8. Let $A, B, C, D$ be four points in space that are not coplanar. With a probability of $\frac{1}{2}$, connect an edge between each pair of points, and whether any two pairs of points are connected is independent. Then the probability that points $A$ and $B$ can be connected by (a single edge or a sequence of edges forming) a spatial polyline is $\qquad$
8. $\frac{3}{4}$.

There are 2 possibilities for whether each pair of points is connected by an edge, resulting in a total of $2^{6}=$ 64 scenarios. Consider the number of scenarios where points $A$ and $B$ can be connected by a polyline.
(1) There is an edge $A B$: There are $2^{5}=32$ scenarios.
(2) There is no edge $A B$, but there is an edge $C D$: In this case, points $A$ and $B$ can be connected by a polyline if and only if point $A$ is connected to at least one of $C$ or $D$, and point $B$ is connected to at least one of $C$ or $D$. The number of such scenarios is
$$
\left(2^{2}-1\right) \times\left(2^{2}-1\right)=9 \text {. }
$$
(3) There is no edge $A B$, and no edge $C D$: In this case, $A C$ and $C B$ being connected has $2^{2}$ scenarios, and $A D$ and $D B$ being connected also has $2^{2}$ scenarios, but the scenario where $A C$, $C B$, $A D$, and $D B$ are all connected is counted twice, so the number of scenarios where points $A$ and $B$ can be connected by a polyline is
$$
2^{2}+2^{2}-1=7 \text {. }
$$

In summary, the total number of scenarios is $32+9+7=48$.
Therefore, the probability that points $A$ and $B$ can be connected by a polyline is $\frac{48}{64}=\frac{3}{4}$.