I refer to the random walk $(S_n)_{n \geq 1}$ where $S_n = X_1 + \cdots + X_n$ and $X_i$ are i.i.d random variables taking values $\pm 1$ with equal probability.
I want to know how to show that
$$\mathbb{P}(T_0 = 2n) = \frac{1}{2n-1} \binom{2n}{n} 2^{-2n},$$
where $T_0 = \inf\{n \geq 1: S_n = 0\}$.
My idea so far is to instead compute $\mathbb{P}(S_{2n}=0) - \mathbb{P}(S_{2n} = 0, T_0 \leq 2n - 1)$, where the first time is an easy binomial computation, and the second term can be summed up as a sum of disjoint events. When I try and do this sum however, I end up in mess.
So, I cannot help but think I as missing some nice symmetry somewhere, or perhaps an easy to solve difference equation...
I think I found an appropriate result in Feller's probability book:
The probability that at epoch $2n$ you return to the origin is $\mathbb{P}(S_{2n}=0)={2n \choose n}2^{-2n}$ (and now I see how you come to your conjecture).
Then, Lemma 1 on page 76 states that the probability that no return to the origin occurs up to and including epoch $2n$ is the same as the probability that a return occurs at epoch $2n$, i.e.,
$$\mathbb{P}(S_1\neq 0, S_2 \neq 0, \dots S_{2n}\neq 0)= \mathbb{P}(S_{2n}=0)$$
Thus, you can write for (see Feller page 77-78)
$$ \mathbb{P}(T_0=2n)=\mathbb{P}(S_{2n-2}=0)-\mathbb{P}(S_{2n}=0)=\frac{1}{2n-1} {2n \choose n}2^{-2n}$$
since the first term is exactly the probability that no return to the origin occurs during the first $2n-2$ epochs.