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.
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.