($V_{A}$). Assuming that the state $\left\vert \Phi
^{+}\right\rangle =\frac{1}{\sqrt{2}}\left( \left\vert H\right\rangle
\left\vert H\right\rangle +\left\vert V\right\rangle \left\vert
V\right\rangle \right) $ was prepared and the settings $a,b$ chosen long before, the agent is also in a position to be certain what these were. The agent is then in a position to use the Born rule to determine the chance of $e_{A}$. But that chance must be relativized not just to a time, but (relativistically) to a space-time point. So consider the following diagram:

As this shows, if the outcome in region $\mathbf{2}$ is of type $V_{B}$ then $Ch_{p}(e_{A})=Ch_{p^{\prime }}(e_{A})=Ch_{r}(e_{A})={\frac12}$, but $Ch_{q}(e_{A})=\cos ^{2}\angle ab$. These are the chances that follow from application of the Born rule to state $\left\vert \Phi
^{+}\right\rangle $, given settings $a,b$. In each case the event $e_{A}$of the next photon detected in $\mathbf{1}$’s registering as vertically polarized has been specified as of type $V_{A}$, and the specification of the reference class at least includes the state and settings. Specifically,$$Ch_{p}(e_{A})=Pr_{a,b}^{\Phi ^{+}}(V_{A})=||\hat{P}^{A}(V)\Phi ^{+}||^{2}={\frac12}.$$$$Ch_{q}(e_{A})=Pr_{a,b}^{\Phi ^{+}}(V_{A}|V_{B})\equiv \frac{Pr_{a,b}^{\Phi
^{+}}(V_{A},V_{B})}{Pr_{b}^{\Phi ^{+}}(V_{B})}=\frac{{\frac12}\cos ^{2}\angle ab}{{\frac12}}=\cos ^{2}\angle ab$$Note that the reference class used in calculating $Ch_{q}(e_{A})$ is narrower: it is further restricted by specification of the outcome as of type $V_{B}$ in region $\mathbf{2}$.

Any agent who accepts 