7. Given real numbers $a_{1}, a_{2}, \cdots, a_{18}$ satisfy:
$$
a_{1}=0,\left|a_{k+1}-a_{k}\right|=1(k=1,2, \cdots, 17) \text {. }
$$

Then the probability that $a_{18}=13$ is $\qquad$
7. $\frac{17}{16384}$.

Let $b_{k}=a_{k+1}-a_{k}$. Then
$$
b_{k}= \pm 1, a_{18}=b_{1}+b_{2}+\cdots+b_{17} \text {. }
$$

Thus, $a_{18}=13$ is equivalent to $b_{1}, b_{2}, \cdots, b_{17}$ having 2
$$
\text { } -1, 15 \text { } 1 \text {s. }
$$

Therefore, the probability that $a_{18}=13$ is
$$
\frac{\mathrm{C}_{17}^{2}}{2^{17}}=\frac{17 \times 8}{2^{17}}=\frac{17}{2^{14}}=\frac{17}{16384}
$$