# AlexMennen comments on Problematic Problems for TDT - Less Wrong

34 29 May 2012 03:41PM

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Comment author: 04 June 2012 11:39:19PM 0 points [-]

But a couple of difficulties arise. The first is that if TDT variants can logically separate from each other (i.e. can prove that their decisions aren't linked) then they won't co-operate with each other in Prisoner's Dilemma. We could end up with a bunch of CliqueBots that only co-operate with their exact clones, which is not ideal.

I think this is avoidable. Let's say that there are two TDT programs called Alice and Bob, which are exactly identical except that Alice's source code contains a comment identifying it as Alice, whereas Bob's source code contains a comment identifying it as Bob. Each of them can read their own source code. Suppose that in problem 1, Omega reveals that the source code it used to run the simulation was Alice. Alice has to one-box. But Bob faces a different situation than Alice does, because he can find a difference between his own source code and the one Omega simulated, whereas Alice could not. So Bob can two-box without effecting what Alice would do.

However, if Alice and Bob play the prisoner's dilemma against each other, the situation is much closer to symmetric. Alice faces a player identical to itself except with the "Alice" comment replaced with "Bob", and Bob faces a player identical to itself except with the "Bob" comment replaced with "Alice". Hopefully, their algorithm would compress this information down to "The other player is identical to me, but has a comment difference in its source code", at which point each player would be in an identical situation.

Comment author: 09 June 2012 11:24:08AM 1 point [-]

You might want to look at my follow-up article which discusses a strategy like this (among others). It's worth noting that slight variations of the problem remove the opportunity for such "sneaky" strategies.

Comment author: 09 June 2012 08:46:14PM 0 points [-]

Ah, thanks. I had missed that, somehow.

Comment author: 06 June 2012 12:12:51PM *  0 points [-]

In a prisoners dilemma Alice and Bob affect each others outcomes. In the newcomb problem, Alice affects Bobs outcome, but Bob doesn't affect Alices outcome. That's why it's OK for Bob to consider himself different in the second case as long as he knows he is definitely not Alice (because otherwise he might actually be in a simulation) but not OK for him to consider himself different in the prisoners dilemma.

Comment author: 25 December 2012 04:13:32PM -1 points [-]

However, if Alice and Bob play the prisoner's dilemma against each other, the situation is much closer to symmetric. Alice faces a player identical to itself except with the "Alice" comment replaced with "Bob", and Bob faces a player identical to itself except with the "Bob" comment replaced with "Alice". Hopefully, their algorithm would compress this information down to "The other player is identical to me, but has a comment difference in its source code", at which point each player would be in an identical situation.

Why doesn't that happen when dealing with Omega?

Comment author: 25 December 2012 08:01:22PM 0 points [-]

Because if Omega uses Alice's source code, then Alice sees that the source code of the simulation is exactly the same as hers, whereas Bob sees that there is a comment difference, so the situation is not symmetric.

Comment author: 25 December 2012 10:21:11PM 0 points [-]

So why doesn't that happen in the prisoner's dilemma?

Comment author: 25 December 2012 10:47:57PM 0 points [-]

Because Alice sees that Bob's source code is the same as hers except for a comment difference, and Bob sees that Alice's source code is the same as his except for a comment difference, so the situation is symmetric.

Comment author: 26 December 2012 01:32:52AM 0 points [-]

Newcomb:

Bob sees that there is a comment difference, so the situation is not symmetric.

Prisoner's Dilemma:

Bob sees that Alice's source code is the same as his except for a comment difference, so the situation is symmetric.

Do you see the contradiction here?

Comment author: 26 December 2012 01:59:23AM *  0 points [-]

Newcomb, Alice: The simulation's source code and available information is literally exactly the same as Alice's, so if Alice 2-boxes, the simulation will too. There's no way around it. So Alice one-boxes.

Newcomb, Bob: The simulation was in the situation described above. Bob thus predicts that it will one-box. Bob himself is in an entirely different situation, since he can see a source code difference, so if he two-boxes, it does not logically imply that the simulation will two-box. So Bob two-boxes and the simulation one-boxes.

Prisoner's Dilemma: Alice sees Bob's source code, and summarizes it as "identical to me except for a different comment". Bob sees Alice's source code, and summarizes it as "identical to me except for a different comment". Both Alice and Bob run the same algorithm, and they now have the same input, so they must produce the same result. They figure this out, and cooperate.

Comment author: 26 December 2012 02:15:28AM *  0 points [-]

Ignore Alice's perspective for a second. Why is Bob acting differently? He's seeing the same code both times.

Comment author: 26 December 2012 02:21:57AM 0 points [-]

Don't ignore Alice's perspective. Bob knows what Alice's perspective is, so since there is a difference in Alice's perspective, there is by extension a difference in Bob's perspective.

Comment author: 26 December 2012 02:25:33AM 0 points [-]

Bob looks at the same code both times. In the PD, he treats it as identical to his own. In NP, he treats it as different. Why?