Two-dimensional random walk First, let's start with the problem in one dimension.
Let's have $R_1, R_2, R_3, ...$ be i.i.d., and $P(R_1 = 1) = P(R_1 = -1) = \frac{1}{2}$. The question would be:  
"How many trajectories $(Y_n)$ there are, such that we start from $(0, 0)$, (where the first slot would be the time, and the second would be the position) and end at $(2n, 0)$, and between time $1$ and $2n$ it doesn't touch the $0$. We can see that all the path like that are $2$ times paths that go from $(1,1)$ to $(2n - 1, 1)$ , and that would we $\binom{2n - 2}{n - 1}$. All the paths that go from $(1, 1)$ to $(2n - 1, 1)$ that have a position $0$, would be $\binom{2n - 2}{n - 2}$, which can we found using reflection principle. So, for $\mathbb{N} =\{1, 2, 3, ...\}$ and $\tau = \text{inf}\{n \in \mathbb{N}: Y_0 = 0\}$, which would just be time before the first instance of our trajcetory going back to $0$, our probability of going to position $0$ at any given $\tau = 2n$ is equal to $P(\tau = 2n) = \dfrac{2(\binom{2n - 2}{n - 1} - \binom{2n - 2}{n - 2})}{2^{2n}}$.
Now, for the two dimensional problem, for $R_1, R_2, ...$ i.i.d. with $P(R_1 = 1) = P(R_1 = -1) = \frac{1}{2}$ and $S_1, S_2, ...$ i.i.d. with $P(S_1 = 1) = P(S_1 = -1) = \frac{1}{2}$, both independent of each other. What would be our $P(X_\tau = 2k)$? 
(edit) Just to be clear, my question is "After $2n$ moves, what is the probability that we are back at the $X$ axis for the first time?". This is the extenstion of the $1d$ problem.
I got to the point where I knew it would be an infinite sum, but I just can't seem to crack this.
Help would be much appreciated.
 A: Fleshing my comment out into an answer: this is not an extension of the 1D problem; it is, in fact, precisely the same as the 1D problem.
You are considering a process $(R_n, S_n)$, where $R_n, S_n$ are both independent 1D random walks. This is a diagonal walk on the integer lattice $\mathbb Z^2$, because the parity of the two positions will always match. Note that the $x$-axis is reached if and only if $R_n = 0$, and that $S_n$ is irrelevant; hence, your question is equivalent to considering when $R_n$ reaches $0$ again for the first time, and you already have your answer.
I want to stress that your random walk on $\mathbb Z^2$ is not quite the traditional one; for instance, it is impossible to reach the state $(1, 0)$. In order to consider the traditional random walk, you can't allow $R_n, S_n$ to be independent of one another, since exactly one of them must move at any given step. The question of the first return to the $x$-axis in the traditional walk is not quite as simple as the solution I outlined above, in that case; instead, when restricting to $R_n$, you now have a lazy random walk (meaning a walk that steps left or right with probability $1/4$ and does nothing with probability $1/2$). You can indeed calculate the first return to $0$ of this lazy random walk, but it isn't the same calculation as the one you've already done.
Final note: you can turn your random walk into the traditional one by rotating it by $45^{\circ}$ and scaling it down by $\sqrt 2$. However, the line you originally called the $x$-axis also rotates and becomes a diagonal line when considering the problem in this way. In other words: your question is equivalent to the expected first return of the traditional random walk to the diagonal set $\{(x, x)\}$.
