On Tue, Aug 26, 2025 at 6:15 PM Quentin Anciaux <[email protected]> wrote:
> Bruce, > > You’re conflating two different levels: the combinatorial existence of 2^N > sequences and their statistical weight. Yes, the Schrodinger equation > produces all 2^N sequences regardless of amplitudes, but the amplitudes > control the measure over those sequences. Changing |a|² and |b|² doesn’t > alter which sequences exist, it alters how observers are distributed among > them. That’s exactly what the Born rule captures: almost all self-locating > observers end up in high-measure branches. Treating all sequences as > equally weighted is equivalent to replacing quantum mechanics with flat > branch counting, which is not what the formalism prescribes. > It seems like you are using the Born rule to assign your weights, or measure, to the sequences. This is illegitimate, because the the Born rule is exactly what is in question in my analysis. I have shown that the Born rule is disconfirmed by the data obtained in the vast majority of sequences. The test of the Born rule is that the proportion of successes in the sequence approximates the Born probability of that success. In the binary case under discussion, the majority of sequences will have approximately equal numbers of zeros and ones as the number of trials, N, becomes large. The Born probability of obtaining a zero for these 50/50 sequences can be any value between zero and one, depending on the original amplitudes. Your model takes no account of this. Bruce -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion visit https://groups.google.com/d/msgid/everything-list/CAFxXSLRrtgGGPefJbTZMy19hfokA1u7Hpa49cOt_-ELfy%3Dy8iA%40mail.gmail.com.
