Concerning Newcomb’s Problem I understand that the dominant position among the regular posters of this site is that you should one-box. This is a position I question.
Suppose Charlie takes on the role of Omega and presents you with Newcomb’s Problem. So far as it is pertinent to the problem Charlie is identical to Omega with the notable exception that his prediction is only %55 likely to be accurate. Should you one-box or two-box in this case?
If you one-box then the expected utility is (.55 1,000,000) $550,000 and if you two-box then it is (.45 1,001,000) $450,450 so it seems you should still one-box even when the prediction is not particularly accurate. Thoughts?
Good question. And with Charlie known to be operating exactly as defined then yes, I would one box. I wouldn't call him Charlie however as that leads to confusion. The significant problem with dealing with someone who is taking the role of Omega is in my ability to form a prediction about them that is sufficient to justify the 'cooperate' response. Once I have that prediction the rest, as you have shown, is just simple math.
I declare this Open Thread open for discussion of Less Wrong topics that have not appeared in recent posts.