Pierre-André_Noël
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Sorry, I have not been specific enough. Each of my 3^^^^3, 3^^^^3-1, 3^^^^3-2, etc. examples are mutually exclusive (but the sofa is part of the "0" case). While they might not span all possibilities (not exhaustive) and could thus sum to less than one, they cannot sum to higher than 1. As I see it, the weakest assumption here is that "more persons/pigs is less or equally likely". If this holds, the "worst case scenario" is epsilon=1/(3^^^^3) but I would guess for far less than that.
Sorry for being out of topic, but has that 3^^^^3 problem been solved already? I just read the posts and, frankly, I fail to see why this caused so much problems.
Among the things that Jaynes repeats a lot in his book is that the sum of all probabilities must be 1. Hence, if you put probabilities somewhere, you must remove elsewhere. What is the prior probability for "me being able to simulate/kill 3^^^^3 persons/pigs"? Let's call that nonzero number "epsilon". Now, I guess that the (3^^^^3)-1 case should have a probability greater or equal than epsilon, same for (3^^^^3)-2 etc. Even with a "cap" at 3^^^^3, this makes epsilon <= 1/(3^^^^3). And... (read more)
Peter de Blanc: You are right and I came to the same conclusion while walking this morning. I was trying to simplify the problem in order to easily obtain numbers <=1/(3^^^^3), which would solve the "paradox". We now agree that I oversimplified it.
Instead of messing with a proof-like approach again, I will try to clarify my intuition. When you start considering events of that magnitude, you must consider a lot of events (including waking up with blue tentacles as hands to take Eliezer's example). The total probability is limited to 1 for exclusive events. Without proof, there is no reason to put more probability there than anywhere else. There is not much... (read more)