我正在嘗試編寫一個程序，以找出在隨機生成的 100 次翻轉的正面和反面列表中出現六個正面或六個反面的條紋的頻率，并重復此 10000 次以查找硬幣翻轉的百分比連續包含六個正面或反面。我的代碼是這樣的，它是某種工作。**我想知道它是否給出了正確的結果：**import randomnumberOfStreaks = 0# Code that creates a list of 100 'heads' or 'tails' values.for experimentNumber in range(10000): results = [] for experiment in range(100): x = random.randint(1, 2) if x == 1: results.append('H') else: results.append('T') # Code that checks if there is a streak of 6 heads or tails in a row. currentStreak = 0 previousResult = results[0] for result in results: if currentStreak == 6: numberOfStreaks += 1 currentStreak = 0 if result == previousResult: currentStreak += 1 previousResult = resultprint('Chance of streak: %s%%' % (numberOfStreaks / 100))
3 回答
holdtom
TA貢獻1805條經驗 獲得超10個贊
當您獲得第 7 次正面或反面重復時，應將其計為連續，因為最后 6 次翻轉確實形成了連續。通過在 6 之后重置計數，您低估了條紋的數量。
從概率的角度來看，您打印的結果不是“機會”的度量（永遠不會超過 100%），而是更接近數學期望（即期望值的總和乘以它們各自的概率）。您的代碼會生成實際嘗試的樣本，但計算沒有意義，除非它們可以與適當的概率數字進行比較。
以下是我將如何實現此采樣并表達結果：
import random
from itertools import accumulate
expCount = 10000
expSize = 100
streakSize = 6
numberOfStreaks = 0
for experimentNumber in range(expCount):
results = [random.choice("HT") for _ in range(expSize)]
consec = [int(a==b) for a,b in zip(results,results[1:])]
streaks = sum( s+1>=streakSize for s in accumulate(consec,lambda s,m:s*m+m))
numberOfStreaks += streaks
ratio = numberOfStreaks / expCount
print(f'Obtained {numberOfStreaks} streaks of {streakSize} head/tail on {expCount} runs of {expSize} flips. On average {ratio:.2f} streaks per run')
慕慕森
TA貢獻1856條經驗 獲得超17個贊
import random
numberOfStreaks = 0
numberOfExperiments = 10000
numberOfAttemps = 100
for experimentNumber in range(numberOfExperiments):
# Code that creates a list of {numberOfAttemps} 'heads' or 'tails' values.
coinFlip = []
for i in range(numberOfAttemps):
if random.randint(0, 1) == 0:
coinFlip.append("H")
else:
coinFlip.append("T")
# Code that checks if there is a streak of 6 heads or tails in a row.
for i in range(len(coinFlip)):
if (["H"] * 6) in [coinFlip[i:i+6]] or (["T"] * 6) in [coinFlip[i:i+6]]:
numberOfStreaks += 1
for n in range(5):
coinFlip.pop(i)
print('Chance of streak: %s%%' % (numberOfStreaks / numberOfExperiments))
輸出：
Chance of streak: 1.5143%
嚕嚕噠
TA貢獻1784條經驗 獲得超7個贊
import random, time
numberOfStreaks = 0
for experimentNumber in range(10000):
# Code that creates a list of 100 'heads' or 'tails' values.
List = []
for value in range(100):
value = random.randint(0,1)
if value == 0:
List.append('H')
if value == 1:
List.append('T')
# print('List = ', end=' ')
# print(List)
# Code that checks if there is a streak of 6 heads or tails in a row.
sixStreakHeads = 0
sixStreakTails = 0
for result in range(len(List)):
if List[result] == 'H':
sixStreakHeads += 1
sixStreakTails = 0
if sixStreakHeads == 6:
sixStreakHeads = 0
numberOfStreaks += 1
if List[result] == 'T':
sixStreakTails += 1
sixStreakHeads = 0
if sixStreakTails == 6:
sixStreakTails = 0
numberOfStreaks += 1
# print('Six Streak Heads: ', sixStreakHeads)
# print('Six Streak Tails: ', sixStreakTails)
# time.sleep(.5)
# print('\n\n\n')
print('Chance of streak: ', numberOfStreaks / 10000 * 100, '%')
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