decision on his best estimate of $Ch_{q}(e_{A})$ in accordance with Ismael’s \[2008\] Ignorance Principle:

> Where you’re not sure about the chances, form a mixture of the chances assigned by different theories of chance with weights determined by your relative confidence in those theories.

Following this principle, Bob should assign $Ch_{q}(e_{A})$ the estimated value ${\frac12}.0+{\frac12}.1={\frac12}$, and base his decision on that. Since measuring polarization with respect to the same axis as Alice would not raise his estimated chance of securing outcome $e_{A}$ in $\mathbf{1}$, he should eliminate this option *whether or not he could execute it*. His estimated value of $Ch_{q}(e_{A})$ were he to measure polarization with respect to an axis rotated $60{{}^\circ}$ from Alice’s is also $\frac12$ (${\frac12}.{\frac14}+{\frac12}.{\frac34}={\frac12}$). Similarly for any other angle. This essentially recapitulates part of the content of the no-signalling theorems, going back to Eberhard \[1978\]. Bell \[2004, pp. 237-8\] shows why manipulation of external fields at $p^{\prime }$ or in $\mathbf{2}$ would also fail to alter Bob’s estimated value of $Ch_{q}(e_{A})$.

But what if Bob had simply arranged for the measurement in $\mathbf{2}$ to have had the different *outcome* $\bar{e}_{B}$? Then $Ch_{q}(e_{A})$ would have been 0 instead of 1. No-one who accepts quantum mechanics can 