双头硬币：连出 10 个正面后的后验概率

用 Bayes 定理。

设：

• $A$：抽到双头硬币；
• $B$：连续 10 次都是正面。

则：

• $P(A)=\frac{1}{1000}$；
• $P(B\mid A)=1$；
• $P(B\mid A^c)=\left(\frac12\right)^{10}=\frac{1}{1024}$。

因此

$$P(A\mid B)=\frac{P(A)P(B\mid A)}{P(A)P(B\mid A)+P(A^c)P(B\mid A^c)} =\frac{\frac{1}{1000}}{\frac{1}{1000}+\frac{999}{1000}\cdot\frac{1}{1024}} =\frac{1024}{2023}.$$

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Original Explanation

Use Bayes’ Theorem.

Let:

• $A$ = event we chose the 2-headed coin
• $B$ = event of getting 10 heads in a row

Then:

• $P(A) = \frac{1}{1000}$
• $P(B \mid A) = 1$ (since the coin has two heads)
• $P(B \mid \text{fair coin}) = \left(\frac{1}{2}\right)^{10} = \frac{1}{1024}$

By Bayes’ Theorem: $$P(A \mid B) = \frac{P(A) \cdot P(B \mid A)}{P(A) \cdot P(B \mid A) + P(\text{not } A) \cdot P(B \mid \text{fair coin})}$$

Plug in the values: $$P(A \mid B) = \frac{\frac{1}{1000} \cdot 1}{\frac{1}{1000} \cdot 1 + \frac{999}{1000} \cdot \frac{1}{1024}} = \frac{1}{1 + \frac{999}{1024}} = \frac{1}{\frac{1024 + 999}{1024}} = \frac{1024}{2023}.$$