有偏差的硬币序列

Biased Coin Sequence

专题: Probability / 概率
难度: L5
来源: OpenQuant

题目详情

你将获得一枚硬币，正面朝上的概率为 $\frac23$ ，反面朝上的概率为 $\frac13$ 。如果你抛这枚硬币，直到出现反面，紧接着出现正面，你预计抛硬币多少次？

You are given a coin with probability $\frac23$ to land on heads and $\frac13$ to land on tails. If you flip this coin until you get a tails immediately followed by heads, how many times do you expect to flip the coin?

解析

我们将看到紧随其后的正面的反面所需的翻转次数称为 $TH$ 。如果我们的第一次翻转是正面，则已添加一次翻转但尚未取得任何进展，因此我们现在预计翻转总数为 $E[TH] + 1$ 。如果我们第一次翻转是反面，我们就完成下一次正面。直到正面的预期翻转次数与 $p = \frac23$ 呈几何关系，因此我们预计会进行额外的 $\frac32$ 翻转，直到我们获得最终正面。考虑到这些结果的概率，我们可以说： $E(TH) = \frac23(E(TH) + 1) + \frac13(\frac32+ 1)$ 我们现在可以求解 $E(TH)$ 。 $E(TH) = \frac23E(TH) + \frac96$ $\frac13E(TH) = \frac96$ $\Rightarrow E(TH) = \boxed{\frac92}$

import random

flip_coin = lambda: 'T' if random.randint(1, 3) == 1 else 'H'

sequence_lengths = []
num_iters = 100_000
for i in range(num_iters):

    sequence = [flip_coin()]
    while(True):

        sequence.append(flip_coin())
        if sequence[-2]=='T' and sequence[-1]=='H':
            sequence_lengths.append(len(sequence))
            break

print(sum(sequence_lengths) / num_iters)

Original Explanation

Let's call the number of flips needed to see a tails immediately followed heads $TH$ . If our first flip is heads, one flip has been added but no progress has been made so we will now expect $E[TH] + 1$ total flips. If our first flip is tails, we finish on the next heads. The expected number of flips until heads is geometric with $p = \frac23$ so we expect an additonal $\frac32$ flips until we get the finishing heads. Given the probabilities these outcomes we can say:

E(TH) = \frac23(E(TH) + 1) + \frac13(\frac32+ 1)

We can now solve for $E(TH)$ . $E(TH) = \frac23E(TH) + \frac96$ $\frac13E(TH) = \frac96$ $\Rightarrow E(TH) = \boxed{\frac92}$

import random

flip_coin = lambda: 'T' if random.randint(1, 3) == 1 else 'H'

sequence_lengths = []
num_iters = 100_000
for i in range(num_iters):

    sequence = [flip_coin()]
    while(True):

        sequence.append(flip_coin())
        if sequence[-2]=='T' and sequence[-1]=='H':
            sequence_lengths.append(len(sequence))
            break

print(sum(sequence_lengths) / num_iters)