No, as described, you have probability (1/2)^n of becoming copy #n, and #99 and the original share (1/2)^99.
The original is copied once—giving 50% #0 and 50% #1. Then #0 is copied again, giving 25% #0, and 25% #2. Then #0 is copied again, giving 12.5% #0, and 12.5% #3, and so forth.
This seems like a useful reductio ad absurdum of this means of calculating subjective expectation.
Yes, I see it now. The dead-end copies function as traps, since they stop your participation in the game. As long as you can consciously differentiate your state as a copy or original, this works.
No, as described, you have probability (1/2)^n of becoming copy #n, and #99 and the original share (1/2)^99.
The original is copied once—giving 50% #0 and 50% #1. Then #0 is copied again, giving 25% #0, and 25% #2. Then #0 is copied again, giving 12.5% #0, and 12.5% #3, and so forth.
This seems like a useful reductio ad absurdum of this means of calculating subjective expectation.
Hmmm.
Yes, I see it now. The dead-end copies function as traps, since they stop your participation in the game. As long as you can consciously differentiate your state as a copy or original, this works.