Name for n-nomial distribution (binomial extended to 3+ possible random values)? I'm trying to work out the required number of samples to have an accurate average for a random series which can take a finite (6) number of values with known probability for each value.

I'm guessing the best mathematical tools would come from a generalisation of binomial distribution studies to a greater number of possible values than 2.

However looking on the web and wikipedia I cannot find the relevant ressources and I assume that I'm just not typing the right keywords to find my answers.

Some my questions:


if it exists how is a n-nomial distribution called?
if known, what is the formula to achieve a given accuracy in relation to number of samples (similar what we have for AWGN): $$meanerror \approx \frac{\sigma}{\sqrt{N}}$$
From $n$ trials and if you have $q(=6)$ variables you can calculate your expected sum by a simple sum of the probability of each outcome multiplied by its value.  For each outcome, define the probability of success $P(O_1)$ and the value of that outcome $O_1$.

Now your expected sum over $n$ trials is given by $$n\sum_{q=1}^6 O_q\times P(O_q)$$

And your expected mean is $$\sum_q O_q\times P(O_q)$$

As for the variance a good estimate is the sum of the individual variances, which are given by $O\times np(1-p)$ so your sum over all $O$ is:

$$\sum_qO_q\times P(O_q)(1-P(O_q))$$

Over sufficiently large number of trials this will approach a normal distribution so you may as well look up on Normal distribution tables - but it will depend on the individual values of $P(O_q)$ and $O_q$ so there's no general solution without plugging those values in and getting the particular mean and variance.
From $n$ trials and if you have $q(=6)$ variables you can calculate your expected sum by a simple sum of the probability of each outcome multiplied by its value.  For each outcome, define the probability of success $P(O_1)$ and the value of that outcome $O_1$.

Now your expected sum over $n$ trials is given by $$n\sum_{q=1}^6 O_q\times P(O_q)$$

And your expected mean is $$\sum_q O_q\times P(O_q)$$

As for the variance a good estimate is the sum of the individual variances, which are given by $O\times np(1-p)$ so your sum over all $O$ is:

$$\sum_qO_q\times P(O_q)(1-P(O_q))$$

Over sufficiently large number of trials this will approach a normal distribution so you may as well look up on Normal distribution tables - but it will depend on the individual values of $P(O_q)$ and $O_q$ so there's no general solution without plugging those values in and getting the particular mean and variance.