# cousin_it comments on Predictability of Decisions and the Diagonal Method - Less Wrong

13 09 March 2012 11:53PM

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Comment author: 10 March 2012 11:06:57AM *  1 point [-]

Namely, the agent will two-box if it predicts that the predictor decides that the agent one-boxes

Then won't the predictor predict that the agent two-boxes?

Comment author: 10 March 2012 11:45:23AM *  1 point [-]

There is a bit of terminological confusion here. Predictor makes a decision to either one-box or two-box, and the decision to one-box could be called "predicting that the agent one-boxes", but it isn't necessarily able to predict so unconditionally, so it's more like "predicting that the agent one-boxes given that the predictor one-boxes".

The predictor is supposed to one-box, but the agent isn't supposed to infer (within M steps) that the predictor one-boxes (just as in finite diagonal step, the objective is to make a decision unpredictable, not impossible). On the other hand, if this plan works out, the agent should be able to infer that the predictor one-boxes conditional on the agent one-boxing.

A common theme is that the diagonal rule could make events unpredictable, but still conditionally predictable. It's interesting whether it's possible to navigate this edge well.

Comment author: 12 March 2012 11:03:53PM 0 points [-]

I don't understand this answer and cousin_it doesn't either (I just asked him).

The predictor is supposed to one-box, but the agent isn't supposed to infer (within M steps) that the predictor one-boxes (just as in finite diagonal step, the objective is to make a decision unpredictable, not impossible).

If the predictor outputs "one-box", then the agent must prove this if it enumerates all proofs within a certain length because there is a proof-by-simulation of this fact, of length proportional to the predictor's run time, right? I don't see how the diagonal step can prevent the agent from finding this proof, unless it makes the predictor not output "one-box".

Comment author: 12 March 2012 11:13:40PM *  2 points [-]

The agent doesn't use enough steps to simulate the predictor, it decides early (because it finds a proof that predictor conditionally one-boxes early), which is also what might allow the predictor to conditionally predict agent's one-boxing within predictor's limited computational resources. The M steps where the agent protects the predictor from being unconditionally predictable is a very small number here, compared to agent's potential capability.