Consider a Bernoulli experiment, such as flipping a not necessarily fair coin, which results in a positive outcome (heads) with probability $p$ and with a negative outcome (tails) with probability $(1-p)$. Now repeat this experiment $n$ times, and let the total number of positive outcomes (heads) be $k$.
Now, given $n$ and $k$ but not $p$ what is the probability that the Bernoulli experiment (flipping the coin) again will result in a positive outcome (head)?
If $n$ and $k$ were 0 this would be an easy task: as there is no further information and the outcomes could be defined the other way around, it is symmetric and thus $p=0.5$.
Otherwise it's a little more complicated (the dependency on n ($|n$) is omitted for clarity). The expected probability is: $\langle p |k\rangle = \int _0 ^1 p\cdot f(p|k) \,\mathrm{d} p$.
Using Bayes' theorem gives $f(p|k) = p(k|p) \frac{f(p)}{p(k)} $ and marginalization gives $p(k)=\int_0^1 p(k|p) \cdot f(p) \,\mathrm{d} p$. The Binomial distribution further tells us that $p(k|p)=\binom{n}{k}\cdot p^k \cdot (1-p)^{n-k}$. But what is the correct term for the prior of $p$, $f(p)$?
Jaynes & co. tell me to use a maximum Entropy prior. To maximize entropy of $p$ ($H(p)=-\int_0^1 f(p) \cdot ln(f(p)) \,\mathrm{d} p$), set $f(p)=1$. But this would not maximize the entropy of $k$; the entropy of $k$ would be maximal if $p$ was $0.5$ thus if $f(p)=0$ but infinity at 0.5 (integrating to 1).
So, what is the correct prior to choose for $f(p)$? Does it depend on $n$?