ksvanhorn comments on Sleeping Beauty Resolved?

ksvanhorn 22 May 2018 18:29 UTC
3 points
What I wrote above may be a bit misleading. The issue isn’t that you have additional terms for conjunctions of the $B_{i}$ , but that the weights $Pr (B_{i} ∣ M)$ sum to more than $1$ . In particular, consider the case when AI Beauty gets exactly one bit of input. Then for $y = 0$ or $1$ ,
$\begin{matrix} Pr (X_{2} (y) ∣ M) & = \frac{1}{2} Pr (H ∣ M) + \frac{3}{4} Pr (not H ∣ M) = \frac{1}{2} \cdot \frac{1}{2} + \frac{3}{4} \cdot \frac{1}{2} = \frac{5}{8} \end{matrix}$
and $5 / 8 + 5 / 8 = 5 / 4 > 1$ . If we try the same decomposition as in my previous comment, then using $Pr (H ∣ X_{2} (y), M) = 1 / 2.5 = 2 / 5$ , we find
$\begin{matrix} Pr (H ∣ M) & = Pr ((H & X_{2} (0)) or (H & X_{2} (1)) ∣ M) = Pr (H & X_{2} (0) ∣ M) + Pr (H & X_{2} (1) ∣ M) - Pr (H & X_{2} (0) & X_{2} (1) ∣ M) = Pr (X_{2} (0) ∣ M) \cdot Pr (H ∣ X_{2} (0), M) + Pr (X_{2} (1) ∣ M) \cdot Pr (H ∣ X_{2} (1), M) - 0 = \frac{5}{8} \cdot \frac{2}{5} + \frac{5}{8} \cdot \frac{2}{5} = \frac{1}{2} \end{matrix}$
and everything is still consistent.