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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 $n\neq j$?

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up vote 3 down vote accepted

Yes. Let $a$ be the number of $+1$ steps and $b$ the number of $-1$ steps in the first $2n$ steps, so that $a+b=2n$. In order to be at $2j$, you must have $a-b=2j$, so just solve the system for $a$ and $b$: $2a=2n+2j$, $a=n+j$, and $b=n-j$. There are $\binom{2n}{n+j}$ ways to pick the $n+j$ steps to the right, so the desired probability is $$\binom{2n}{n+j}2^{-2n}\;.$$

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