M chooses either M or !M. I get either $1.1 or $0.1 depending on Omega’s whims
D(M) = false, D(!M) = false, E = false
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.
If Omega is just infallible then when D(M) = false, !E just never happens and I get either $0.1M or $1.1M depending on Omega’s whims. Since I’m being a smart ass I probably get $0.1M. So I use D(M) = M and get expected payout of $0.91M.
If Omega resolves “I am wrong” to “I give maximum payout” then I choose D(M) = false and get $1.1M (or sometimes either $1.1 or $0.1).
If Omega resolves “I am wrong” to “I give minimum payout” then I once again get $0.1M when D(M) = false and E.
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.
Let:
M be ‘There is $1 in the big box’
When:
D(M) = true, D(!M) = true, E = true
Omega fails.
D(M) = true, D(!M) = true, E = false
Omega chooses M or !M. I get $1M or 0.
D(M) = true, D(!M) = false, E = true
Omega chooses M=false. I get $0.1.
D(M) = true, D(!M) = false, E = false
Omega chooses M=true. I get $1M.
D(M) = false, D(!M) = false, E = true
M chooses either M or !M. I get either $1.1 or $0.1 depending on Omega’s whims
D(M) = false, D(!M) = false, E = false
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.
If Omega is just infallible then when D(M) = false, !E just never happens and I get either $0.1M or $1.1M depending on Omega’s whims. Since I’m being a smart ass I probably get $0.1M. So I use D(M) = M and get expected payout of $0.91M.
If Omega resolves “I am wrong” to “I give maximum payout” then I choose D(M) = false and get $1.1M (or sometimes either $1.1 or $0.1).
If Omega resolves “I am wrong” to “I give minimum payout” then I once again get $0.1M when D(M) = false and E.
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.