Symmetric Random walk on $\mathbb {Z}^d$ 

Consider the symmetric random walk on $\mathbb{Z}^d $. Symmetric means that the probability of going into any of the $2^d$ directions is $1/2^d$. Starting in 0, what is the probability of returning to 0 in n steps?



My solution is
$$
p_{00}^{(n)}=\binom {n}{n/2}^d \frac{1}{2^{dn}}.
$$
Because for each dimension the steps in the one direction have to be the same as in the opposite direction and for each dimension there a $\binom {n}{n/2}$ possibilities to do so.
Am I right?
 A: No. This time, there are $2d$ directions to choose from. 
From the formula for $d=2$, you will have to change the binomial coefficient to
$$
\frac {n!}{\left(a_1!\right)^2 
\left(a_2!\right)^2\dots \left(a_{d}!\right)^2 
}$$for every  choice of $a_1\dots a_{d}$ so that $\sum_{i=1}^{d} a_i = n/2$ ($a_i$ is the number of moves made in the direction $+ e_i$) and the probability replaced accordingly by
$$
\prod_{i=1}^{d} \left(\frac1{2d}\right)^{2a_i}
$$
Then the global answer is
$$
p_{d,n} =  \sum_{\sum_{i=1}^{d} a_i=n/2} 
\frac {n!}{\left(a_1!\right)^2 
\left(a_2!\right)^2\dots \left(a_{d}!\right)^2 
}
\prod_{i=1}^{d} \left(\frac1{2d}\right)^{2a_i}
$$
A: Your formula works correctly for $d=1$ (of course) and also for $d=2$.  The way to see $d=2$ is that each step changes $x+y$ by $\pm 1$ each with probability $\frac{1}{2}$ and also, each step changes $x-y$ by $\pm 1$ each with probability $\frac{1}{2}$, and those results are independent.  So it is valid to multiply the two individual probabilities of ending at the origin, which is what your formula does.
Unfortunately, for $d >2$ (random walk in more than 2 dimensions), there is no similar clever reason why your formula would work.  And in fact, for a 2-step random walk in 3 dimensions, the probability of ending at the origin is $\frac{1}{6}$ whilst your formula gives an answer of $\frac{1}{8}$.
(I've always wanted to use the word "whilst" in a sentence.)
