问：在 Python 中连续 N 次正面朝上的预期抛硬币次数。我的代码给出的答案与已发布的正确答案不匹配，但不确定原因

Question

我正在尝试编写 Python 代码来查看平均需要抛多少次硬币才能连续出现 N 个正面朝上的序列。

让我感到困惑的是，我的代码生成的答案与在线给出的答案不匹配，例如这里（以及许多其他地方）https://math.stackexchange.com/questions/364038/expected-number-of-coin-tosses-to-get-five-consecutive-heads

据此，我需要连续获得各种正面朝上的预期投掷次数为：E(1) = 2, E(2) = 6, E(3) = 14, E(4) = 30，E(5) = 62。但我没有得到这些答案！例如，我得到 E(3) = 8，而不是 14。下面的代码运行s 给出了答案，但您可以更改 n 以测试连续出现的其他目标数量。

出了什么问题？估计是我的代码逻辑有问题，但我承认我搞不清楚是什么。

您可以在此处查看运行并修改我的代码副本：https://trinket.io/python/17154b2cbd

下面是代码本身，在运行nable trinket.io 页面之外。如果您能帮助找出问题所在，我们将不胜感激！

非常感谢，

拉吉 P.S。我能找到的最接近的相关问题是这个：但是，据我所知，该问题中的代码实际上并没有测试两个 consecutive 头，而是测试一个以头开始然后在稍后的序列，可能是非连续的，时间得到另一个头。

# Click here to run and/or modify this code:
# https://trinket.io/python/17154b2cbd

import random
# n is  the target number of heads in a row
# Change the value of n, for different target heads-sequences
n = 3  
possible_tosses = [ 'h', 't' ]
num_trials = 1000
target_seq = ['h' for i in range(0,n)]
toss_sequence = []
seq_lengths_rec = []

for trial_num in range(0,num_trials):

    if (trial_num % 100) == 0:
        print 'Trial num', trial_num, 'out of', num_trials
        # (The free version of trinket.io uses Python2)

    target_reached = 0
    toss_num = 0

    while target_reached == 0:

        toss_num += 1
        random.shuffle(possible_tosses)
        this_toss = possible_tosses[0]
        #print([toss_num, this_toss])
        toss_sequence.append(this_toss)
        last_n_tosses = toss_sequence[-n:]
        #print(last_n_tosses)
    if last_n_tosses == target_seq:
        #print('Reached target at toss', toss_num)
        target_reached = 1
        seq_lengths_rec.append(toss_num)

print 'Average', sum(seq_lengths_rec) / len(seq_lengths_rec)

Answer 1

您不会为每个实验重新初始化 toss_sequence，因此您使用预先存在的正面序列开始每个实验，在第一次尝试时有二分之一的机会击中目标序列每个新实验。

在外循环中初始化 toss_sequence 将解决您的问题：

import random
# n is  the target number of heads in a row
# Change the value of n, for different target heads-sequences
n = 4
possible_tosses = [ 'h', 't' ]
num_trials = 1000
target_seq = ['h' for i in range(0,n)]
seq_lengths_rec = []

for trial_num in range(0,num_trials):

    if (trial_num % 100) == 0:
        print('Trial num {} out of {}'.format(trial_num, num_trials))
        # (The free version of trinket.io uses Python2)

    target_reached = 0
    toss_num = 0
    toss_sequence = []

    while target_reached == 0:

        toss_num += 1
        random.shuffle(possible_tosses)
        this_toss = possible_tosses[0]
        #print([toss_num, this_toss])
        toss_sequence.append(this_toss)
        last_n_tosses = toss_sequence[-n:]
        #print(last_n_tosses)
        if last_n_tosses == target_seq:
            #print('Reached target at toss', toss_num)
            target_reached = 1
            seq_lengths_rec.append(toss_num)

print(sum(seq_lengths_rec) / len(seq_lengths_rec))

您可以稍微简化您的代码，使其不易出错：

import random
# n is  the target number of heads in a row
# Change the value of n, for different target heads-sequences
n = 3
possible_tosses = [ 'h', 't' ]
num_trials = 1000
seq_lengths_rec = []

for trial_num in range(0, num_trials):

    if (trial_num % 100) == 0:
        print('Trial num {} out of {}'.format(trial_num, num_trials))
        # (The free version of trinket.io uses Python2)

    heads_counter = 0
    toss_counter = 0

    while heads_counter < n:
        toss_counter += 1

        this_toss = random.choice(possible_tosses)

        if this_toss == 'h':
            heads_counter += 1
        else:
            heads_counter = 0
    seq_lengths_rec.append(toss_counter)


print(sum(seq_lengths_rec) / len(seq_lengths_rec))

Answer 2

我们可以通过运行每个实验足够长（理想情况下是无限次）的次数来消除一个额外的循环，例如，每次抛硬币 n=1000 次。现在，很可能在每次这样的试验中都会出现 5 个正面朝上的序列。如果确实出现了，我们可以称该试验为有效试验，否则我们可以拒绝该试验。

最后，我们可以取平均需要的投掷次数w.r.t。有效试验的次数（通过 LLN 它将近似于预期的投掷次数）。考虑以下代码：

N = 100000 # total number of trials
n = 1000 # long enough sequence of tosses
k = 5 # k heads in a row
ntosses = []
pat = ''.join(['1']*k)
effective_trials = 0
for i in range(N): # num of trials
    seq = ''.join(map(str,random.choices(range(2),k=n))) # toss a coin n times (long enough times)
    if pat in seq:
        ntosses.append(seq.index(pat) + k)
        effective_trials += 1
print(effective_trials, sum(ntosses) / effective_trials)
# 100000 62.19919

请注意，如果 n 较小，结果可能不正确，因为它试图近似无限次抛硬币（要找到预期的抛硬币次数以获得连续 5 次正面朝上，n=1000 是可以的，因为实际预期值为 62)。

问：在 Python 中连续 N 次正面朝上的预期抛硬币次数。我的代码给出的答案与已发布的正确答案不匹配，但不确定原因

Q: Expected number of coin tosses to get N heads in a row, in Python. My code gives answers that don't match published correct ones, but unsure why

python

probability

montecarlo