The probability that a simple random walker is at 0 after $2n$ steps is $P(S_{2n}=0)=\binom{2n}{n}2^{-2n}$. What is the probability that a random walker is at integer $2j$?
Well, I understand that since steps are either $+1$ or $-1$, to be at $2j$,
$2j\leq 2n$. After $2n$ steps, if $2n=2j$ you have $2n$ of $+1$ and 0 of $-1$. Then, $n=j$ and $2n=(n+j)$. So would you have $(n+j)$ steps of $+1$ in all cases, even when $n\neq j$?