1000 枚硬币里找双头币

设 $A$ 为“选中双头币”，$B$ 为“10 连正”。

$$P(A)=\frac{1}{1000},\quad P(B\mid A)=1,\quad P(B\mid A^c)=\left(\frac12\right)^{10}=\frac{1}{1024}.$$

贝叶斯：

$$P(A\mid B)=\frac{P(B\mid A)P(A)}{P(B\mid A)P(A)+P(B\mid A^c)P(A^c)} =\frac{\frac{1}{1000}}{\frac{1}{1000}+\frac{999}{1000}\cdot\frac{1}{1024}}\approx 0.506.$$

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Original Explanation

Let $A$ = “the chosen coin is the two-headed coin,” and $B$ = “10 heads in a row.”

• $P(A) = \frac{1}{1000}, \quad P(A^c) = \frac{999}{1000}.$
• $P(B \mid A) = 1,$
• $P(B \mid A^c) = \left(\frac{1}{2}\right)^{10} = \frac{1}{1024}.$

By Bayes’ formula: $$P(A \mid B) = \frac{P(B \mid A)\,P(A)}{P(B \mid A)\,P(A) + P(B \mid A^c)\,P(A^c)} = \frac{\frac{1}{1000}}{\frac{1}{1000} + \frac{999}{1000}\cdot \frac{1}{1024}} \approx 0.5.$$