找出硬币条纹的百分比

Question

我正在尝试编写一个程序，以找出在随机生成的 100 次翻转的正面和反面列表中出现六个正面或六个反面的条纹的频率，并重复此 10000 次以查找硬币翻转的百分比连续包含六个正面或反面。

我的代码是这样的，它是某种工作。**我想知道它是否给出了正确的结果：**

import random

numberOfStreaks = 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 = result

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

Answer 1

当您获得第 7 次正面或反面重复时，应将其计为连续，因为最后 6 次翻转确实形成了连续。通过在 6 之后重置计数，您低估了条纹的数量。从概率的角度来看，您打印的结果不是“机会”的度量（永远不会超过 100%），而是更接近数学期望（即期望值的总和乘以它们各自的概率）。您的代码会生成实际尝试的样本，但计算没有意义，除非它们可以与适当的概率数字进行比较。以下是我将如何实现此采样并表达结果：import randomfrom itertools import accumulateexpCount   = 10000expSize    = 100streakSize = 6numberOfStreaks = 0for 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 += streaksratio = numberOfStreaks / expCountprint(f'Obtained {numberOfStreaks} streaks of {streakSize} head/tail on {expCount} runs of {expSize} flips. On average {ratio:.2f} streaks per run')

Answer 2

import randomnumberOfStreaks = 0numberOfExperiments = 10000numberOfAttemps = 100for 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))输出：Chance of streak: 1.5143%

Answer 3

import random, timenumberOfStreaks = 0for 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, '%')