8. Let $A, B, C, D$ be four non-coplanar points in space. 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 of each other. Then the probability that $A$ and $B$ can be connected by (a single edge or a sequence of edges forming a) spatial polyline is $\qquad$ .
Notice that the connections among $A, B, C, D$ have a total of $2^{6}=64$ methods. Consider the cases where $A, B$ cannot be connected by a space broken line.
(1) If $C D$ is connected by a line segment, at this point, it can only be $A C, A D$ connected or $B C, B D$ connected. Since both choices include a scenario where one line segment is also not connected (i.e., the graph contains only the line segment $C D$) which is counted repeatedly, there are $4+4-1=7$ methods;
(2) If $C D$ is not connected by a line segment, there are two scenarios. One is $A C, B D$ connected or $A D, B C$ connected. Since both choices include a scenario where one line segment is also not connected (i.e., the graph has no line segments connected) which is counted repeatedly, there are $4+4-1=7$ methods; The other is simultaneously connecting $A C, A D$ or simultaneously connecting $B C, B D$, which has 2 methods. In summary, there are 16 methods where $A, B$ cannot be connected by a space broken line. Let $E=$ “$A, B$ can be connected by a space broken line”, then $P(E)=1-P(\bar{E})=1-\frac{16}{64}=\frac{3}{4}$.