At the end of the day, what do you do with Bayesian Estimates? I have often heard that in certain instances, it can be more beneficial to use Bayesian based methods because they provide "a distribution of possible answers" (i.e. the posterior distribution) instead of a single answer (as done in the frequentist case). However, it seems that at the end of the day, the analyst is still required to transform this "distribution of possible answers" into a single answer.
For example : if a Bayesian model is used to estimate the posterior distribution of "mu", the analyst is still required to either take the MAP or the Expectation of this distribution to return a final answer.
Is this the main benefit of Bayesian models? If the priors are correctly specified, the credible intervals associated with the expectation of the posterior distribution (of the parameter of interest) are more reliable?

Suppose your prior distribution is that a coin may be biased so it has success probability $\theta$ near $1/3.$ Specifically, you consider that $\theta\sim\mathsf{Beta}(2,4),$ so that $E(\theta)=1/3,$ $P(\theta < 1/2) = 0.8125,$
$P(0.0527 <\theta < 0.716) = 0.96,$
and $\theta$ has density function
$f(\theta) = K\theta^{2-1}(1-\theta)^{4-1},$ where $K$ is the norming constant.
[Computations in R.]
pbeta(.5, 2,4)
[1] 0.8125
qbeta(c(.025,.975),2,4)
[1] 0.05274495 0.71641794

Then you are allowed to toss the coin $n = 30$ times, obtaining $x = 9$ heads. Thus, your likelihood function
is $g(x|\theta) \propto \theta^9(1-\theta)^{21},$ where the
symbol $\propto$ indicates that the norming constant
has been omitted.
Finally, by
Bayes' Theorem, the posterior distribution is
$$g(\theta|x) \propto f(\theta)\times g(x|\theta)\\
\propto \theta^{2-1}(1-\theta)^{4-1} \times
\theta^9(1-\theta)^{21}\\
\propto \theta^{11-1}(1-\theta)^{25-1}.$$
where we recognize the last line as the kernel (density without norming constant) of $\mathsf{Beta}(11,25).$
Thus the posterior mean of $\theta$ is $E(\theta|x)=11/36= 0.3056,$ slightly smaller
than the prior $E(\theta) = 0.333$ because you have information from your 30 tosses.
Also, a 95% posterior credible interval for $\theta$ is
$(0.169, 0.463).$
qbeta(c(.025,.975),11,25)
[1] 0.1685172 0.4630446

The posterior distribution has information from your initial hunch that the coin may be biased towards Tails and also the results of your thirty-toss experiment with the coin.
The interpretation of this Bayesian interval estimate
differs from the 'long run' interpretation of a frequentist
confidence interval. Because the prior and likelihood both
describe the particular coin at hand, we can say that
the interval $(0.169, 0.463)$ also applies to the coin at hand. In particular we are pretty sure the coin is not fair.
