Bruce, Your conclusion hinges on rejecting the binomial distribution in the Everettian context, but this misrepresents how the Born rule is tested experimentally. In actual experiments, observers find themselves in a single branch and estimate probabilities from relative frequencies within that branch. The binomial model is exactly what captures those frequencies. Everett does not magically change the combinatorics.
Your reasoning assumes equal weighting of all fine-grained sequences and uniform observer sampling across them. But that assumption is yours, not Everett's. In Everett's framework, the squared amplitudes define the measure over branches, so typical observers overwhelmingly find frequencies matching the Born rule even though all sequences exist. The lottery analogy makes this clearer: having 1,000,000 tickets but 400,000 duplicates of the same number does not make every ticket equally predictive of what a typical draw looks like. Existence ≠ equal weight. Dismissing amplitudes as irrelevant is exactly where your derivation breaks down. Quentin All those moments will be lost in time, like tears in rain. (Roy Batty/Rutger Hauer) Le jeu. 28 août 2025, 06:49, Bruce Kellett <[email protected]> a écrit : > On Thu, Aug 28, 2025 at 2:30 PM Brent Meeker <[email protected]> > wrote: > >> On 8/27/2025 5:07 PM, Bruce Kellett wrote: >> >> On Thu, Aug 28, 2025 at 5:16 AM Brent Meeker <[email protected]> >> wrote: >> >>> I think some specificity would help this debate. Suppose N=6, so there >>> are 64 different sequences in 64 different worlds. The number of observers >>> is irrelevants; we can suppose the results are recorded mechanically in >>> each world. Further suppose that a=b so there is no question of whether >>> amplitudes are being respected. Then in one of the worlds we have 011000. >>> Per the Born rule its probability is 0.2344. In MWI it is 1/64=0.0156. >>> The difference arises because the observers applying the Born rule looks at >>> it as an instance of 2 out of 6 successes. >>> >> >> The trouble with this is that you are treating this as an instance of >> Bernoulli trials with probability p= 0.5. When every outcome occurs with >> every trial we no longer have a Binomial distribution The Binomial >> distribution assumes that you have x successes out of N trials. In the >> Everettian case you have one success on every trial. >> >> So your probability above for 2 successes applies to Bernoulli trials >> with one as the 'success'. The thing is that the probability of getting a >> zero is also 1/2, so we also have four successes out of six trials in your >> example. The Binomial probability for this result is also 0.2344. Actually, >> if we regard this experiment as a test of the Born rule, we have four zeros >> in 6 trials, which gives an estimate of the probability as 4/6 = 0.667, or >> as two ones in 6 trials which gives an estimate of the probability as 2/6 = >> 0.333, neither estimate is equivalent to |a|^2 = 0.5. The difference >> becomes more pronounced as N increases. The problem with your analysis is >> that you are assuming a binomial distribution. and we do not have any such >> distribution. >> >> That's how an experimenter will compute the probability of 2 successes in >> 6 trials given p=0.5. Yes, the probability estimate is 0.33 given 2 in >> 6. The probability of the result given p is not generally the same as the >> estimate of p given the result. The former is binomial distributed. The >> estimates of p aren't part of a distribution since they don't add up to 1. >> > > You are still assuming that the results follow a binomial distribution. > This is not the case. As I said, in the Everett situation, when every > outcome is realized in every trial, you have one success in every trial, > and the resulting distribution is not binomial. The normal approximation > for large N, as Jesse seems to assume, simply does not hold, since the > distribution is not binomial. > > 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/CAFxXSLQ_N1_wYg7aQBMabdigBEnR4uRJuArSC0Kiv26brfzDOg%40mail.gmail.com > <https://groups.google.com/d/msgid/everything-list/CAFxXSLQ_N1_wYg7aQBMabdigBEnR4uRJuArSC0Kiv26brfzDOg%40mail.gmail.com?utm_medium=email&utm_source=footer> > . > -- 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/CAMW2kAqXiARGPGT5jrNo7qt55xyo0edvTtViGCfrbNGsc3zpRw%40mail.gmail.com.
