Moving particle

This is just the two-dimensional simple random walk, viewed at a $45^\circ$ angle. It is a well-known result that the $n$ -dimensional simple random walk is recurrent for $n= 1,2$ , and transient for $n \ge 3$ . Thus this random walk is recurrent, which makes the probability of eventual return to the origin equal to $1 = \boxed{100}\%$ .

Here is a detailed proof in this case.

Let $X_n$ be the random variable that is equal to $1$ if the particle has returned to the origin after $n$ steps, and $0$ otherwise (note that $X_n = 0$ if $n$ is odd. If $p_n$ the the probability of returning to the origin after $n$ steps, then $p_n = E[X_n]$ . If $X$ is the random variable equal to the total number of returns to the origin, then $X = \sum_{n=1}^\infty X_n$ , and hence $E[X] \; = \; \sum_{n=1}^\infty E[X_n] \; = \; \sum_{n=1}^\infty p_n$ Suppose that $\rho$ is the probability of ever returning to the origin. Then $P[X \ge k] \; = \; \rho^k$ for all $k \ge 1$ , and hence we deduce that $E[X] \; = \; \sum_{k=1}^\infty P[X \ge k] \; = \; \left\{ \begin{array}{lll} \frac{\rho}{1-\rho} & \hspace{1cm} & \rho < 1 \\ \infty & & \rho = 1 \end{array}\right.$ At each step of this random walk, the particle moves either left or right one step, and at the same times moves either up or down one step. Moves of left and right are equally likely, as are moves of up and down. Moreover, the left and right moves are independent of the up and down moves. Thus, in order to return to the origin after $2n$ moves, there have to be $n$ left and $n$ right moves, and $n$ up and $n$ down moves. Thus $p_{2n} \; = \; \left(\frac{1}{2^{2n}}\binom{2n}{n}\right)^2 \hspace{2cm} n \ge 1$ Stirling's approximation tells us that $p_{2n} \sim \frac{1}{\pi n} \hspace{2cm} n \to \infty$ and hence that $E[X] \; = \; \sum_{n=1}^\infty p_n \; = \; \infty$ which implies that $\rho = 1 = \boxed{100}\%$ .