三重硬币

Triple Coin

专题: Probability / 概率
难度: L3
来源: OpenQuant

题目详情

你有 3 个硬币。一枚总是正面朝上，一枚总是反面，一枚只是一枚普通的公平硬币。如果你选择一枚硬币，抛两次，两次都出现正面，那么再次抛同一枚硬币，再次出现正面的概率是多少？

You have 3 coins. One always comes up heads, one always comes up tails, and one is just a regular fair coin. If you pick one coin, flip it twice and get heads twice, what's the probability of flipping the same coin again and getting heads again?

解析

设随机变量 $C$ 表示选到的硬币：

$c_1$ ：永远正面
$c_2$ ：永远反面
$c_3$ ：公平硬币

观察到两次都为正面（记为 $HH$ ）后，只可能是 $c_1$ 或 $c_3$ 。

先用贝叶斯更新选中硬币的后验概率：

P(HH|c_1)=1,\quad P(HH|c_2)=0,\quad P(HH|c_3)=\left(\frac12\right)^2=\frac14,

且先验 $P(c_1)=P(c_2)=P(c_3)=\frac13$ 。因此

P(HH)=\frac13\cdot 1+\frac13\cdot 0+\frac13\cdot \frac14=\frac{5}{12},

P(c_1|HH)=\frac{\frac13}{\frac{5}{12}}=\frac45,\quad P(c_3|HH)=\frac{\frac{1}{12}}{\frac{5}{12}}=\frac15.

第三次为正面的概率为

P(H_3|HH)=1\cdot\frac45+\frac12\cdot\frac15=\boxed{\frac{9}{10}}.

Original Explanation

Lets call $H_i$ the result of having a heads on the $i$ -th throw, and $C$ the random variable corresponding to the choice of the coin, with three possible values: $c_1$ is the coin which always comes up heads, $c_2$ is the the coin which always comes up tails, and $c_3$ is the fair coin.

= P(H_3 | H_1H_2) = P(H_1 H_2 H_3 | H_1 H_2) = \frac{P(H_1H_2H_3)}{P(H_1H_2)} = \frac{\sum_CP(H_1H_2H_3,C)}{\sum_CP(H_1H_2,C)}

Lets compute the numerator: $\sum_CP(H_1H_2H_3,C) = P(H_1H_2H_3|c_1)P(c_1) + P(H_1H_2H_3|c_2)P(c_2) + P(H_1H_2H_3|c_3)P(c_3) \\ = 1 \times 1/3 + 0 \times 1/3 + 1/8 \times 1/3 = 9/24$

Lets compute the denominator:

\sum_CP(H_1H_2,C) = P(H_1H_2|c_1) P(c_1) + P(H_1H_2|c_2)P(c_2) + P(H_1H_2|c_3)P(c_3) \\ = 1 \times 1/3 + 0 \times 1/3 + 1/4 \times 1/3 = 5/12\\

Putting these two together we get:

P(H_3|H_1H_2) = \frac{9/24}{5/12} = \frac{9}{10}