Case 1: expected value of playing the game with 4 bullets = -4/6($10+$5) = -$10, 3 bullets: -3/6($10+$5) = -$7.5, delta = $2.5
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So you should pay up to $2.5 to take the option in Case 1
This does not follow. Rather, you should be willing to pay an amount $X such that -3/6($10+$5) - 3/6 $X > -$10. That means that you are willing to pay as much as $5 in Case 1.
I hate to admit this, but you seem to be right. My mistake was that I did not allow for the fact that when you pay, the stake you are risking gets reduced by the amount you have paid.
This leads to bizarre follow-ons... for example, you should pay only half as much to remove the last bullet (1/6 --> 0/6) as you would to remove 1 bullet from 4 (4/6 --> 3/6)... and you'd pay 6 times as much to remove the first bullet (6/6 --> 5/6) as you would to remove the last bullet (1/6 --> 0/6). Which incidentally contradicts the original version of the parad...
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?