Interesting probability rule for predicting the outcome of a trial I remember a friend of mine claiming something a bit funny. It went along the lines of:
Given that one has observed $n$ trials of an experiment and $s$ of them have been successes (as opposed to failures) and given that we have no further knowledge about the process at hand, the optimal formula for predicting the outcome of the next trial is: $p(s|n)=(s+1/n+2)$
Now, he couldn't show where he got all this from, but he claimed that the statement can be proven from reasonable axioms and that is has merit. It must be that $(s+1/n+2)$ is some kind of a median value in the distribution of the possible probability mass functions regarding the experiment.
But why isn't the formula/median simply $p(s|n)=(s/n)$ ? For enough many trials and successes, there is no significant difference between the two, but for smaller values, there is. What am I missing? Also, where is this rule from and is it even correct? Perhaps someone here has the alleged proof from the "reasonable axioms"? 
Please keep the answer understandable for my feeble mind. 
Thanks :)
 A: Suppose we conduct $n$ independent, identically-distributed Bernoulli trials where, prior to commencing our experiments, we choose the probability of success on each trial (which we label $\rho$) uniformly from the interval $(0,1)$. Then,
$$
\begin{eqnarray*}
&\mathbf{P}\left((n+1)^{st}\mbox{ trial is a succcess}\,\bigg|\,\,s\mbox{ successes in }n\mbox{ trials}\right)&\newline
&&\newline
{}={}&\dfrac{
\mathbf{P}\left((n+1)^{st}\mbox{ trial is a succcess}\,\wedge\,s\mbox{ successes in }n\mbox{ trials}\right)}{
\mathbf{P}\left(s\mbox{ successes in }n\mbox{ trials}\right)}&\newline
&&\newline
{}={} &\dfrac{\displaystyle\int\limits_{0}^{1}{n\choose s}\rho^{s+1}\left(1-\rho\right)^{n-s}\,\mathrm d\rho}{\displaystyle\int\limits_{0}^{1}{n\choose s}\rho^{s}\left(1-\rho\right)^{n-s}\,\mathrm d\rho}{}={}\dfrac{\displaystyle\int\limits_{0}^{1}\rho^{s+1}\left(1-\rho\right)^{n-s}\,\mathrm d\rho}{\displaystyle\int\limits_{0}^{1}\rho^{s}\left(1-\rho\right)^{n-s}\,\mathrm d\rho}&\newline
&&\newline
{}={}&\dfrac{(s+1)!\,(n-s)!}{(n+2)!}\dfrac{(n+1)!}{s!\,(n-s)!}&\newline
&&\newline
{}={}&\dfrac{s+1}{n+2}\,.
\end{eqnarray*}
$$
