What is the probability that $\langle u, v \rangle = \langle u, w \rangle$? Consider three vectors $u$, $v$, $w$ of dimension $n$. All the vector entries are either $-1$ or $1$ with equal probability and are i.i.d. and all three vectors are independent.  I am trying to work out the probability that $\langle u, v \rangle = \langle u, w \rangle$.
Clearly $\langle u, v \rangle$ and $\langle u, w \rangle$ are both  distributed as the location of a symmetric random walk with $n$ steps. The probability of two independent symmetric random walks being at the same position after $n$ steps is $\binom{2n}{n}\frac{1}{2^{2n}}.$

Is  $\binom{2n}{n}\frac{1}{2^{2n}}$ the probability that $\langle u, v \rangle = \langle u, w \rangle$?

 A: Yes. Here is another way to show that. It suffices to calculate the probability when $u$ is fixed. Observe that 
$$\langle u,v\rangle = |\{i:u_i=v_i\}|-|\{i:u_i\neq v_i\}| = 2|\{i:u_i=v_i\}|-n$$ and so $\langle u,v\rangle = \langle u,w\rangle$ if and only if $v$ and $w$ have the same number of entries in common with $u$.
Let $X_k$ be the set of elements of $\{-1,1\}^n$ that have exactly $k$ entries in common with $u$. There are exactly $n\choose k$ elements of $X_k$ and so the probability of a random element of $\{-1,1\}^n$ being in $X_k$ is $\frac{n\choose k}{2^n}$. 
We are looking for the probability that there is some $k\in\{0,\ldots,n\}$ such that $v\in X_k$ and $w\in X_k$, and since these are independent, this is then
$$\sum_{k=0}^n\left(\frac{n\choose k}{2^n}\right)^2 = \frac{1}{2^{2n}}\sum_{k=0}^n {n\choose k}^2 = \frac{{2n}\choose n}{2^{2n}}.$$
Edit: I used the fact that $$\sum_{k=0}^n {n\choose k}^2= {{2n}\choose n}.$$ To see this, it is perhaps easier to rewrite it as $${{2n}\choose n}=\sum_{k=0}^n {n\choose k}{n\choose{n-k}}.$$ Given a collection of $2n$ balls, paint half of them red and the other half black. Then taking an arbitrary subset of $n$ of these balls is equivalent to taking $k$ red balls and $n-k$ black balls for some $k$.
