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wuthefwasthat comments on Zeckhauser's roulette - Less Wrong
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Comments (45)
I also reject the claim that C and B are equivalent (unless the utility of survival is 0, +infinity, or -infinity). If I accepted their line of argument, then I would also have to answer the following set of questions with a single answer.
Question E: Given that you're playing Russian Roulette with a full 100-shooter, how much would you pay to remove all 100 of the bullets?
Question F: Given that you're playing Russian Roulette with a full 1-shooter, how much would you pay to remove the bullet?
Question G: With 99% certainty, you will be executed. With 1% certainty you will be forced to play Russian Roulette with a full 1-shooter. How much would you pay to remove the bullet?
Question H: Given that you're playing Russian Roulette with a full 100-shooter, how much would you pay to remove one of the bullets?
You reject the claim, but can you point out a flaw in their argument?
I claim that the answers to E, F, and G should indeed be the same, but H is not equivalent to them. This should be intuitive. Their line of argument does not claim H is equivalent to E/F/G - do the math out and you'll see.
Actually my revised opinion, as expressed in my reply to Tyrell_McAllister, is that the authors' analysis is correct given the highly unlikely set-up. In a more realistic scenario, I accept the equivalences A~B and C~D, but not B~C.
I really don't know what you have in mind here. Do you also claim that cases A, B, C are equivalent to each other but not to D?
Oops, sorry! I misread. My bad. I would agree that they are all equivalent.