> 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] <mailto:[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. Bruno > > 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] > <mailto:[email protected]>. > To view this discussion on the web visit > https://groups.google.com/d/msgid/everything-list/CAFxXSLSngViLiKWmM4BdrAdpSzmOzPq9zzu1zRpRATRFqCwMZA%40mail.gmail.com > > <https://groups.google.com/d/msgid/everything-list/CAFxXSLSngViLiKWmM4BdrAdpSzmOzPq9zzu1zRpRATRFqCwMZA%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 on the web visit https://groups.google.com/d/msgid/everything-list/ADCAB003-4E49-4B04-BD00-EC7D6E53CA23%40ulb.ac.be.
