How do you measure how "typical" a set of events are? Given a set of events and their probability of occuring, where they all add to 1. Edit: the set of events can be of any size, the example below is only A and B with 50/50 probabilities for simplicity of example.
Such as A=0.5 and B=0.5.
Given a group of events, where order doesn't matter:
Group 1: AAABBB
Group 2: AAAAAABBBBBB
Group 3: AABBBB
Group 4: AAAABBBBBBBB
Event groups 1 and 2 are 50% A, and 50% B. When I measure the probability of getting Group 1 vs Group 2, Group 2 is less likely because getting exactly 50/50 in 12 trials is lower than in 6 trials, because the possible combinations that aren't 50/50 goes up. But both are exactly the expected theoretical distribution. How would I measure how typical a distribution while being independent of sample size? Where the output for G1 = G2, and output for G3 = G4. 
 A: Since you are looking at only two possible event outcomes, $A$ and $B$, we can model the event outcome as a Bernoulli random variable, $X$, with parameter $p$. This $p$ is the probability of a "success", which we can define as the probability of event $A$. The occurrence of event $B$, is then a failure and this happens with probability $1-p$. In your example, you set $p=0.5$, but if you'd like to generalize it can take any value between $0$ and $1$.
To look at groups, you need to know the probability of the outcome for a series of Bernoulli trials. This is given by a Binomial distribution. For $n$ trials, the probability of $k$ successes ($A$'s) is given by:
$$P(X=k) = {n\choose k} p^k (1-p)^{n-k}. $$
We can test this on your example groups.
Group 1 has probability 
$$ P(X=3) = {6\choose 3} (0.5)^3 (1-0.5)^{6-3} = 5/16$$
while Group 2 has probability
$$ P(X=6) = {12\choose 6} (0.5)^6 (1-0.5)^{12-6} = 231/1024$$
It seems that you are interested not in these count probabilities, but rather in the probabilities of seeing some proportion of event outcomes. There are several methods to find confidence intervals for these observed proportions as described here, but ultimately they will all depend on the sample size.
