Automate the boring stuff - Chapter 4: Coin Flip Solution

I'm just learning with the same book too. I found this easier to treat as a string, rather than a list. I believe that the problem is asking you to find whether a streak of 6 or more consecutive heads or a streak of 6 or more consecutive tails occurs in 100 tosses of a coin. For each 100 tosses, the result is 0 or 1 (I don't think you count multiple streaks). The 10,000 runs is to gain a reasonable accuracy and avoid sampling error.

import random number_of_streaks = 0 for experiment_number in range(10000): # Code that creates a list of 100 'heads' or 'tails' values coin_list = '' for i in range(100): if random.randint(0, 1) == 0: coin_list = coin_list + 'H' # heads else: coin_list = coin_list + 'T' # tails # Code that checks if there is a streak of 6 heads or tails in a row if 'HHHHHH' in coin_list or 'TTTTTT' in coin_list: number_of_streaks +=1 print('Chance of streak: %s%%' % (number_of_streaks / 100)) 

Further to advice in comments from @rchome, I added a simple timer to this original code (https://stackoverflow.com/a/1557584/17555691) and it gave the following response:

Chance of streak: 80.06% --- 0.8480644226074219 seconds ---

Next, I made a small change, using append to a list and then converting to a string:

# Code that creates a list of 100 'heads' or 'tails' values coin_list = [] for i in range(100): if random.randint(0, 1) == 0: coin_list.append('H') # heads else: coin_list.append('T') # tails # Code that checks if there is a streak of 6 heads or tails in a row coin_string = ''.join(coin_list) if 'HHHHHH' in coin_string or 'TTTTTT' in coin_string: number_of_streaks +=1 

Output from this version:

Chance of streak: 80.26% --- 0.739051342010498 seconds ---

Then made a change, using recommendation from @Alain T:

# Code that creates a list of 100 'heads' or 'tails' values flips = "".join(random.choice('HT') for _ in range(100)) # Code that checks if there is a streak of 6 heads or tails in a row if 'HHHHHH' in flips or 'TTTTTT' in flips: number_of_streaks +=1 

Output from this version: Chance of streak: 80.74% --- 0.4248924255371094 seconds ---

The above results were quite typical on repeated executions.

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) 

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.

#Defining Inputs and importing modules import random numberOfStreaks = 0 mylist=[] countH=0 countT=0 conuterTail=[] conuterHead=[] for experimentNumber in range(10000): # Code that creates a list of 10000 'heads' or 'tails' values. rndnum=random.randint(0,1); if rndnum==0: countH=0 # if the random number is "0" then "HEAD" counts will be reset mylist.append(['T']) countT+=1 else: countT=0 # if the random number is "1" then "TAIL" counts will be reset mylist.append(['H']) countH+=1 # Code that checks if there is a streak of 6 heads or tails in a row. if countT==6: conuterTail.append(countT) elif countH==6: conuterHead.append(countH); numberOfStreaks=len(conuterHead)+len(conuterTail) print('Chance of streak: %s%%' % (numberOfStreaks / 100)) 

I understood the question to be asking: what is the chance that a streak of 6 consecutive like coin tosses would be contained within a set of 100 random coin tosses. I thought this should include instances of multiple streaks within the same set of tosses. Whenever a streak of values reached six in a row, the count was reset to zero, in order to capture the next streak. Some iterations contained no streaks, while others contained multiple. A large number of iterations were averaged to smooth variances. The average value returned was 158-160%, meaning very likely.

import random numberOfStreaks = 0 countH = 0 countT = 0 iterations_of_experiment = 10000 for experimentNumber in range(iterations_of_experiment): #Code that creates a list of 100 'heads' or 'tails' values setOfHundred = [] #set list to empty each iteration for i in range(100): if random.randint(0, 1) == 0: countT = 0 #Set Tails count to zero on head flip setOfHundred.append('H') countH += 1 if countH == 6: numberOfStreaks += 1 countH = 0 #Reset counter to zero to capture - else: #multiple streaks countH = 0 setOfHundred.append('T') countT += 1 if countT == 6: numberOfStreaks += 1 countT = 0 print('In ' + str(iterations_of_experiment) + \ ' iterations of 100 coin tosses, there were ' + str(numberOfStreaks) + \ ' streaks of 6 consecutive like coin flips.') print('Chance of streak: ' + str(numberOfStreaks / iterations_of_experiment * 100) + '%')