I think this is the time when I would like to ACTUALLY understand what you are talking about...
I think this is important, but you lost me on nimimi: > N!/M!*M! Would appreciate any examples from personal-life-perspecitve too! > On 9 Mar 2020, at 22:08, Bruce Kellett <[email protected]> wrote: > > On Tue, Mar 10, 2020 at 8:54 AM Russell Standish <[email protected] > <mailto:[email protected]>> wrote: > On Sun, Mar 08, 2020 at 10:10:23PM +1100, Bruce Kellett wrote: > > > > > > In order to infer a probability of p = 0.5, your branch data > > must > > have > > > > approximately equal numbers of zeros and ones. The number of > > branches > > > with > > > > equal numbers of zeros and ones is given by the binomial > > coefficient. For > > > large > > > > even N = 2M trials, this coefficient is N!/M!*M!. Using the > > Stirling > > > > approximation to the factorial for large N, this goes as > > 2^N/sqrt > > (N) > > > (within > > > > factors of order one). Since there are 2^N sequences, the > > proportion with > > > n_0 = > > > > n_1 vanishes as 1/sqrt(N) for N large. > > > > > > > > This is the nub of the proof you wanted. > > No - it is simply irrelevant. The statement I made was about the > proportion of strings whose bit ratio lies within certain percentage > of the expected value. > > After all when making a measurement, you are are interested in the > value and its error bounds, eg 10mm +/- 0.1%, or 10mm +/- 0.01mm. We > can never know its exact value. > > > If you are using experimental data to estimate a quantity (and a p value is a > quantity in the required sense), then you are interested in the confidence > interval, not an absolute or percentage error. And the confidence interval > for a given probability of including the true value decreases with the number > of trials (since the standard error decreases with N). > > 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/CAFxXSLQZ6-y7W70ZDRDEqbcVY2agyrAB8SpovtgfiHMGQxDMqA%40mail.gmail.com > > <https://groups.google.com/d/msgid/everything-list/CAFxXSLQZ6-y7W70ZDRDEqbcVY2agyrAB8SpovtgfiHMGQxDMqA%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/76F1634F-6B3B-40C8-A460-FA6EC139E599%40gmail.com.
