On 8/27/2025 9:49 PM, Bruce Kellett wrote:
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.
I said exactly that above: "In MWI it is 1/64=0.0156." But the
distribution of results in our single world, per the Born rule is binomial.
Brent
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/37ff01d1-b191-4998-be2b-c36a8fb8312f%40gmail.com.
