= \frac18(2-\sqrt2),\\
{\mathcal P}(-\pm-) &= {\mathcal P}\{6,8\} = \frac12 \cos^2(3\pi/8) = \frac18(2-\sqrt2),
             \tag{AC}\\
{\mathcal P}(+\pm-) &= {\mathcal P}\{2,4\} = \frac12 \sin^2(3\pi/8) = \frac18(2+\sqrt2)\\
{\mathcal P}(-\pm+) &= {\mathcal P}\{5,7\} = \frac12 \sin^2(3\pi/8) = \frac18(2+\sqrt2).\end{aligned}$$

Let $[a=b]$ be the event that the measurements for orientations $\vec A, \vec B$ coincide, and $[a\ne b]$ be the complementary event, that these measurements are distinct. Define events $[b=c], [b\ne c], [a=c]$ and $[a\ne c]$ similarly. We have $$\begin{aligned}
[a=b] &= (++\pm)\cup(--\pm) = \{1,2,7,8\}, 
&&{\mathcal P}[a=b]= 1/2,\\
[a\ne b] &= (+-\pm)\cup(-+\pm) = \{3,4,5,6\}, 
&&{\mathcal P}[a\ne b]= 1/2,\\
[b=c] &= (\pm++)\cup(\pm--) = \{1,4,5,8\}, 
&&{\mathcal P}[b=c] = \cos^2(\pi/8) = \frac14(2+\sqrt2),\\
[a=c] &= (+\pm+)\cup(-\pm-) = \{1,3,6,8\}, 
&&{\mathcal P}[a=c]= \sin^2(\pi/8) = \frac14(2-\sqrt2).\end{aligned}$$

Suppose that a possibly-signed probability distribution $Q$ extends ${\mathcal P}$ to all events. Let $U$ be the nonobservable event $\{3,6\}$. The following computation shows that $Q$ cannot be traditional. $$\begin{aligned}
2Q(U) &= \big[Q(U) + Q([a=c]-U)\big] + \big[Q(U) + Q([a\ne b]-U)\big] \\
&\qquad- \big[Q([a=c]-U) + Q([a\ne b]-U)\big]\\
&= {\mathcal P}[a=c] + {\mathcal P}[a\ne b] - {\mathcal P}[b=c] = \frac14(2-\sqrt2) +\frac12 -\frac14(2+\sqrt2)\\ &= \frac12(1-\sqrt2)\\
Q(U) &= \frac14(1-\sqrt2) < 0.\end{aligned}$$

Now let’s address the question whether there is any solution of the Extension Problem in our case.

Consider the transformation $g$ 