# Question about simple inference on Bernoulli trials

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?

• Just to clarify, you don't think that getting 10 out of 10 heads is in some way evidence against the coin being fair? – jbowman Nov 11 '19 at 15:53
• @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. – rosas Nov 11 '19 at 16:06
• Which one is more likely under an alternative such as $p=0.9$? – jbowman Nov 11 '19 at 16:18

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.
• 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. – Dave Nov 11 '19 at 16:50