Random walk problem in the plane Let a particle in the plane $R^2$ executes random jumps at discrete times $t= 1, 2, ...$. At each step, the particle jumps from the point it is a distance of lenght one. The angle of any new jump (say, with the $x$ axis) is uniformly distributed in $[0,2\pi]$. 
Question: If initially ($t=0$) the particle is at the origin, what is the probability that it gets back to the unit disk of the plane for each time $t= 2, 3, 4, ...$? In particular, what is the value of this probability for $t=3$?
Thanks for any help!
 A: $\newcommand{\+}{^{\dagger}}%
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The probability density for the step $n$ is given by
$\ds{{\rm p}\pars{\vec{r}_{n}} \equiv {\delta\pars{r_{n} - 1} \over 2\pi}}$. The probability density $\pp_{N}\pars{\vec{r}}$ of arriving at $\vec{r}$ after $N$ steps is given by:

\begin{align}
\pp_{N}\pars{\vec{r}}
&\equiv
\int\dd^{2}\vec{r}_{1}\,{\rm p}\pars{\vec{r}_{1}}\ldots
\int\dd^{2}\vec{r}_{N}{\rm p}\pars{\vec{r}_{N}}
\delta\pars{\vec{r} - \sum_{\ell = 1}^{N}\vec{r}_{\ell}}
\\[3mm]&=
\int\dd^{2}\vec{r}_{1}\,{\rm p}\pars{\vec{r}_{1}}\ldots
\int\dd^{2}\vec{r}_{N}{\rm p}\pars{\vec{r}_{N}}
\int\exp\pars{\ic\vec{k}\cdot\bracks{\vec{r} - \sum_{\ell = 1}^{N}\vec{r}_{\ell}}}
\,{\dd^{2}\vec{k} \over \pars{2\pi}^{2}}
\\[3mm]&=
\int\expo{\ic\vec{k}\cdot\vec{r}}
\bracks{\int\dd^{2}\vec{R}\,{\rm p}\pars{\vec{R}}\expo{-\ic\vec{k}\cdot\vec{R}}}^{N}
\,{\dd^{2}\vec{k} \over \pars{2\pi}^{2}}
=
\int\expo{\ic\vec{k}\cdot\vec{r}}
\bracks{\int_{0}^{2\pi}\expo{-\ic k\cos\pars{\theta}}\,{\dd\theta \over 2\pi}}^{N}
\,{\dd^{2}\vec{k} \over \pars{2\pi}^{2}}
\end{align}

However
$$
\int_{0}^{2\pi}\expo{-\ic k\cos\pars{\theta}}\,{\dd\theta \over 2\pi}
=
\int_{-\pi}^{\pi}\expo{\ic k\cos\pars{\theta}}\,{\dd\theta \over 2\pi}
=
{1 \over \pi}\int_{0}^{\pi}\expo{\ic k\cos\pars{\theta}}\,\dd\theta
=
{\rm J}_{0}\pars{k}
$$
where ${\rm J}_{\nu}\pars{k}$ is the
$\nu$-$\it\mbox{order Bessel Function of the First Kind}$.

\begin{align}
\pp_{N}\pars{\vec{r}}
&=
\int\expo{\ic\vec{k}\cdot\vec{r}}{\rm J}_{0}^{N}\pars{k}
\,{\dd^{2}\vec{k} \over \pars{2\pi}^{2}}
=
{1 \over 2\pi}\int_{0}^{\infty}\dd k\,k\,{\rm J}_{0}^{N}\pars{k}\int_{0}^{2\pi}
\expo{\ic kr\cos\pars{\theta}}\,{\dd\theta \over 2\pi}
\end{align}
$$
\color{#ff0000}{\pp_{N}\pars{\vec{r}}
=
{1 \over 2\pi}\int_{0}^{\infty}{\rm J}_{0}\pars{kr}{\rm J}_{0}^{N}\pars{k}k
\,\dd k}
$$

The probability ${\rm P}_{N{\Huge\circ}}$ that it returns to the unit circle after $N$ steps is given by:
\begin{align}
\color{#0000ff}{\large{\rm P}_{N{\Huge\circ}}}
&=
\int_{r\ <\ 1}\pp_{N}\pars{\vec{r}}\,\dd^{2}\vec{r}
=
\int_{0}^{\infty}\overbrace{\bracks{\int_{0}^{1}{\rm J}_{0}\pars{kr}r\,\dd r}}
^{{\rm J}_{1}\pars{k}/k}
{\rm J}_{0}^{N}\pars{k}k\,\dd k
\\[3mm]&=
\color{#0000ff}{\large\int_{0}^{\infty}{\rm J}_{1}\pars{k}
{\rm J}_{0}^{N}\pars{k}\,\dd k}
\end{align}
We compute a few values with Wolfram Alpha:
$$
\begin{array}{rclrcl}
{\rm P}_{0{\Huge\circ}} =\int_{0}^{\infty}{\rm J}_{1}\pars{k}
\,\dd k & = & 1\,,\quad
{\rm P}_{1{\Huge\circ}} =\int_{0}^{\infty}{\rm J}_{1}\pars{k}
{\rm J}_{0}\pars{k}\,\dd k & = & \half
\\
{\rm P}_{2{\Huge\circ}} =\int_{0}^{\infty}{\rm J}_{1}\pars{k}{\rm J}_{0}^{2}\pars{k}
\,\dd k & = & {1 \over 3}\,,\quad
{\rm P}_{3{\Huge\circ}} =\int_{0}^{\infty}{\rm J}_{1}\pars{k}
{\rm J}_{0}^{3}\pars{k}\,\dd k & = & {1 \over 4}
\\
{\rm P}_{4{\Huge\circ}} =\int_{0}^{\infty}{\rm J}_{1}\pars{k}{\rm J}_{0}^{4}\pars{k}
\,\dd k & = & {1 \over 5}\,,\quad
{\rm P}_{5{\Huge\circ}} =\int_{0}^{\infty}{\rm J}_{1}\pars{k}
{\rm J}_{0}^{5}\pars{k}\,\dd k & = & {1 \over 6}
\end{array}
$$
It $\tt\large seems$ the exact result is
$\ds{%
{\rm P}_{N{\Huge\circ}}
=
\int_{0}^{\infty}{\rm J}_{1}\pars{k}{\rm J}_{0}^{N}\pars{k} \,\dd k
=
{1 \over N + 1}}$
Mathematica can solve this integral and it yields the $\ul{\mbox{exact result}}$ $\color{#0000ff}{\Large{1 \over N + 1}}$ when $\color{#ff0000}{\large\Re N > -1}$.
A: This is a classic problem, first solved by the Dutch mathematician J.C. Kluyver in 1905. He derives the result given in the answer of Felix Marin, that the probability to return to the unit disc after $n$ steps is $1/(n+1)$:


For a purely geometric derivation, see arXiv:1007.4870. (There the result is attributed to Rayleigh, but it appears Rayleigh only considered the large-$n$ limit.) See also this MO posting.
A: You can create a probility dristribution for the location after one step (the unit circle with a certain constant prbability). After two steps, I think that should be doable too. You can also carculate for each point the probability of entering de unit disk in the next step. Then, by combining the probability distribution ($P_1(x,y)$) and the probability of ending in the disk $P_2(x,y)$, you get
$$
\int P_1(x,y)P_2(x,y) \,dx\,dy
$$
A: This is not a complete answer, but is kind of long for a comment.  
For $t=2$, draw a picture.  It's pretty easy to see the answer is $1/3$.
For the $t=3$ case, let the angles be $\theta_1$, $\theta_2$, and $\theta_3$.  By symmetry, you can assume without loss of generality that $\theta_1 = 0$.  So you need to find the probability that
$\|\langle 1, 0 \rangle +  \langle \cos\theta_2, \sin\theta_2 \rangle  +\langle \cos\theta_3,\sin\theta_3\rangle \|<1$.  A little computation shows that this is equals 
$P(1+\cos\theta_2+\cos\theta_3+\cos(\theta_2-\theta_3)<0)$, where $\theta_2$ and $\theta_3$ are chosen independently from the uniform distribution on $[0,2\pi]$.  That's as far as I got.
This is the probability that the particle is in the unit disc at time $t=3$ (not $t$ equals 2 or 3).
A: This addresses the t=3 case:
Assume you start at origin ($P_0$), and without loss of generality, assume first move is to $P_1=(1, 0)$. Your second move will be to $P_2=(1 + cos(\theta),  sin(\theta))$, where theta is uniform [0, $\pi$] (can ignore the [$\pi,2\pi$] range due to symmetry). Density function of $\theta$ is just $f(\theta)=\frac{1}{\pi}$
Draw unit-circles around $P_0$ and $P_2$, and let X be the second intersection point of these two circles (aside from $P_1$).
Now draw the quadrilateral  $\overline{P_0P_1P_2X}$. Turns out this is actually a rhombus, with each side of length 1. From geometry of parallel-lines, one can see that the angle $\angle P_1P_2X$ equals theta. Also, this angle $\theta$ subtends the arc of the $P_2$ circle that is inside the $P_0$ circle.
Therefore, for any given $\theta$, the probability that a chosen point on the $P_2$ circle will be inside the $P_0$ circle is simply $\frac{\theta}{2\pi}$.
Putting this all together, the probability that we return to the unit-circle after 3 moves is:
$\int^\pi_0(1/\pi) * (\theta/2\pi) d\theta = 1/4$
