\tfrac{1}{2 \pi}\tfrac{1}{r}$$
which has the shape of a kind of  chinese hat shape with an infinite spike at its center (see Appendix 1 for explanations)  
The vector sum $\vec{OM_1}+\vec{M_1M_2}$ gives a convolution $f$ of two "chinese hat" distributions. This convolution also has, in a natural manner,  a rotational symmetry, meaning that it can be expressed under the following form:
$$f(r,\theta)=K\psi(r)$$ for a certain normalizing constant $K$, independently of $\theta.$
It remains to find $f$ as the convolution of $\varphi(r)$ with itself.
The last step will be the computation of:
$$\tag{3}\int_{r=0}^1 \int_{\theta=0}^{2 \pi}f(r)\color{red}{r} dr d\theta$$ where $\color{red}{r}$ is the classical jacobian for polar coordinates.
Remark 1 : I have done an extensive simulation using (1). It gives a probability close to $0.777$.
Remark 2 : there is a connection with Radon Transform ; let us recall that this transform amounts to a convolution $f \to f*1/r$ with kernel $1/r$. Thus what we are looking for is the kernel equivalent to applying twice the Radon transform.
Appendix 1: proof of formula (3).
Recall that we deal with points $P$ that are distributed in the unit disk with a uniform distribution for their distance to origin $R=OP$ and a uniform distribution for their polar angle. 
Consider the elementary annulus defined by $r_0<r<r_0+dr$ ; its area is $\pi(r_0+dr)^2-\pi(r_0)^2=2 \pi r_0$ (term $dr^2$ has been dropped); the probability that a point is in this annulus is the volume $2 \pi r_0 \varphi(r_0)$ of the corresponding  elementary cylinder under the surface defined by the density. But this quantity must be a constant $K$ independent of position $r_0$. As a consequence $\varphi(r_0)=\tfrac{K}{r_0}$. The fact that $K=\tfrac{1}{2 \pi}$ is due to the normalizing condition $\int_{\theta=0}^{2 \pi} \int_{r=0}^{r=1} \varphi(r)rdrd\theta = 1$.
Appendix 2: References for formula (2) (let us recall that we don't use it here): through the following link (https://math.stackexchange.com/q/934688), I had the indication of a very interesting book by A.M. Mathai "An introduction to geometrical probability" (formula (2.3.63) p.205) ; this book is downloadable (see the last comment). In a non-expected way, (2) is the same as formula that is used in confocal laser technology (see (ftp://ftp.wiley.com/sci_tech_med/electro-optical/problems.pdf)).
