The game is texas holdem, you get dealt two cards out of a 52 card deck. Hence there are 1326 possible combinations of two cards you can get.
Let's say our opponent has oppened the pot with a raise 5 out 10 possible times (50%). Can we calculate the probability of this being his true raising frequenzy?
Edit: Sorry for not being clear. What i'm looking to do is determine how many hands you need on a opponent to have trustworthy statistics. If a opponent has raised first in 50%, we will assume it's hands in the top 50% of starting hands (equity wise). Can we make a 95% confidence interval of his raising frequency?
What you're asking might be better off at the statistics stackexchange site, but I'll answer with the best of my (probably slightly hazy) statistical knowledge.
You're asking for a interval with which you can estimate the opponents preflop raising percentage. While in your example it is remarkably high (50% of all hands would include some pretty rubbish hands), the incredibly small sample of 10 hands is almost certainly the reason for this.
Let p = proportion of hands the opponent raises preflop with. You don't know p, you only have an estimate, which we can call $q$.
$q$ has an expected value, of $q$ (this seems self explanatory and i'll explain why this needs to be said). $q$ is not the real proportion, it is only an estimate based on a random sample. Therefore if you had seen a different sample of 10 hands, your $q$ would be different. The standard deviation of your $q$ is given by:
$$ \sqrt{\frac{q(1-q)}{n}} $$
A 95% confidence interval is 1.96 standard deviations either side of the mean, so your confidence interval is $q$ plus or minus 1.96 times this standard deviation. I'll let you substitute your values in, but you'll find you'll need a bigger sample to get a good answer of the opponents aggression.