string comparison in a while loop

Do both strings have the same lenght? otherwise you should consider using something like:

while count < min(len(seqA), len(seqB)):

Also, the zip function might come in handy here to pair off the letters in each word. It is a python builtin. e.g.

def letter_score(s1, s2):
    score = 0
    previous_match = False

    z = zip(s1, s2)
    for pair in z:
        if pair[0] == pair[1]:
            if previous_match:
                score += 3
            else:
                score += 1
                previous_match = True
        else:
            score -= 1
            previous_match = False
    return score

The error message says it: You're trying to access a character beyond the boundaries of your string. Consider this:

>>> s = "hello"
>>> len(s)
5
>>> s[4]
'o'

Now when (at the start of the first while loop) count is 1 below len(seqA), then you're incrementing count and then you're doing seqA[count] which will throw this exception.

Let's assume you're calling pairwisescore("a", "a"):

score = 0
count = 0
while count < len(seqA):                   # 0 < 1 --> OK 
    if seqA[count] == seqB[count]:         # "a" == "a" --> OK
        score = score + 1                  # score = 1
        count = count + 1                  # count = 1
        while seqA[count] == seqB[count]:  # seqA[1] doesn't exist!

In the second while loop you must test that count is less than len(seqA):

while count < len(seqA) and seqA[count] == seqB[count]:
   ...

and, also there's possibly other bug: If the length of seqB is less than length of seqA, you'll again see runtime exception. so, you should change every occurence of count < len(seqA) with count < min(len(seqA), len(seqB)).

def pairwiseScore(seqA, seqB):
    count = 0
    score = 0
    isOne = False
    isTwoOrMore = False
    while count < min(len(seqA), len(seqB)):
        if seqA[count] == seqB[count]:
            if isTwoOrMore:
                score = score + 3
                count = count + 1
            else:
                if isOne:
                    isTwoOrMore = True
                score = score + 1
                count = count + 1
                isOne = True
        elif seqA[count] != seqB[count]:
            score = score - 1
            count = count + 1
            isOne = False
            isTwoOrMore = False
    return score

a = 'apple'
b = 'aplle'
print(pairwiseScore(a, b))

I think this one solve the problem, I added a "counting" bool variable. And to respond to the question, your program didn't compare the length of the second string.

while count < min(len(seqA), len(seqB)):

The problem is that you do this:

    count = count + 1

Both before the inner while loop and at the end of it. But you then continue using seqA[count] before checking it against len(seqA) again - so, once it goes too high, Python will try to read past the end of seqA, and you'll get that error (when, if the condition had been checked again after incrementing, the loop would have ended).

Using a Python for loop will get around bugs like this, since Python will manage count for you:

for a,b in zip(seqA, seqB):
     if a == b:
        score += 1

You can implement the extra-points bit easily in this by keeping track of whether the previous character is a match, rather than trying to work out how many after this one are. A boolean variable last_matched that you keep updated would help with this.