Example 7 A snail crawls along the grid lines on a sufficiently large square grid paper with a side length of 1, starting from a certain intersection. Each time it randomly chooses one of the four directions: up, down, left, or right, and crawls a unit length at a uniform speed. Find the probability that the snail returns to the starting point exactly after 2016 crawls.
【Analysis】Let the random variables $x_{1}, x_{2}, x_{3}, x_{4}$ represent the values of the snail moving up, down, left, and right, respectively. It is clear that the random vector $\xi=\left(x_{1}, x_{2}, x_{3}, x_{4}\right)$ follows a multinomial distribution, with the generating function given by
$$
\begin{array}{l}
\varphi\left(x_{1}, x_{2}, x_{3}, x_{4}\right) \\
=\left(\frac{1}{4} x_{1}+\frac{1}{4} x_{2}+\frac{1}{4} x_{3}+\frac{1}{4} x_{4}\right)^{2016} .
\end{array}
$$

In the expansion of the above expression, the factors $x_{1}^{n_{1}}, x_{2}^{n_{2}}, x_{3}^{n_{3}}, x_{4}^{n_{4}}$ represent the snail moving up, down, left, and right by $n_{1}, n_{2}, n_{3}, n_{4}$ units, respectively. For the snail to return to the starting point after 2016 moves, it must be that $n_{1}=n_{2}, n_{3}=n_{4}$.
In the generating function, let
$$
x_{1}=x, x_{2}=\frac{1}{x}, x_{3}=y, x_{4}=\frac{1}{y} \text {. }
$$

Then $x_{1}^{n_{1}} x_{2}^{n_{2}} x_{3}^{n_{3}} x_{4}^{n_{4}}=x^{n_{1}-n_{2}} y^{n_{3}-n_{4}}=1$.
Thus, the generating function becomes
$$
\begin{array}{l}
\frac{1}{4^{2016}}\left(x+\frac{1}{x}+y+\frac{1}{y}\right)^{2016} \\
\quad=\frac{1}{4^{2016}}\left(\frac{x y+1}{x}+\frac{x y+1}{y}\right)^{2016} \\
\quad=\frac{1}{4^{2016}} \times \frac{(x y+1)^{2016}(x+y)^{2016}}{x^{2016} y^{2016}} .
\end{array}
$$

To find the probability that the snail returns to the starting point after 2016 moves, we need to find the coefficient of $x^{2016} y^{2016}$ in the numerator $(x y+1)^{2016}(x+y)^{2016}$.
Notice that,
$$
\begin{array}{l}
(x y+1)^{2016}(x+y)^{2016} \\
=\left(\sum_{k=0}^{2016} \mathrm{C}_{2016}^{k} x^{2016-k} y^{2016-k}\right) \cdot \\
\left(\sum_{t=0}^{2016} \mathrm{C}_{2016}^{t} x^{2016-t} y^{t}\right),
\end{array}
$$

In the first bracket of the above expansion, the exponents of $x$ and $y$ are equal in each term. Therefore, in the second bracket, we must also find terms where the exponents of $x$ and $y$ are equal, which can only be when $2016-t=t$, i.e., $t=1008$, corresponding to the term $\mathrm{C}_{2016}^{1008} x^{1008} y^{1008}$. At this point, in the first bracket, we can only take $k=1008$, corresponding to the term $\mathrm{C}_{2016}^{1008} x^{1008} y^{1008}$, and their product is $\left(\mathrm{C}_{2016}^{1008}\right)^{2} x^{2016} y^{2016}$.
Thus, the required probability is $p=\frac{\left(\mathrm{C}_{2016}^{1008}\right)^{2}}{4^{2016}}$.