Calculate Probability Distribution I would like to calculate an expected distribution curve telling me the probabilities of allele frequency calls for DNA sequencing data. I think the problem is easiest to understand with coin flips so i'll explain that way.
So I have $10^6$ coins each with heads and tails. With infinite flips of each coin the ratio of heads to tails would be $0.5$.
I am flipping each coin $\sim50\times$.
I would like to calculate a curve predicting the outcomes of the coinflips such that I can ask things like 95% of the coins will have a ratio less than $0.6$.
Hopefully that makes sense.
 A: The number of "heads" you get from one flip is $0$ or $1$, each with probability $1/2$.  That distribution has expected value $1/2$ and standard deviation $1/2$.
With $1$ million independent tosses, the expected value is $1$ million times that, and the variance is $1$ million times the variance with one toss, which is $(1/2)^2$.
(That is why Abraham de Moivre in the $18$the century introduced standard deviation, rather than the more obvious mean absolute deviation, as a measure of dispersion: the variances of independent observations can be simply added up like that.)
Thus with $1$ million tosses, the expected number of "heads" is  $5{,}000{,}000$ and the variance is $250{,}000$; hence the standard deviation is $500$.
You can approximate the probability distribution of the number of "heads" is $1$ million tosses with the normal distribution with expected value $5{,}000{,}000$ and standard deviation $500$.
For example, the probability that you get more than $5{,}000{,}500$ "heads" is the probability that a normally distributed random variable is more than $1$ standard deviation above the mean, i.e. about $0.16$.
(Since $>5{,}000{,}500$ is the same as $\ge5{,}000{,}501$, it is usual in this sort of thing to find the probability of exceeding $5{,}000{,}500.5$. That is a "continuity correction", used because you're using a continuous approximation to a discrete distribution.  But when the number of trials is so big, that makes little difference.  When the number of trials is small, use a continuity correction.)
A: So the problem is
1) predict whether $50$  fair coin tosses yield at least $60\%$ of tails. In other words, you get a random variable $T$ following binomial distribution $B_{50,1/2}$ and want the probability that this random variable is greater than $0.6\times 50 =30$, i.e. $ p=P[T\ge 50]$. The explicit formula can be too difficult to use here (even though CAS software is perfectly capable of doing that), but you can use the central limit theorem to get an approximation (or, if you want a simplier estimation, Chebyshev inequality).
2) Now having established the model for one coin, you can pass to the whole collection. Each  coin will have two states:  "ratio above 0.6" with probability $p$ and "ratio strictly less than $0.6$" with probability $1-p$. Again, the fact that you have $95\%$ of coins in the state "above", means, from probability point of view, that a random variable following the binomial distribution $B_{10^6, p}$ takes values greater than $0.95*10^6$. This probability, as previously, can either be found with computer algebra system software or approximated with the help of central limit theorem (or Chebyshev inequality).
If you need clarifications, ask in comments.
On a side note, my intuation says that the probability you want to find will be vanishingly small.
