On Mon, Feb 17, 2020 at 4:13 PM 'Brent Meeker' via Everything List <
[email protected]> wrote:
> On 2/16/2020 2:17 PM, Bruce Kellett wrote:
>
> No,that argument is mistaken, as Kent's general argument in terms of the
> binomial expansion shows. All 2^N persons will use the frequency operator
> to conclude that their probabilities are the correct ones. Some will be
> seriously wrong,
>
> But almost all will intersubjectively agree that p is near 0.5. Science
> theories are based on intersubjective agreement...not personal exepriences.
>
This is a point that has come up several times in this discussion of Adrian
Kent's argument that all observers in an Everett world will think that
their observed probabilities are the correct ones, and that other observers
will agree. Brent has argued that all will agree that p is near 0.5.
This argument has worried me, so I thought that some serious calculations
were in order. If you have an even number of trials, N = 2M, the the number
of binary strings that have equal numbers of ones and zeros (both equal to
M) is given, from the binomial distribution, as N!/M!*M!. Using the
Stirling expansion for the factorial, as N gets large,
N! ~ sqrt(N) N^N.
So (2M)!/M!^2 ~ 1/sqrt(N).
In other words, the proportion of trials with equal numbers of zeros and
ones decreases as 1/sqrt(N) as N becomes large. This means that it is
simply not true that the majority of observers will find a probability at
or near p = 0.5. In fact, the proportion who find any specific probability
decreases as N increases. This makes sense, since the number of binary
strings increases exponentially, as 2^N, but the number of instances with
any particular proportion of zeros and ones increases only linearly with N.
Bruce
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