What is this notion in statistics? Question and request
I'm sure the following idea is known, and must be basic, in statistics. I suspect it's related to the confidence interval (CI). However, CI is a statistical notion notorious of frequently being misinterpreted, so I cannot be sure. I hope you can point out what the following idea is, and also point out a rigorous account which supports your answer. The more rigorous the better.
The idea
Given a coin whose output set is $\{+,-\}$. Assume the distribution of the outcome is fixed. Assume moreover that after tossing it $10000$ times, exactly $7000$ times were $+$ and exactly $3000$ times were $-$.
Of course, we cannot assert that the underlying distribution is 70%:30%. However, given any possible distribution $p:(1-p)$ in terms of $p \in [0,1]$, we can calculated the probability, denoted by $\phi(p)$, of that "Exactly 7000 positive $+$ shows up in an 10000-time toss trial."
Define the subset $S:= \{p\in[0,1] \,|\, \phi(p) \geq 95\% \}$. What is this subset called?
 A: It is known as the 'empty set'.
Joking aside: the probability of getting EXACTLY 7000 out of 10000 is extremely small, even in the best case where the each coin independently has $p = .7$ chance of landing on $+$. Your function $\phi$ never exceeds 0.01 (in fact doesn't even get that high) so there are no values in its domain where it reaches $.95%$.
Related: suppose you want to test the hypothesis that the coin is fair (so that $p = .5$). Then of course tossing it 10000 times is a great way to settle this matter. However: the fact that $\phi(.5)$ is very small does not normally count as evidence against the Null hypothesis that $p = .5$. What would count as evidence is that the probability $\psi(p)$ of finding $70000$ or more $+$'s in 10000 tosses is still small for $p = .5$.
The set $S$ but with $\psi$ in the role of $\phi$ would be a (one-sided) confidence interval for the unknown true value of $p$ (though not the only possible choice of confidence interval!) after observing the data - the claim that we can reject our Null-hypothesis of $p$ being $0.5$ is equivalent to $0.5$ not lying in that interval.
By contrast: the fact that 0.5 does not lie in your interval $S$ proves nothing about the fairness of the coin since NO value of $p$ lies in your $S$.
I hope this is helpful. If you tell us some more about what intended use you had for $S$ I could perhaps give a more useful answer.
EDITED IN LATER: while your interval $S$ does not have its own name, it might be useful to note that your function $\phi$ does. It is called the likelihood of $p$. If you draw the graph of $\phi$ you see it reaches its highest value exactly where you expect it: at $p = 0.7$. Intuitively it is clear that the best estimate for $p$ you can give after seeing this data is indeed $\hat{p} = 0.7$. Now the fact that your function $\phi$ takes its largest value there helps you to justify this choice of estimator by using a pre-existing name for it: the maximum likelihood estimator.
Maximum likelihood estimators are widely used as estimators in statistics, also outside the context of coin tosses. They even have their own abbreviation: MLE.
You are looking for ways, it seem, to express how flat vs peaked the graph of $\phi$ is around its maximum. Confidence intervals do that, but not quite in the way you propose. I will think a bit about measures that are closer to your proposal. The conventional wisdom is of course that larger sample size (more coin tosses) give you more reliable estimates, so narrower intervals. This is true for confidence intervals. But in your case this might be tricky. It seems that picking the right value that may play the role of the .95 might itself depend on the sample size, making everything a bit messy. I will think a bit more about it.
