On Thu, Feb 20, 2020 at 10:32 PM Bruno Marchal <[email protected]> wrote:
> On 20 Feb 2020, at 06:12, Bruce Kellett <[email protected]> wrote: > > 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. > > > > That makes not much sense. To say that the probability of having a girl or > a boy is roughy 1/2 (slightly different in China) does not mean that the > probability of having exactly 5 girls and 5 boys is one, in a row of ten > kids, as you argue correctly above. In fact that is (binomial 5 10) times > (1/2)^10 = 0,246…, and get lower and lower when repeating the trials. > > You just give the good argument for inferring P=1/2 in the (iterated) > self-duplication. It seems to me. > You have missed the point in a spectacular fashion. I calculated the number of times in the 2^N binary strings that there are equal numbers of zeros and ones -- not the probability that you will get equal numbers. If you multiply that probability by the number of strings, you get the number I calculated, which is N!/M!*M! (for N=2M). It is the fact that this number goes to zero as N goes to infinity that shows that equal probabilities do not dominate the complete set of binary strings. This complete the proof that assuming p = 1/2 in the iterated WM-duplication case is completely unjustified by the data. 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 on the web visit https://groups.google.com/d/msgid/everything-list/CAFxXSLRk_NJqse18RMdhdpRb%2B0nZWWuYDejfoHrwOJkgbaq24g%40mail.gmail.com.
