Python - Predicting the probability of 6 Heads or Tails in a row from a set sample size
Eliminating useless code
Enumerating a range is really pointless
In [6]: sample_size = 5

In [7]: for i, experimentNumber in enumerate(range(sample_size)):
   ...:     print(i, experimentNumber)
   ...:     
0 0
1 1
2 2
3 3
4 4

So we can easily replace one by the other. We do not even need to replace as experimentNumber is not used anywhere. Next we notice that i is also used only once where we can replace results[i-5:] by superior construct results[-6:].
We also eliminate the superfluous exception handling. So far this is already covered by @ppperys answer.
Efficiency
you create a complet list of length sample_size of random values in memory. This is not required and may be a problem on big sample sizes. As you always need the last 6 values only you could go for collections.deque which can maintain a maxlen.
from collections import deque
results = deque(maxlen=6)

For the evaluation made easy we do not use ('H', 'T') but numbers. We do not need to comare with a streak any more but do it arithmetically. Here is the only pitfall - we must check if the queue is filled completely to not accidentally accept a short sequence of zeros.
for _ in range(sample_size):
    results.append(random.choice((0, 1)))
    if len(results) == 6 and sum(results) in (0, 6):
        numberOfStreaks += 1

This not only saves memory but we also get rid of a temporary temp and the predifined head_streak and tail_streak. We notice the magic number 6 appearing multiple times - use a variable. We also make a testable function. We end up with
import random
from collections import deque

def streak_probability(streak_len, sample_size):
    results = deque(maxlen=streak_len)
    numberOfStreaks = 0
    for _ in range(sample_size):
        results.append(random.choice((0, 1)))
        if len(results) == streak_len and sum(results) in (0, streak_len):
            numberOfStreaks += 1
    return numberOfStreaks / sample_size


print('Chance of streak: %s%%' % (streak_probability(6, 1000000))

Algorithm
This simulation will give good results for big numbers of sample_size. However if the sample size was smaller than 6 it will always return 0. As you divide the final streak count by the sample size you indicate, that you would like to get the probability of a streak per &quot;additional&quot; coin toss. So we should fill the queue before starting to count. That way an average of a large number of runs with a small sample size would match a single run of a large sample size. If we prefill we do not have to check the fill state of the queue (yes I filled to the max while one less would be sufficient).
def prefilled_streak_probability(streak_len, sample_size):
    results = deque((random.choice((0, 1)) for _ in range(streak_len)), maxlen=streak_len)
    numberOfStreaks = 0
    for _ in range(sample_size):
        results.append(random.choice((0, 1)))
        if sum(results) in (0, streak_len):
            numberOfStreaks += 1
    return numberOfStreaks / sample_size

Now test the difference - we compare the original sample size of 1.000.000 to 100.000 repetitions of sample size 10
s=10
n=100000
print('no prefill')
print('Single big sample - Chance of streak: %s%%' % (streak_probability(6, s*n)))
probs = [streak_probability(6, s) for _ in range(n)]
print('Multiple small samples - Chance of streak: %s%%' % (sum(probs)/len(probs)))

print('with prefill')
print('Single big sample - Chance of streak: %s%%' % (prefilled_streak_probability(6, s*n)))
probs = [prefilled_streak_probability(6, s) for _ in range(n)]
print('Multiple small samples - Chance of streak: %s%%' % (sum(probs)/len(probs)))

we get
no prefill
Single big sample - Chance of streak: 0.031372%
Multiple small samples - Chance of streak: 0.01573599999999932%
with prefill
Single big sample - Chance of streak: 0.031093%
Multiple small samples - Chance of streak: 0.031131999999994574%