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Suppose we have these two sequences of 10 independent coin flips:

  • HTTHHTTHHT
  • HHHHHHHHHH

And suppose we want to test wether the coin is balanced or not (let $p=P(H)$).

Under $p=\frac{1}{2}$ both sequences have the same probability of $$ \bigg(\frac{1}{2}\bigg)^{10}. $$

I don't understand why, even if these two sequences have the same probability under $p=0.5$, the second one seems more against the hypothesis $p=0.5$ than the first one.

Aside from the Binomial likelihood where combinatorics comes into play to count the number of $H$, why will the these two equally likely sequences lead to different inference?

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  • $\begingroup$ Just to clarify, you don't think that getting 10 out of 10 heads is in some way evidence against the coin being fair? $\endgroup$ – jbowman Nov 11 '19 at 15:53
  • $\begingroup$ @jbowman I do! I just don't understand how, given that under $p=0.5$ they are both equally likely, the second one will be more against $p=0.5$ than the first one. $\endgroup$ – rosas Nov 11 '19 at 16:06
  • $\begingroup$ Which one is more likely under an alternative such as $p=0.9$? $\endgroup$ – jbowman Nov 11 '19 at 16:18
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Your observation is correct that under $p=0.5$ every observed sequence of heads and tails has the same probability. This means that the precise sequence of heads and tails is not useful as a test statistic for discriminating between $p=0.5$ and $p\neq 0.5$.

That's why the number of heads $k$ (or the proportion of heads $k/n$) is used as a test statistic. Its distribution under $p=0.5$ is known (binomial distribution with $p=0.5$) and you can compute the acceptance region $[k_1,k_2]=[np-\epsilon, np+\epsilon]$ for $p=0.5$ as $$P(|k-np|\leq\epsilon) \approx 1-\alpha$$ where $\alpha$ is the confidence significance level.

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    $\begingroup$ Good answer, +1. My complaint is calling $\alpha$ the confidence level. We can define anything to be anything we wish, but $\alpha$ has a pretty well established meaning in statistics of being the complement of the confidence level. $\endgroup$ – Dave Nov 11 '19 at 16:50
  • $\begingroup$ @dave Thanks for pointing this out. I have corrected the term "confidence" to "significance". $\endgroup$ – cdalitz Nov 11 '19 at 18:16

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