On 3/5/2020 2:45 AM, Bruce Kellett wrote:
Now sequences with small departures from equal numbers will still give probabilities within the confidence interval of p = 0.5. But this confidence interval also shrinks as 1/sqrt(N) as N increases, so these additional sequences do not contribute a growing number of cases giving p ~ 0.5 as N increases. So, again within factors of order unity, the proportion of sequences consistent with p = 0.5 decreases without limit as N increases. So it is not the case that a very large proportion of the binary strings will report p = 0.5. The proportion lying outside the confidence interval of p = 0.5 is not vanishingly small -- it grows with N.
I agree with you argument about unequal probabilities, in which all the binomial sequences occur anyway leading to inference of p=0.5. But in the above paragraph you are wrong about the how the probability density function of the observed value changes as N->oo. For any given interval around the true value, p=0.5, the fraction of observed values within that interval increases as N->oo. For example in N=100 trials, the proportion of observers who calculate an estimate of p in the interval (0.45 0.55) is 0.68. For N=500 it's 0.975. For N=1000 it's 0.998.
Confidence intervals are constructed to include the true value with some fixed probability. But that interval becomes narrower as 1/sqrt(N). So the proportion lying inside and outside the interval is relatively constant, but the interval gets narrower.
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