from observable events to all events.

4.  If such a signed extension exists then find one.

The Fat-outcomes proviso of the previous section makes the Extension Problem trivial in the case of a single ensemble: the given ${\mathcal P}$ is already defined on all events. If there are exactly two ensembles $A$ and $B$, imposing partitions $\Pi_A = \{A_1, \dots, A_m\}$ and $\Pi_B = \{B_1, \dots, B_n\}$, and if every intersection $A_i\cap B_j\ne\emptyset$, then there is a traditional solution for the Extension Problem: set ${\mathcal P}(A_i\cap B_j) = {\mathcal P}(A_i)\cdot{\mathcal P}(B_j)$.

The observation frame of the scenario is described in the previous Example. It has exactly three ensembles. It remains only to specify the probability distribution ${\mathcal P}$. It is given by the following table. $$\begin{aligned}
{\mathcal P}\{00,01\}=0\quad &{\mathcal P}\{10,11\}=1\\
{\mathcal P}\{00,10\}=0\quad &{\mathcal P}\{01,11\}=1 \\
{\mathcal P}\{00,11\}=0\quad &{\mathcal P}\{01,10\}=1 \end{aligned}$$

The following simple theorem will turn out to be useful.

\[thm:sym\] Let $G$ be a group of automorphisms of an observation space $O=(\Omega,{\ensuremath{\mathrm{CO}}},{\mathcal P})$. If $Q$ is a possibly-signed extension of ${\mathcal P}$ to all events, then so is the average $$R(e) = \frac1{|G|} \sum_{g\in G} Q(ge).$$

If event $e$ is observable then $Q(ge)
= Q(e) = {\mathcal P}(e)$ for 