先验

后验为 $p\mid H^n\sim \mathrm{Beta}(n+1,1)$。

因此

$$\mathbb{P}(\text{下次正面}\mid H^n)=\mathbb{E}[p\mid H^n]=\boxed{\frac{n+1}{n+2}}.$$

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英文解析

Denote by $A_{k}$ the event that all of the first $k$ outcomes are heads and by $B_{k}$ the event that the $k$ - th outcome is a head. We need to find $\mathbb{P}\left(B_{n + 1}\mid A_n\right)$ . Note that $A_{n}\cap B_{n + 1} = A_{n + 1}$ . Then,

$$\begin{array}{r}\mathbb{P}\left(B_{n + 1}\mid A_n\right) = \frac{\mathbb{P}\left(A_n\cap B_{n + 1}\right)}{\mathbb{P}\left(A_n\right)}\\ = \frac{\mathbb{P}\left(A_{n + 1}\right)}{\mathbb{P}\left(A_n\right)} \end{array}$$

Denote by $F$ the cumulative distribution function of the uniform random variable on [0, 1]. We apply the law of total probability and condition on the random variable $p$ that takes a uniform value in the interval [0, 1]. We obtain that

$$\begin{array}{l}\mathbb{P}\left(A_{n}\right) = \int_{0}^{1}\mathbb{P}\left(A_{n}\mid p = t\right)dF(t)\\ \displaystyle = \int_{0}^{1}t^{n}dt = \frac{1}{n + 1}. \end{array}$$

From (2.244) and (2.245), it follows that

$$\mathbb{P}\left(B_{n + 1}\mid A_n\right) = \frac{n + 1}{n + 2}.$$