Yes. Suppose you had a gun with one trillion chambers. 1 is empty, 2 is empty, 3 contains a "negotiable" bullet, and 4 through one trillion all have non-negotiable bullets. How much would you pay to remove the bullet from 3?
Your first reaction is that forget it, it doesn't matter, you're quite certainly going to die anyway. Your second reaction is eh, what the heck, spend all your money on it, you're going to die anyway and you can't take it with you.
The only thing preventing you from spending all your money on bullet 3 is the infinitesimal chance that the gun will land on chamber 1 or 2, and then you'll regret having to live the rest of your life in poverty when you would have survived in any case. So although you don't really care because you're overwhelmingly likely to die either way, you balance your tiny chance of landing on bullet 3 and regretting you didn't buy it off against your tiny chance of landing on bullets 1 or 2 and regretting that you did.
This calculation remains the same whether we decrease the non-negotiable chambers to 0 or increase them to 3^^^3.
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?