Today I came a cross an old math thing I did some years ago. I remembered that I didn't find a similar result from anywhere at that time but forgot the thing altogether. So now I would like to ask if the following is a valid result and if it is unknown one? - Sorry if this not the right place to ask..

When working with a small enough samples of a large enough populations one should use the following formula for estimating probability $P(A)$ for the whole population instead of sample mean:

$\frac{n+1}{n+m+2}$, where $n$ is the number of experiments where $A$ is true and $m$ is the number of experiments where $A$ is false (sample mean would be $\frac{n}{n+m}$).

The "proof":

Lets assume that $P(A|B(n,m))$ is the probability for $A$ when we have observed $n$ number of positive experiments and $m$ number of negative experiments - $B(n,m)$ is the probability to observe those things in the sample.

$P(A|B(n,m)) = \frac{P(A \wedge B(n,m)}{P(B(n,m))}$

Now lets also assume that when we have no sample or evidence at all about $A$. In this case the probability is:

$P(A) = \int_{0}^{1}xdx$

and the probability to observe $n$ positive examples and $m$ negative examples is:

$P(B(n,m)) = \int_{0}^{1} { n+m \choose n } x^n (1-x)^m dx$

From this we can say:

$P(A|B(n,m)) = \frac{P(A \wedge B(n,m)}{P(B(n,m))}$

$= \frac{\int_{0}^{1} x { n+m \choose n } x^n (1-x)^m dx}{\int_{0}^{1} { n+m \choose n } x^n (1-x)^m dx}$

$= \frac{\int_{0}^{1} x^{n+1} (1-x)^m dx}{\int_{0}^{1} x^n (1-x)^m dx}$

$= \frac{\Gamma(n+2)\Gamma(m+1)}{\Gamma(m+n+3)} / \frac{\Gamma(n+1)\Gamma(m+1)}{\Gamma(m+n+2)}$

Because $n$ and $m$ are positive integers $\Gamma(k) \rightarrow (k-1)!$

$= \frac{(n+1)!m!}{(m+n+2)!} / \frac{n!m!}{(m+n+1)!}$

$= \frac{n+1}{m+n+2}$