Comment author: thrawnca 18 August 2016 02:34:22AM *  0 points [-]

Sorry, but I'm not in the habit of taking one for the quantum superteam. And I don't think that it really helps to solve the problem; it just means that you don't necessarily care so much about winning any more. Not exactly the point.

Plus we are explicitly told that the coin is deterministic and comes down tails in the majority of worlds.

Comment author: lolbifrons 18 August 2016 12:09:15PM *  0 points [-]

Sorry, but I'm not in the habit of taking one for the quantum superteam.

If you're not willing to "take one for the team" of superyous, I'm not sure you understand the implications of "every implementation of you is you."

And I don't think that it really helps to solve the problem;

It does solve the problem, though, because it's a consistent way to formalize the decision so that on average for things like this you are winning.

it just means that you don't necessarily care so much about winning any more. Not exactly the point.

I think you're missing the point here. Winning in this case is doing the thing that on average nets you the most success for problems of this class, one single instance of it notwithstanding.

Plus we are explicitly told that the coin is deterministic and comes down tails in the majority of worlds.

And this explains why you're missing the point. We are told no such thing. We are told it's a fair coin and that can only mean that if you divide up worlds by their probability density, you win in half of them. This is defined.

What seems to be confusing you is that you're told "in this particular problem, for the sake of argument, assume you're in one of the worlds where you lose." It states nothing about those worlds being over represented.

In response to The Allais Paradox
Comment author: lolbifrons 18 August 2016 12:23:34AM *  0 points [-]

So here's why I prefer 1A and 2B after doing the math, and what that math is.

1A = 24000
1B = 26206 (rounded)
2A = 8160
2B = 8910

Now, if you take (iB-iA)/iA, which represents the percent increase in the expected value of iB over iA, you get the same number, as you stated.

(iB-iA)/iA = .0919 (rounded)

This number's reciprocal represents the number of times greater the expected value of iA is than the marginal expected value of iB

iA/(iB-iA) = 10.88 (not rounded)

Now, take this number and divide it by the quantity p(iA wins)-p(iB wins). This represents how much you have to value the first $24000 you receive over the next $3000 to pick iA over iB. Keep in mind that 24/3 = 8, so if $1 = 1 utilon in all cases, you should pick iA only when this quotient is less than 8.

1A/(1B-1A)/[p(1A wins)-p(1B wins)] = 369.92
2A/(2B-2A)/[p(2A wins)-p(2B wins)] = 1088

I have liabilities in excess of my assets of around $15000. That first $15000 is very important to me in a very quantized, thresholdy way, but it is not absolute. I can make the money some other way, but not needing to - having it available to me right now because of this game - represents more utility than a linear mapping of dollars to utility suggests, by a large factor.

The next threshold like this in my life that I can think of is "enough money to buy a house in Los Angeles without taking out a mortgage," of which $3000 is a negligible portion.

I'd say that the utility I assign the first $24000 because of this lies between 370 and 1080 times the utility I assign the next $3000. This is why I take 1A and 2B given that this entire thing is performed only once. Once my debts are paid, all bets (on 1A) are off.

If we're dealing with utilons rather than dollars, or I have repeated opportunity to play (which is necessary for you to "money pump" me) iB is the obvious choice in both cases.

Comment author: thrawnca 17 August 2016 07:52:19AM 0 points [-]

I think that what really does my head in about this problem is, although I may right now be motivated to make a commitment, because of the hope of winning the 10K, nonetheless my commitment cannot rely on that motivation, because when it comes to the crunch, that possibility has evaporated and the associated motivation is gone. I can only make an effective commitment if I have something more persistent - like the suggested $1000 contract with a third party. Without that, I cannot trust my future self to follow through, because the reasons that I would currently like it to follow through will no longer apply.

MBlume stated that if you want to be known as the sort of person who'll do X given Y, then when Y turns up, you'd better do X. That's a good principle - but it too can't apply, unless at the point of being presented with the request for $100, you still care about being known as that sort of person - in other words, you expect a later repetition of the scenario in some form or another. This applies as well to Eliezer's reasoning about how to design a self-modifying decision agent - which will have to make many future decisions of the same kind.

Just wanting the 10K isn't enough to make an effective precommitment. You need some motivation that will persist in the face of no longer having the possibility of the 10K.

Comment author: lolbifrons 17 August 2016 01:30:00PM 0 points [-]

It seems to me the answer becomes more obvious when you stop imagining the counterfactual you who would have won the $10000, and start imagining the 50% of superpositions of you who are currently winning the $10000 in their respective worlds.

Every implementation of you is you, and half of them are winning $10000 as the other half lose $100. Take one for the team.

Comment author: Cyan2 06 February 2008 10:46:20PM 1 point [-]

Here's an example which doesn't bear on Conservation of Expected Evidence as math, but does bear on the statement,

"There is no possible plan you can devise, no clever strategy, no cunning device, by which you can legitimately expect your confidence in a fixed proposition to be higher (on average) than before."

taken at face value.

It's called the Cable Guy Paradox; it was created by Alan Hรกjek, a philosopher the Australian National University. (I personally think the term Paradox is a little strong for this scenario.)

Here it is: the cable guy is coming tomorrow, but cannot say exactly when. He may arrive any time between 8 am and 4 pm. You and a friend agree that the probability density for his arrival should be uniform over that interval. Your friend challenges you to a bet: even money for the event that the cable guy arrives before noon. You get to pick which side of the bet you want to take -- by expected utility, you should be indifferent. Here's the curious thing: if you pick the morning bet, then almost surely there will be times in the morning when you would prefer to switch to the afternoon bet.

This would seem to be a situation in which "you can legitimately expect your confidence in a fixed proposition to be higher (on average) than before," even though the equation P(H) = P(H|E)*P(E) + P(H|~E)*P(~E) is not violated. I'm not sure, but I think it's due to multiple possible interpretations of the word "before".

Comment author: lolbifrons 17 August 2016 11:53:12AM *  0 points [-]

If you count the amount of "wanting to switch" you expect to have because the cable guy hasn't arrived yet, it should equal exactly the amount of "wishing you hadn't been wrong" you expect to have if you pick the second half because the cable guy arrived before your window started.

I'm not sure how to say this so it's more easily parseable, but this equality is exactly what conservation of expected evidence describes.