It seems I was solving an equivalent problem. In the formulation you are using, the weighted average should reflect the number of wakeups.
What this results means is that SB should expect with probabilty 1⁄3, that if she were shown the results of the coin toss, she would observe that the result was heads.
No, it shouldn’t—that’s the point. Why would you think it should?
Note that I am already taking observer-counting into account—among observers that actually exist in each coin-outcome-scenario. Hence the fact that P(heads) approaches 1⁄3 in the many-shot case.
There are always 2 coin flips, and the results are not known to SB. I can’t guess what you mean, but I think you need to reread Bostrom’s paper.
It seems I was solving an equivalent problem. In the formulation you are using, the weighted average should reflect the number of wakeups.
What this results means is that SB should expect with probabilty 1⁄3, that if she were shown the results of the coin toss, she would observe that the result was heads.
No, it shouldn’t—that’s the point. Why would you think it should?
Note that I am already taking observer-counting into account—among observers that actually exist in each coin-outcome-scenario. Hence the fact that P(heads) approaches 1⁄3 in the many-shot case.