= 27d567bc2d48d9f40e9ccc20ea370b15)
Arran_Stirton comments on A (very) tentative refutation of Pascal's mugging - Less Wrong
You are viewing a comment permalink. View the original post to see all comments and the full post content.
= 783df68a0f980790206b9ea87794c5b6)
Loading…= b715d02e85f7b3b4ae65ca3d3e450868)
Subscribe to RSS Feed
Powered by Reddit
= f037147d6e6c911a85753b9abdedda8d)
You are viewing a comment permalink. View the original post to see all comments and the full post content.
Comments (34)
The fraction of states will diverge. The closest you're likely to get is to weight the programs by length. If you do so, the weighted portion will be slightly more than if you just weight the shortest ways of stating it.
Also, consider that you could take any possible program that works, and add "Also, 3^^^^3 people are killed" on top of it.
Thank-you for the feedback!
But would the number of states diverge?
Say we're talking about a target and we're predicting that at least one arrow will strike it in the next five minutes; in that case the space for five light-minutes around that target would have to be in one of a set of initial states that would result in there being at least one arrow in the target after five minutes.
Then if we predict that at least two arrows will strike in the next five minutes, then we're narrowing our set of states by only considering the ones were two or more arrows strike. This is necessarily lower than the set for one arrow as all the states in the one-arrow-set are in the two-arrow-set except for all the states that result in just one arrow striking.
Then again for three arrows and so on. Wouldn't this lead to a converging fraction of states?
I'm not sure how taking any possible program and adding "Also, 3^^^^3 people are killed" would effect this, could you elaborate at all?
Thanks again.
It wouldn't converge for each arrow. If you did the limit by program length, it might converge, but there's no obvious reason that there must be a certain order to count them in. If you can count them in any order, you can alternate between states where the arrow hits and states where it does not, and make it look like it has a 50% chance of hitting.
Given a state, you could add another non-interacting (or just interacting based on one guy's decision) universe where 3^^^^3 people die. This universe is constant complexity, and small complexity compared to 3^^^^3. Let's call the complexity of that part k.
Given a possible universe with complexity n, there is a possible universe where Pascal's mugger is telling the truth with complexity n+k. When you get to n+k, there are 2^k times as many possibilities to choose from, so it's 2^k times less likely. This is much more likely than the 1/3^^^^3 times less likely needed for Pascal's mugger to not be credible enough.
Well for any finite amount of time that you predict into the future you've also got a finite amount of space to consider, as anything too far away wouldn't be able to travel fast enough to effect the outcome of the thing being predicted about. Each state of the universe would really be the state of this finite area of space which would be expressed in binary. One way to order the states would be in terms of how large a binary number they form, from smallest to largest.
I'm not sure how making it look like the arrow had a 50% chance of hitting would make any difference to anything though?
Given a state, you could also add another non-interacting (or just interacting based on one guy's decision) universe where 3^^^^3 people lives are saved. I don't know if this is the right terminology, but it seems to me that when you start adding extra possible universes on, their outcomes become causally decoupled from the original decision to give/not-give the mugger $5.
You have an entire universe to consider. You don't deal with just possible universes just began. You deal with all possible universes. There are simple universes that eventually come out to this, but no simple ones that start this way. Also, a limited speed of light is not guaranteed. As far as we can tell, it's limited, but it might not be.
If you can always calculate it so it's 50%, along with any other probability, you're clearly doing something wrong. It should only calculate to one value.
You can also add on modified versions that are coupled, or things like that. It's a bit more complicated than I said, but there's still a good chance (as in more than 1/3^^^^3) that the mugger isn't bluffing.