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I am really struggling with the following question. I'll describe the situation.

If we do x coin tosses, then the chance that we have more heads than tails (H>T) is described by equation 1 below.

If we do x coin tosses, then the chance that we have an equal number of heads and tails (H=T) is described by equation 2 below.

Finally, if we do x coin tosses, then the chance that we have more tails than heads (H<T) is the same as for equation 1; shown in equation 3 below.

Now I have a situation where this entire situation is repeated y times, but with different x. For example, let's say y=7 and I know x for each y. For example, x∈{1,4,4,5,10,14,17}. For this example, if I see, respectively H>T, H=T, H>T, H>T, H>T, H>T, H<T, then how can I statistically test the hypothesis that in this situation H>T is more likely to occur than expected by chance?

I am hoping there might be a way to generalise this, because I have multiple different y's, so I am hoping for a general answer that I can apply to all my different y's.

I hope my explanation is clear. Thank you. enter image description here

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    $\begingroup$ Is the same coin thrown in each of the $y=7$ trials? $\endgroup$ Commented Aug 11, 2020 at 11:27
  • $\begingroup$ Thank you for asking. Yes, the same coin. $\endgroup$ Commented Aug 11, 2020 at 12:09

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Use the total number of trials, $x=\sum_{i=1}^y x_i$, and the total number H of heads and T of tails. Disregard the fact that the original data come from multiple series of trials.

Why is this the best you can do? Turn the situation around. Suppose you have only one long series of $x$ trials with associated H and T. If there were a way to gain more information out of multiple series of trials $(x_i)$, then you could use your single series and cut it up into chunks $(x_i)$ after the fact to gain additional information. But any such "posterior chunking" cannot generate information that is not already present in the initial single series.

(The answer would be different if your trials had some inherent grouping structure, e.g., such that each series were done by a different coin, which is why I asked.)

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    $\begingroup$ That is the most powerful test if the information is available. If not, and you only have the $H\gt = \lt T$ information, you might imagine ignoring $H=T$ results and seeing whether there are a significantly larger number of $H \gt T$ compared to $H < T$. In both cases you would end up with a binomial test, and yours would have a larger effective sample size $\endgroup$
    – Henry
    Commented Aug 11, 2020 at 13:33
  • $\begingroup$ @stephankolassa and Henry Thank you both for your answers. They are very insightful! I will go ahead as you suggested, ignoring H=T cases and using a binomial test. Many thanks! $\endgroup$ Commented Aug 11, 2020 at 21:30

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