Probability of random walk returning to 0 Given a symmetric 1-dimensional random walk starting at 0 -- what is the probability of the walk returning $k$ times to 0 after $2N$ steps?
I know that the total number of paths it can take is $2^{2N}$. My problem is finding the total number of paths returning to 0. Also, since you can only return at even $k$'s, the no. of possible hitting points is $ {{N}\choose{k}}.  $ 

Let $T$ denote the first return time to $0$. The generating function $E_0(s^T)$ of $T$ for the random walk starting at $0$ is such that $$E_0(s^T)=sE_1(s^T),$$ where $E_1(s^T)$ is the generating function of $T$  for the random walk starting at $1$. By the same first-step argument, $$E_1(s^T)=\frac12s(1+E_2(s^T)),$$ where $E_2(s^T)$ is the generating function of $T$ for the random walk starting at $2$. By the Markov property, $$E_2(s^T)=E_1(s^T)^2.$$ Solving this system    in $E_0(s^T)$, $E_1(s^T)$ and $E_2(s^T)$ yields $$E_0(s^T)=1-\sqrt{1-s^2}.$$ Thus, the generating function of the $k$th return to $0$ starting from $0$ is $$E_0(s^T)^k=\left(1-\sqrt{1-s^2}\right)^k.$$
These functions have series expansions in powers of $s^2$. The probability that the $k$th return to $0$ occurs after exactly $2N$ steps is the coefficient of $s^{2N}$ in $E_0(s^T)^k$, that is,
$$
(-1)^N\sum_{i=1}^k(-1)^i{k\choose i}{i/2\choose N}.
$$
