D e s c r i p t i o n \mathscr{Description} Description
S o l u t i o n \mathfrak{Solution} Solution
- 先考虑简单的再来一瓶问题
- 设买一瓶可乐有 p p p概率再来一瓶, E ( ξ ) = x E(\xi)=x E(ξ)=x
- x = 1 + p x ⟹ x = 1 1 − p x=1+px\Longrightarrow x=\frac{1}{1-p} x=1+px⟹x=1−p1
- 同理
- 设当前选了 x x x,之后还能选的数期望为 f x f_x fx
- f x = 1 ∗ ∑ i = 1 x − 1 p i + p x ( 1 + f x ) + ∑ i = x + 1 n p i ( 1 + f i ) \,\,\,\,\,\,\,\,\,\,\,\,f_x=1*\mathop{\sum}\limits^{x-1}_{i=1}p_i+p_x(1+f_x)+\mathop{\sum}\limits^{n}_{i=x+1}p_i(1+f_i) fx=1∗i=1∑x−1