Comment author:Benquo
30 October 2007 11:49:14AM
5 points
[-]

What if it were a repeatable choice?

Suppose you choose dust specks, say, 1,000,000,000 times. That's a considerable amount of torture inflicted on 3^^^3 people. I suspect that you could find the number of times equivalent to torturing each of thoes 3^^^3 people 50 years, and that number would be smaller than 3^^^3. In other words, choose the dust speck enough times, and more people would be tortured effectually for longer than if you chose the 50-year torture an equivalent number of times.

If that math is correct, I'd have to go with the torture, not the dust specks.

Likewise, if this was iterated 3^^^3+1 times(ie 3^^^3 plus the reader),it could easily be 50*3^^^3 (ie > 3^^^3+1) people tortured. The odds are if it's possible for you to make this choice, unless you have reason to believe otherwise they may too, making this an implicit prisoner's dilemma of sorts. On the other side, 3^^^3 specks could possibly crush you, and/or your local cluster of galaxies into a black hole, so there's that to consider if you consider the life within meaningful distance of of every one of those 3^^^3 people valuable.

Comment author:Benquo
03 October 2013 01:15:49PM
*
2 points
[-]

I'm not sure I follow your argument.

I'm going to assume that for a single person, 3^^3 dust specks = 50 years of torture. (My earlier figure seems wrong, but 3^^3 dust specks over 50 years is a little under 5,000 dust specks per second.) I'm going to ignore the +1 because these are big numbers already.

If this were iterated 3^^^3 times, then we have the choice between:

TORTURE: 3^^^3 people are each tortured for 50 years, once.

DUST SPECKS: 3^^^3 people are tortured for 50 years, repeated (3^^^3)/(3^^3)=3^(3^^3-3^3) times.

The probability I'm the only person person selected out of 3^^^3 for such a decision p(i) is less than any reasonable estimate of how many people could be selected, imho. Let's say well below 700dB against. The chances are much greater that some probability fo those about to be dust specked or tortured also gets this choice (p(k)). p(k)*3^^^3 > p(i) => 3^^^3 > p(i)/p(k) => true for any reasonable p(i)/p(k)

So this means that the effective number of dust particles given to each of us is going to be roughly (1-p(i))p(k)3^^^3.

I'm going to assume any amount of dust larger in mass than a few orders of magnitude above the Chandrasekhar limit (1e33 kg) is going to result in a black hole. I can even assume a significant error margin in my understanding of how black holes work, and the reuslts do not change.

The smallest dust particle is probably a single hydrogen atom(really everything resoles to hydrogen at small enough quantities, right?). 1 mol of hydrogen weighs about 1 gram. So (1-p(i))(p(k)3^^^3 * (1 gram/mol)*(6e-23 'specks'/mol) * (1e-3 kg/g) * (1e-33 kg/black hole) = roughly ( 3^^^3 ) (~1e-730) = roughly 3^^^3 black holes.

In conclusion, I think at this level, I would choose 'cancel' / 'default' / 'roll a dice and determine the choice randomly/not choose' BUT would woefully update my concept of the sizee of the universe to contain enough mass to even support a reasonably infentessimal probability of some proportion of 3^^^3 specks of dust, and 3^^^3 people or at least some reasonable proportion thereof.

The question I have now is how is our model of the universe to update given this moral dillema? What is the new radius of the universe given this situation? It can't be big enough for 3^^^3 dust specks piled on the edge of our universe outside of our light cone somewhere. Either way I think the new radius ought to be termed the "Yudkowsky Radius".

I suppose you could view the utility as a meaninful object in this frame and abstract away the dust, too, but in the end the dust-utility system is going to encompaps both anyway so solving the problem on either level is going to solve it on both.

## Comments (599)

OldWhat if it were a repeatable choice?

Suppose you choose dust specks, say, 1,000,000,000 times. That's a considerable amount of torture inflicted on 3^^^3 people. I suspect that you could find the number of times equivalent to torturing each of thoes 3^^^3 people 50 years, and that number would be smaller than 3^^^3. In other words, choose the dust speck enough times, and more people would be tortured effectually for longer than if you chose the 50-year torture an equivalent number of times.

If that math is correct, I'd have to go with the torture, not the dust specks.

Likewise, if this was iterated 3^^^3+1 times(ie 3^^^3 plus the reader),it could easily be 50*3^^^3 (ie > 3^^^3+1) people tortured. The odds are if it's possible for you to make this choice, unless you have reason to believe otherwise they may too, making this an implicit prisoner's dilemma of sorts. On the other side, 3^^^3 specks could possibly crush you, and/or your local cluster of galaxies into a black hole, so there's that to consider if you consider the life within meaningful distance of of every one of those 3^^^3 people valuable.

*2 points [-]I'm not sure I follow your argument.

I'm going to assume that for a single person, 3^^3 dust specks = 50 years of torture. (My earlier figure seems wrong, but 3^^3 dust specks over 50 years is a little under 5,000 dust specks per second.) I'm going to ignore the +1 because these are big numbers already.

If this were iterated 3^^^3 times, then we have the choice between:

TORTURE: 3^^^3 people are each tortured for 50 years, once.

DUST SPECKS: 3^^^3 people are tortured for 50 years, repeated (3^^^3)/(3^^3)=3^(3^^3-3^3) times.

The probability I'm the only person person selected out of 3^^^3 for such a decision p(i) is less than any reasonable estimate of how many people could be selected, imho. Let's say well below 700dB against. The chances are much greater that some probability fo those about to be dust specked or tortured also gets this choice (p(k)). p(k)*3^^^3 > p(i) => 3^^^3 > p(i)/p(k) => true for any reasonable p(i)/p(k)

So this means that the effective number of dust particles given to each of us is going to be roughly (1-p(i))p(k)3^^^3.

I'm going to assume any amount of dust larger in mass than a few orders of magnitude above the Chandrasekhar limit (1e33 kg) is going to result in a black hole. I can even assume a significant error margin in my understanding of how black holes work, and the reuslts do not change.

The smallest dust particle is probably a single hydrogen atom(really everything resoles to hydrogen at small enough quantities, right?). 1 mol of hydrogen weighs about 1 gram. So (1-p(i))(p(k)3^^^3 * (1 gram/mol)*(6e-23 'specks'/mol) * (1e-3 kg/g) * (1e-33 kg/black hole) = roughly ( 3^^^3 ) (~1e-730) = roughly 3^^^3 black holes.

ie 3^(3

1^32^33^...^37e13 -730) = roughly 3^(31^32^33^...^37e13)ie 3

1^32^33^...^37e13 - 730 = roughly 31^32^33^...^37e13.In conclusion, I think at this level, I would choose 'cancel' / 'default' / 'roll a dice and determine the choice randomly/not choose' BUT would woefully update my concept of the sizee of the universe to contain enough mass to even support a reasonably infentessimal probability of some proportion of 3^^^3 specks of dust, and 3^^^3 people or at least some reasonable proportion thereof.

The question I have now is how is our model of the universe to update given this moral dillema? What is the new radius of the universe given this situation? It can't be big enough for 3^^^3 dust specks piled on the edge of our universe outside of our light cone somewhere. Either way I think the new radius ought to be termed the "Yudkowsky Radius".

I don't really care what happens if you take the dust speck literally; the point is to exemplify an extremely small disutility.

I suppose you could view the utility as a meaninful object in this frame and abstract away the dust, too, but in the end the dust-utility system is going to encompaps both anyway so solving the problem on either level is going to solve it on both.