wedrifid comments on A problem with Timeless Decision Theory (TDT) - Less Wrong
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Let:
When:
Omega fails.
Omega chooses M or !M. I get $1M or 0.
Omega chooses M=false. I get $0.1.
Omega chooses M=true. I get $1M.
M chooses either M or !M. I get either $1.1 or $0.1 depending on Omega's whims
Omega has no option. I make Omega look like a fool.
So, depending on how 'Omega is wrong' is resolved I use either D(M) = M or D(M) = false.
These are the conclusions of Wedrifid-Just-Works-It-Out Decision Theory. It should match TDT when TDT is formulated right (and I don't make a mistake).
No, but it seems that way because I neglected in my OP to supply some key details of the transparent-boxes scenario. See my new edit at the end of the OP.
So, with those details, that resolves to "I get $0". This makes D(M) = !M the unambiguous 'correct' decision function.