Hazing as Counterfactual Mugging?
post by SilasBarta · 2010-10-11T14:17:09.201Z · LW · GW · Legacy · 8 commentsContents
8 comments
In the interest of making decision theory problems more relevant, I thought I'd propose a real-life version of counterfactual mugging. This is discussed in Drescher's Good and Real, and many places before. I will call it the Hazing Problem by comparison to this practice (possibly NSFW – this is hazing, folks, not Disneyland).
The problem involves a timewise sequence of agents who each decide whether to "haze" (abuse) the next agent. (They cannot impose any penalty on previous agent.) For all agents n, here is their preference ranking:
1) not be hazed by n-1
2) be hazed by n-1, and haze n+1
3) be hazed by n-1, do NOT haze n+1
or, less formally:
1) not be hazed
2) haze and be hazed
3) be hazed, but stop the practice
The problem is: you have been hazed by n-1. Should you haze n+1?
Like in counterfactual mugging, the average agent has lower utility by conditioning on having been hazed, no matter how big the utility difference between 2) and 3) is. Also, it involves you having to make a choice from within a "losing" part of the "branching", which has implications for the other branches.
You might object the choice of whether to haze is not random, as Omega’s coinflip is in CM; however, there are deterministic phrasings of CM, and your own epistemic limits blur the distinction.
UDT sees optimality in returning not-haze unconditionally. CDT reasons that its having been hazed is fixed, and so hazes. I *think* EDT would choose to haze because it would prefer to learn that, having been hazed, they hazed n+1, but I'm not sure about that.
I also think that TDT chooses not-haze, although this is questionable since I'm claiming this is isomorphic to CM. I would think TDT reasons that, "If n's regarded it as optimal to not haze despite having been hazed, then I would not be in a position of having been hazed, so I zero out the disutility of choosing not-haze."
Thoughts on the similarity and usefulness of the comparison?
8 comments
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comment by Vladimir_Nesov · 2010-10-11T17:31:40.006Z · LW(p) · GW(p)
This is much more like PD than CM, TDT-wise. Under similar assumptions to those necessary to cooperate in PD, TDT will not haze in this problem.
TDT considers the current possible world, as explicitly depending on agent's decisions. As a result, any site where agent's decision influences the current possible world, in the past of in the future, generates a collection of counterfactual worlds that TDT agent also takes into account. In particular, on Newcomb's problem, agent's decision is located in the past, and opens both possibilities for consideration. In Newcomb's with transparent boxes, the agent should update on the state of the second box and exclude the possibility of its alternative state, but because its decision site is still located in the past, this decision site allows to regenerate the alternative with the different state of the second box, and so TDT agent acts correctly on transparent Newcomb's.
Likewise, in the hazing problem, if our agent is considered sufficiently similar to the previous agent to cooperate on PD with it, then the previous agent constitutes a decision point in the past. As a result, the counterfactual where the previous agent performs the opposite action (doesn't haze) is open for consideration, as the outcome of a possible decision of our agent stemming from (in particular) that past decision point. (Furthermore, strictly speaking the counterfactual resulting from the decision to not haze would have all agents in the sequence not hazing the next one, not just the previous agent not hazing with all the agents from 1 to n-2 still hazing.)
Replies from: Manfred↑ comment by Manfred · 2010-10-14T17:03:19.137Z · LW(p) · GW(p)
That seems silly. I'd think the time difference thoroughly breaks the symmetry that the TDT answer to the prisoner's dilemma that I've seen relies on. It's like if one of the prisoners got to hear the other prisoners' answer before deciding.
But I suppose your answer does increase the usefulness of this problem as something to think about.
Replies from: Vladimir_Nesov↑ comment by Vladimir_Nesov · 2010-10-14T18:03:28.490Z · LW(p) · GW(p)
TDT agents do what they'd want to precommit to doing. Thus, in PD where one prisoner sees the other's decision, TDT agents still cooperate (since if the second player doesn't cooperate, then it won't benefit from first player's cooperation, but it wants to, hence it does cooperate).
Replies from: Manfred↑ comment by Manfred · 2010-10-14T22:38:23.066Z · LW(p) · GW(p)
And now it's Parfit's hitchhiker! :D
Except instead of having psychic powers and only accepting people who intend to pay, now the driver accepts everyone and is a mute.
Hm, so perhaps hazing isn't quite like this asymmetric PD, since the current generation cannot give anything back to the older generation. But still, it's interesting to talk about, so:
Would the second player cooperate even if the first player defected, since that's what they would do if they'd precommitted?
Replies from: Vladimir_Nesov↑ comment by Vladimir_Nesov · 2010-10-15T09:43:51.876Z · LW(p) · GW(p)
Would the second player cooperate even if the first player defected, since that's what they would do if they'd precommitted?
If the first player is even a cooperating rock, the second player should defect. The argument only applies to a "sufficiently similar" TDT agent as the first player, which won't defect, hence your new problem statement doesn't satisfy that condition.
Replies from: Manfredcomment by [deleted] · 2013-01-15T22:40:10.608Z · LW(p) · GW(p)
This seems a lot like the problem of choosing whether to procrastinate or not, with "haze" corresponding to "procrastinate" and the negation of the former corresponding to the negation of the latter. Your greatest preference is for your past self to have already not procrastinated, second is your current self procrastinates (given that your past self already has), and your least preference is your current self breaks the habit. The reflectively consistent move is to not procrastinate/haze, unconditionally (though with others, it should be based upon whether your decision correlates with theirs rather than unconditionally).
comment by cousin_it · 2010-10-12T02:08:23.129Z · LW(p) · GW(p)
I think standard game theory applies here. If an agent isn't hazed, does it prefer to haze or not? If it slightly prefers not-hazing, the only Nash equilibrium is the one where everyone hazes iff they were hazed. So if the chain has a beginning, then the first agent doesn't haze (because it wasn't hazed) and no one ever hazes anyone. But if there's a slight preference for hazing, the only Nash equilibrium is the one where everyone hazes unconditionally. This discontinuity hints that your problem formulation is an unstable point. I don't completely understand what this implies for UDT's answer.