Calculating the joint probability of n events from a sample sequence of occurrences
I'm just going to look at cond_probs here. You'll see that there's plenty in this function for one answer.

1. Comments on your code


The line return probs has the wrong indentation. Copy/paste error?
There are no docstrings. What do these functions do and how do I call them? (The text from your question would be a good starting-point for docstrings.)
There are no test cases. This is the kind of code that would benefit from some unit tests and/or doctests.
You write, "cond_probs() produces a dictionary {'AT': n1, 'AG': n2, 'AA': n3, 'AC': n4, ... , 'TC': n16}" but when I run it the dictionary has tuples for keys, not strings:

&gt;&gt;&gt; cond_probs(['AAC'])
{('A', 'C'): 0.5, ('A', 'A'): 0.5}

I don't understand the behaviour when reverse=True. You write, "In reverse, the transitional probabilities are calculated with the sequences in reverse" but this doesn't seem to be the case:

&gt;&gt;&gt; cond_probs(['CAA'], reverse=True)
{('A', 'A'): 1.0}


What has happened to C? This looks wrong to me.

If I were writing this, I'd implement cond_probs to run only in the forward direction; to do the reverse direction I would pass in reversed sequences.
Prefer using True and False for Booleans, not 1 and 0. So the second parameter to cond_probs should be reverse=False.
Instead of:

counts = {}
# ...
counts[X] = counts.get(X, 0) + 1


use collections.Counter and write:

counts = Counter()
# ...
counts[X] += 1


(But see below for how to use collections.Counter.update.)
Instead of:

range(len(seq))[::-1][:-1]


write:

range(len(seq) - 1, 0, -1)


(But see below for how to avoid this.)
Prefer iteration over elements of sequences instead of iteration over their indexes. So instead of:

for i in range(len(seq) - 1):
    counts[seq[i],seq[i + 1]] = counts.get((seq[i], seq[i + 1]), 0) + 1


use zip:

for t in zip(seq, seq[1:]):
    counts[t] += 1


or, even more simply, using collections.Counter.update:

counts.update(zip(seq, seq[1:]))


(If you are concerned about memory use, then you could use  itertools.islice and write islice(seq, 1, None) instead of seq[1:] to avoid the copy.)
In this code:

for s in states:
    sCounts = dict((k,v) for (k,v) in counts.items() if k[0] == s)


you iterate over the whole of counts for every state, collecting the ones you want. It would be more efficient to iterate only over the transitions from s. See below for how this would look.


2. Revised code

def cond_probs_2(sequences):
    """Calculate 1-step transitional probabilities from sequences.
    Return dictionary mapping transitions to probabilities.

    &gt;&gt;&gt; sequences = ['ACABC', 'ABCAB']
    &gt;&gt;&gt; sorted(cond_probs_2(sequences).items())
    ... # doctest: +NORMALIZE_WHITESPACE
    [(('A', 'B'), 0.75),
     (('A', 'C'), 0.25),
     (('B', 'C'), 1.0),
     (('C', 'A'), 1.0)]

    """
    counts = Counter()
    entries = set()
    for seq in sequences:
        entries.update(seq)
        for a, b in zip(seq, islice(seq, 1, None)):
            counts[a, b] += 1
    probs = {}
    for a in entries:
        suma = float(sum(counts[a, b] for b in entries))
        if suma != 0:
            probs.update(((a, b), counts[a, b] / suma) for b in entries
                         if counts[a, b])
    return probs


This is about 30% faster than your code:

&gt;&gt;&gt; from timeit import timeit
&gt;&gt;&gt; from random import choice
&gt;&gt;&gt; data = ''.join(choice('ACGT') for _ in range(1000000))
&gt;&gt;&gt; cond_probs([data]) == cond_probs_2([data])
True
&gt;&gt;&gt; timeit(lambda:cond_probs([data]), number=1) # yours
1.4706195190083236
&gt;&gt;&gt; timeit(lambda:cond_probs_2([data]), number=1) # mine
0.9721288200234994


3. Rewriting in NumPy

Here's how I'd go about this kind of task in NumPy. NumPy works best with fixed-size numeric data, so I'd encode the input as small integers:

&gt;&gt;&gt; import numpy as np
&gt;&gt;&gt; data = 'TTAGCAGTCAGTTCAGCGTTCGACTGA'
&gt;&gt;&gt; distinct = set(data)
&gt;&gt;&gt; coding = {j:i for i, j in enumerate(distinct)}
&gt;&gt;&gt; coding
{'A': 0, 'C': 1, 'T': 2, 'G': 3}
&gt;&gt;&gt; coded_data = np.fromiter((coding[i] for i in data), dtype=np.uint8)
&gt;&gt;&gt; coded_data
array([2, 2, 0, 3, 1, 0, 3, 2, 1, 0, 3, 2, 2, 1, 0, 3, 1, 3, 2, 2, 1, 3, 0,
       1, 2, 3, 0], dtype=uint8)


(I've chosen the type np.uint8 here because there are only four different characters in the input strings. You might need to choose a different type.)

Then you can encode adjacent pairs of numbers from 0–3 as a single number from 0–15, count the number of occurrences of each pair using numpy.bincount, and decode the result using numpy.reshape:

&gt;&gt;&gt; n = len(distinct)
&gt;&gt;&gt; pairs = coded_data[:-1] + n * coded_data[1:]
&gt;&gt;&gt; counts = np.bincount(pairs, minlength=n*n).reshape(n, n)
&gt;&gt;&gt; counts
array([[0, 3, 1, 2],
       [1, 0, 3, 2],
       [0, 1, 3, 3],
       [4, 2, 1, 0]])


This array represents the transition table:

          source 
       │ A  C  T  G
     ──┼───────────
     A │ 0, 3, 1, 2
dest C │ 1, 0, 3, 2
     T │ 0, 1, 3, 3
     G │ 4, 2, 1, 0


You can then compute the transition probabilities by dividing each element by the sum of its column:

&gt;&gt;&gt; counts / counts.sum(axis=0)
array([[ 0.        ,  0.5       ,  0.125     ,  0.28571429],
       [ 0.2       ,  0.        ,  0.375     ,  0.28571429],
       [ 0.        ,  0.16666667,  0.375     ,  0.42857143],
       [ 0.8       ,  0.33333333,  0.125     ,  0.        ]])


Here's the above as a function:

from itertools import chain
import numpy as np

def cond_probs_np(sequences):
    """Calculate 1-step transitional probabilities from sequences.
    Return dictionary mapping transitions to probabilities.

    &gt;&gt;&gt; sequences = ['ACABC', 'ABCAB']
    &gt;&gt;&gt; sorted(cond_probs_np(sequences).items())
    ... # doctest: +NORMALIZE_WHITESPACE
    [(('A', 'B'), 0.75),
     (('A', 'C'), 0.25),
     (('B', 'C'), 1.0),
     (('C', 'A'), 1.0)]

    """
    distinct = set(chain.from_iterable(sequences))
    n = len(distinct)
    assert(n &lt; 256)             # so that they will fit in an np.uint8
    coding = {j:i for i, j in enumerate(distinct)}
    counts = np.zeros((n, n))
    for seq in sequences:
        coded_seq = np.fromiter((coding[i] for i in seq), dtype=np.uint8)
        pairs = coded_seq[:-1] + n * coded_seq[1:]
        counts += np.bincount(pairs, minlength=n*n).reshape(n, n)
    totals = counts.sum(axis=0)
    totals[totals == 0] = 1     # avoid division by zero
    probs = counts / totals
    return {(a, b): p for a, b in product(distinct, repeat=2) 
            for p in (probs[coding[b], coding[a]],) if p}


And this is about seven times faster than your original code:

&gt;&gt;&gt; cond_probs([data]) == cond_probs_np([data])
True
&gt;&gt;&gt; timeit(lambda:cond_probs_np([data]), number=1)
0.2206433709943667