P ( P($1M box is empty | you one-box) = P($1M box is empty | you two-box) ) >= P(Omega cannot violate causality)
For what it’s worth: this is incorrect. If P(omega cannot violate causality) = 1, that still wouldn’t mean P ( P($1M box is empty | you one-box) = P($1M box is empty | you two-box) ) = 1. If If P(omega cannot violate causality) = 1, P($1M box is empty | you one-box) < P($1M box is empty | you two-box) due to Omega and the agent being subjunctively dependent on the agent’s decision procedure. (Unless, of course, the agent doesn’t believe there is subjunctive dependence going on.) Subjunctive dependence doesn’t violate causality.
That equation you quoted is in branch 2, “2. Omega is a “nearly perfect” predictor. You assign P(general) a value very, very close to 1.” So it IS correct, by stipulation.
For what it’s worth: this is incorrect. If P(omega cannot violate causality) = 1, that still wouldn’t mean P ( P($1M box is empty | you one-box) = P($1M box is empty | you two-box) ) = 1. If If P(omega cannot violate causality) = 1, P($1M box is empty | you one-box) < P($1M box is empty | you two-box) due to Omega and the agent being subjunctively dependent on the agent’s decision procedure. (Unless, of course, the agent doesn’t believe there is subjunctive dependence going on.) Subjunctive dependence doesn’t violate causality.
That equation you quoted is in branch 2, “2. Omega is a “nearly perfect” predictor. You assign P(general) a value very, very close to 1.” So it IS correct, by stipulation.