In the former case you'd pay infinity (or all you have) either way. In the latter case you'd pay zero either way. I don't see how that contradicts Landsburg.
You're right and I'm being stupid, thanks. But what if you value both weasels (proportional to the probability of survival) and candy bars (proportional to remaining money in case of survival)? Then each bullet destroys a fixed number of weasels and no candy bars, so you should pay 2x more candy bars to remove two bullets instead of one, no?
Imagine you're playing Russian roulette. Case 1: a six-shooter contains four bullets, and you're asked how much you'll pay to remove one of them. Case 2: a six-shooter contains two bullets, and you're asked how much you'll pay to remove both of them. Steven Landsburg describes an argument by Richard Zeckhauser and Richard Jeffrey saying you should pay the same amount in both cases, provided that you don't have heirs and all your remaining money magically disappears when you die. What do you think?