any $g\in G$; hence $R(e) = {\mathcal P}(e)$. It remains to check that $R$ is a possibly-signed probability distribution. Since $\Omega$ is observable, $R(\Omega) = \frac1{|G|} \sum_{g\in G}
{\mathcal P}(\Omega) = {\mathcal P}(\Omega) = 1$. Further, if events $e_1,e_2$ are disjoint and coobservable then $$R(e_1\cup e_2) 
= \frac1{|G|} \sum_{g\in G} Q(g(e_1\cup e_2))
 = \frac1{|G|} \sum_{g\in G} [Q(ge_1) + Q(ge_2)]
 = R(e_1) + R(e_2).$$

Bell’s theorem and negative probabilities {#sec:bell}
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Quantum theory is contextual in the sense that the value of an observable $O$, measured as a part of one context, may differ from the value of $O$ measured as a part of another context. Attempts to avoid contextuality may lead to negative probabilities. This will be illustrated in the present section. We start with two Gedankenexperiments exhibiting the contextuality of quantum mechanics.

Prepare the state $\frac1{\sqrt2} \Big({\ensuremath{|01\rangle}} - {\ensuremath{|10\rangle}}\Big)$, known as the spin singlet state, of a pair of spin 1/2 particles, e.g., electrons. Here 0 represents spin up in the $z$ direction, and 1 represents spin down. Choose an arbitrary direction $\bf a$ and measure spin in direction $\bf a$ on both particles, getting $+\frac12$ if the spin is up or $-\frac12$ if the spin is down. According 