翻到第一张 A 的期望张数

把 4 张 A 视作把其余 48 张牌切分成 5 段。由于对称性，每段期望长度相同，且总长度为 48，因此第一段期望为 $48/5$。

第一张 A 出现在第一段之后的下一张，因此期望翻牌张数为

$$\frac{48}{5}+1=\frac{53}{5}=10.6.$$

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Original Explanation

The four aces are dispersed throughout the deck and cut the 48 remaining cards into 5 distinct sections, each of some random length $X_{i} \in [0, 48]$ where $1 \leq i \leq 5$; that is, $\displaystyle \sum_{i=1}^5 X_i = 48$. Furthermore, by symmetry, $E[X_1] = E[X_2] = \dots = E[X_5]$, as in the absence of any additional information, none of the sections is expected to be any larger or smaller than any other. Thus, by linearity of expectation:

$$E\left[\displaystyle \sum_{i=1}^5 X_i\right] = 5\times E[X_1] = 48 \Rightarrow E[X_1] = \frac{48}{5}$$

We have found that the expected number of cards in the first section is $\frac{48}{5}$, so we will observe the first ace on the next card. Hence, we expect to flip $\frac{48}{5}+1 = 10.6$ cards to observe the first ace.