Automate the boring stuff - Chapter 4: Coin Flip Solution-CodePudding

I am struggling with the solution of the 'Coin Flip' practice project at the end of chapter 4 in 'Automate the boring stuff' for python programming.
I have two solutions, both yielding a totally different result (first one is clearly false). I am not sure, what is the right solution to the answer.

Solution 1:

import random
nextFlip = []
numberOfStreaks = 0

# Code that creates a list of 10000 'heads' or 'tails' values.

for expNum in range(10000):                
    select = random.randint(0,1)
    if select == 0:
        nextFlip.append('H')
    elif select == 1:
        nextFlip.append('T')
    
# Code that checks if there is a streak of 6 heads or tails in a row.

    for i in range(0,(len(nextFlip)-6)):
        if nextFlip[i] == nextFlip[i 1] == nextFlip[i 2] == nextFlip[i 3] == nextFlip[i 4] == nextFlip[i 5] != nextFlip[i 6]:
            numberOfStreaks  =1


print('Chance of streak: %s%%' % ((numberOfStreaks / 10000)*100))

Solution 2:

import random
nextFlip = []
hlist = 0
tlist = 0
numberOfStreaks = 0

# Code that creates a list of 10000 'heads' or 'tails' values.

for expNum in range(10000):                
    select = random.randint(0,1)
    if select == 0:
        nextFlip.append('H')
    elif select == 1:
        nextFlip.append('T')

    
# Code that checks if there is a streak of 6 heads or tails in a row.

    for i in range(0,(len(nextFlip)-6)):
        if nextFlip[i] == 'H':
            hlist  = 1
            if hlist == 6:
                numberOfStreaks  =1
            
        elif nextFlip[i] == 'T':
            tlist  = 1
            if tlist == 6:
                numberOfStreaks  =1

print('Chance of streak: %s%%' % ((numberOfStreaks / 10000)*100))

Maybe someone can help me and tell me what I did wrong.

CodePudding user response：

This seems to work:-

import random

N = 10_000
S = 6
HEAD = 'H'
TAIL = 'T'
T = [HEAD if random.randint(0, 1) else TAIL for _ in range(N)]
c = T[0]
s = 1
STREAKS = 0
for t in T[1:]:
    if t == c:
        s  = 1
        if s == S:
            STREAKS  = 1
            s = 0
    else:
        c = t
        s = 1
print(STREAKS)

CodePudding user response：

You could generate the random flips using a comprehension and store it in a string to make processing of the streaks easier. Since streaks can overlap, you need to examine subranges starting at every position:

flips   = "".join(random.choice("HT") for _ in range(1000))

streaks = sum(flips[i:i 6] in ('HHHHHH','TTTTTT') for i in range(1000))

The sum() function will convert boolean values to 1 or zero (True is 1), so adding up the result of the streak conditions produces the total.