Thanks for the clarification, that helps a lot. I can now say safely that the problem you're interested in (without natural selection) is definitely a Markov chain (or actually a set of Markov chains). So for this problem, I highly recommend reading up on those, there is a lot of knowledge about them that you can find and will help you.
So if I understand correctly, in the problem without natural selection, every generation each letter (I'm going to stick with letters for now) has a probability to change. Each change (if it occurs) draws a new value from a set of letters of size a completely randomly. From my point of view, this means that the probability distributions of each of the letters are completely independent of each other: all you need to know is the probability P(v) that one of the letters has the right value at a certain generation v. Then you just raise that probability to the power n of letters that you have. Do you agree on that?
This probability P(v) is straightforward to calculate from the theory of Markov chains. I'll try to do the translation as best as I can. Suppose we have as our letters A T G and C and we are going to follow what happens to a particular nucleotide as it mutates. At any given generation v, we will denote the probabilities that the nucleotide is a particular flavor by a vector:
{PA(v), PT(v), PC(v), PG(v)},
where:
PA(v) + PT(v) + PC(v) + PG(v) == 1
(because it's definitely one of the four). So e.g., PA(v) is the probability that the nucleotide has value A at generation v. The vector
{PA(v), PT(v), PC(v), PG(v)}
is called the state vector of the Markov chain at step v. Next, we want to do something with this vector: we want to advance it one step of time. This we do with a matrix that contains the transition probabilities to go from one state to another. For our 4-letter alphabet, this matrix is:
transitionMatrix[m_] =
{
{1 - m, m/3, m/3, m/3},
{m/3, 1 - m, m/3, m/3},
{m/3, m/3, 1 - m, m/3},
{m/3, m/3, m/3, 1 - m}
}
Take the first row of this matrix: these are the transition probabilities if the nucleotide is an A right now. The 1 - m entry (which is on the diagonal of the matrix) is the probability that the nucleotide doesn't change: if it was an A, it stays a A. The off-diagonal m/3 entries are the probabilities that it changes into T (2nd entry in the row: m/3), C (3rd entry in the row: also m/3) or G (4th entry in the row: again m/3). The remaining rows are the transition probabilities for when the nucleotide is currently T, C or G respectively. Note that every row sums to 1.
Now here comes the crucial bit: to get the state
{PA(v+1), PT(v+1), PC(v+1), PG(v+1)}
for the next generation, you perform a left-side matrix multiplication:
{PA(v+1), PT(v+1), PC(v+1), PG(v+1)} =
{PA(v+1), PT(v+1), PC(v+1), PG(v+1)} . transitionMatrix[m]
Judging from what you wrote down, you already understand the basic idea here, though make sure you really understand this step.
Bear with me for a bit longer, we're nearly there. You assumed the letters at the first generation (v=1) are random, so
{PA(1), PT(1), PC(1), PG(1)} = {1/4, 1/4, 1/4, 1/4}
Note that this is a left-eigenvector of transitionMatrix[m] (with eigenvalue 1):
{1/4, 1/4, 1/4, 1/4}.transitionMatrix[m] == {1/4, 1/4, 1/4, 1/4 }
This means that the probability vector doesn't change: at every step, the probabilities for A, T, C and G remain 1/4, no matter the value of m!
If you now want the state vector at a given v for any given starting state, you use DiscreteMarkovProcess in Mathematica. For example, try:
FullSimplify[
PDF[
DiscreteMarkovProcess[{1,0,0,0}, transitionMatrix[m]][v],
k]
]
Here, {1,0,0,0} is the starting state (the nucleotide starts as A with probability 1). v is the generation again, and k is the index of the letters: k = 1 represents A, k = 2 represents T, k = 3 represents C, and k = 4 represents G. If you evaluate this code, you'll see that you get a probability distribution that exponentially decays to the equilibrium state vector {1/4, 1/4, 1/4, 1/4} with increasing v.
So there you have it. I think this should answer your problem, though if I misinterpreted part of the question you should let me know. At the very least this should put you on the right track, I hope.