import random

def flip_a_fair_coin(N):
    coin = ['H', 'T']
    
    record = []
    for i in range(N):
        record.append(random.choice(coin))
        
    # record is a list of H and T; turn it into a string by using string.join.
    record = "".join(record)
    return record

flip_a_fair_coin(5)

# count the number of times you see a "run" of a given pattern, e.g. HTTHHT, in a given pattern of coinflips
def count_runs(run, coinflips):
    return coinflips.count(run)
    
flips = flip_a_fair_coin(10000)
the_count = count_runs('HTHHTT', flips)

# expected:
p = 1.0/2**6   # length of the 'HTHHTT' string, above
expected_count = p * len(flips)
print 'expected', expected_count, '; got:', the_count, '; diff: ', abs(the_count - expected_count)

# CTB -- the main problem is this: 'count' does not count overlapping runs.
# so, for example:
print count_runs("HT", "HTHTHTHT")
# works fine, but:

print count_runs("HH", "HHHH")

# does not -- there should be THREE pairs of heads in there, but .count is only finding the one at position 0 and the one at position 2.
# So basically any runs that could contain overlaps don't work.

# There are a couple different ways to do this.  Perhaps the easiest is this:
def count_runs2(run, coinflips):
    the_count = 0
    for n in range(len(coinflips)):
        if coinflips[n:n + len(run)] == run:
            the_count += 1
    return the_count

print count_runs2("HH", "HHHH")


# One important point for everyone -- string.find doesn't work the way you think it does!
s = "HH"
if s.find("HH"):
    print 'Found it!'
else:
    print 'Didn\'t find it...'

# this is because string.find returns -1 when it *doesn't* find the string... and that evaluates to True in an if statement.

# REMEMBER: ALWAYS TEST YOUR CODE WITH SOMETHING SIMPLE TO SEE IF IT'S WORKING!