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This is compatible with my ##χ^2## Test (posts #2 and #9). However, it seems to lead to the wrong answer.Number Nine said:Another way of doing #2

If the sequence was generated by a coin toss, then it is ##Binomial(199,0.5)##. Since each flip is independent, the probability that flip ##n+1## is identical to flip ##n## is ##0.5##, so the number of repetitions is ##Binomial(198,0.5)##.

Let ##k## be the number of repetitions. We want to estimate ##p = 0.5##.

The first sequence contains 117 repetitions, which gives (under a uniform prior) a posterior distribution of ##Beta(1 + 117, 198 - 117 + 1)##, which is greater than ##.5## with probability ##>0.99##.

The second contains 96 repetitions, which gives (under a uniform prior) a posterior distribution of ##Beta(1 + 96, 198 - 96 + 1)##, which is compatible with ##p = 0.5##.