A frequentist confidence interval. A commonly used frequentist 95% confidence interval for
$P(\mathrm{Heads}) = p$ due to Agresti and Cooil is
of the following form:
The point estimate is $\tilde p = (x+2)/(n+4) = (80+2)/(100+4) = 0.7885.$
Then,
the CI is $\tilde p \pm 1.96{\sqrt{\tilde p(1-\tilde p)/(n+4)}}.$
which computes to $(0.71, 0.87).$
[Agresti's idea of artificially adding two Head and two Tails to the data has some theoretical basis, and has been shown to improve the actual coverage of confidence intervals, especially when the number of trials is below a few hundred.]
A Bayesian probability interval. If you take a Bayesian approach, beginning with a prior
belief that you have no idea at all wither the coin is really fair, then you might use a prior distribution $p \sim \mathsf{Unif}(0,1) \equiv \mathsf{Beta}(1,1).$ [It is natural to use a beta distribution to model a probability because its support is $(0,1).]$
Then the binomial likelihood function corresponding to 80 heads in 100 would be proportional to $p^{80}(1-p)^{20}.$
Then the Bayesian posterior distribution would be proportional to
$p^{81}(1-p)^{21},$ which is the kernel (density without norming constant) of $\mathsf{Beta}(80,20).$ The 95% Bayesian posterior probability interval for $p$ is then obtained by cutting probability 0.25 from each tail of the posterior distribution: from R statistical software, we get $(0.72, 0.87),$ which is numerically much the same as the frequentist CI above.
qbeta(c(.025,.975), 80, 20)
[1] 0.7166324 0.8720153
According to a Bayesian interpretation of this posterior
interval, you might say you think the probability the next
coin will come up Heads with probability between 72% and 87$.
However, still using a Bayesian approach, suppose you had an
opportunity to examine the coin in advance. It looks like a genuine US quarter to you, with not signs of tampering.
Then you go into the discussion with a prior coin that the
coin is nearly biased. That prior opinion might correspond
to the prior distribution $\mathsf{Beta}(10,10)$ with density function proportional of $p^{9}(1-p)^{9},$ and you're
pretty sure it it will come up Heads between 29% and 71% of
the time. [In effect, your preliminary examination of the coin is equivalent to about 20 tosses.]
qbeta(c(.025,.975), 10,10)
[1] 0.2886432 0.7113568
Then, multiplying the prior density by the binomial likelihood of the observed results, you would have the Bayesian posterior
distribution $\mathsf{Beta}(10+80, 10+20),$ and your 95 % posterior probability interval would be $(.67, .82)$.
qbeta(c(.025,.975), 90, 30)
[1] 0.6691131 0.8229970
Note: A hard core probabalist, who thinks all coins are fair, all tosses are independent and doesn't put much faith in experimental data, might ascribe the 100 tosses to a run of unusual luck, and say the $101$st coin toss will be have
a 50-50 chance of being Heads.
