Manfred comments on Problematic Problems for TDT - Less Wrong

36 Post author: drnickbone 29 May 2012 03:41PM

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Comment author: lackofcheese 24 May 2012 07:51:17AM *  2 points [-]

Let's say that TDT agents can be divided into two categories, TDT-A and TDT-B, based on a single random bit added to their source code in advance. Then TDT-A can take the strategy of always picking the first box in Problem 2, and TDT-B can always pick the second box.

Now, if you're a TDT agent being offered the problem; with the aforementioned strategy, there's a 50% chance that the simulated agent is different than you, netting you $1 million. This also narrows down the advantage of the CDT agent - now they only have a 50% chance of winning the money, which is equal to yours.

Comment author: Manfred 24 May 2012 07:31:23PM 3 points [-]

Actually, the way the problem is specified, Omega puts the money in box 3.

Comment author: drnickbone 24 May 2012 07:58:11PM 0 points [-]

The argument is that the simulation is either TDT-A in this case, or TDT-B. Either way, the simulated agent will pick a single favourite box (1 or 2) with certainty, so the money is in either Box 2 or Box 1,

Though I can see an interpretation which leads to Box 3. Omega simulates a "new-born" TDT (which is neither -A nor -B) and watches as it differentiates itself to one variant or the other, each with equal probability. So the new-born picks boxes 1 and 2 with equal frequency over multiple simulations, and Box 3 contains the money. Is that what you were thinking?

Comment author: Manfred 24 May 2012 08:00:53PM *  0 points [-]

Is that what you were thinking?

Yes. I was thinking that Omega would have access to the agent's source code, and be running the "play against yourself, if you pick a different number than yourself you win" game. Omega is a jerk :D

Comment author: lackofcheese 24 May 2012 08:17:25PM *  2 points [-]

If it's your own exact source being simulated, then it's probably impossible to do better than 10%, and the problem isn't interesting anymore.