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.
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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.
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."
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.
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.
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.