wedrifid comments on A problem with Timeless Decision Theory (TDT) - Less Wrong

36 Post author: Gary_Drescher 04 February 2010 06:47PM

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Comment author: wedrifid 05 February 2010 07:04:26AM *  0 points [-]

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).

Comment author: Gary_Drescher 05 February 2010 01:46:43PM 1 point [-]

When:

D(M) = true, D(!M) = true, E = true

Omega fails.

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.

Comment author: wedrifid 05 February 2010 05:44:44PM 0 points [-]

So, with those details, that resolves to "I get $0". This makes D(M) = !M the unambiguous 'correct' decision function.