# Return probability random walk

I'm struggling with this question. Any help would be appreciated!

A single particle starts at a given initial site and moves across the one-dimensional lattice, (a chain of sites with an equidistant spacing): At time $t_n$ the particle has taken $2n$ random steps from its initial site.

I.e., for time $t_1$ the particle has made $2$ random steps which are either

(i)LL = $2$ sites to the left,

(ii) RR = $2$ sites to the right,

(iii) LR = back to initial site, and

(iv) RL = back to initial site.

For time $t_2$ the particle takes $4$ random steps which gives $2^4 = 16$ different paths.

(a) Find the probability $P_n$ that the particle has returned to its initial site at time $t_n$, given for time $t_1$ the probability is $P_1 = \frac24 = \frac12$, since $2$ particles end up at the initial site.

(b) If $N$ particles start at the same initial site and move completely independently. How does the probability that all the particles return to this initial site at the same time behave for large times?

• – Adrian Tam Jul 18 '18 at 4:31

We can find the number of possibilities of a particle returning by considering that if the particle returns at time $t_{2n}$, the particle has moved left $n$ times and right $n$ times. Then we choose, which exactly the $n$ right moves are among the $2n$ moves, so the rest are left moves. There are $\binom{2n}{n}$ possibilities.
Therefore the probability of a particle returning is $$\frac{\binom{2n}n}{2^n}\text.$$ For $k$ particles, the probability of them all returning at time $t_n$ is $$\left(\frac{\binom{2n}n}{2^n}\right)^k \text.$$