抽到第一张 A 的期望

答案是 $\frac{53}{5}=10.6$。

把 4 张 A 的位置视为把其余 48 张牌平均分到 5 个“缝隙”里，第一张 A 前的非 A 牌期望为 $48/5$，再加上翻到那张 A 本身的 1 张：

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

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Original Explanation

53/5

Solution

Define X1,X2…X48, such that Xi = 1 if ith card turns over before 4 aces, 0 otherwise. Thus, total number of cards turned to see first ace = 1 + sum (Xi) using linearity of expectation, E(X) = 1 + sum ( E(Xi)) Now consider the ith card and the four aces, all the orders are equally likely: X,A,A,A,A Or A,X,A,A,A, or 3 others. They are equally likely when we have no knowledge about position of these Aces with respect of the deck. Hence we have 5 equally likely slots for ith card: 1 A 2 A 3 A 4 A 5. We are interested in 1st slot. Hence E(Xi) = 1/5 for all i = 1 to 48 Thus E(X) = 1 + 48/5 = 53/5 = 10.6

This can be done by recursive equation f(n) = (4/n) * 1 + ((n-4)/n) * (1+f(n-1)), where f(n)=expected cards to flip in a deck of n cards, but I definitely can't solve this without a computer.

A shorter explanation is to consider the 52 cards uniformly distributed over (0,1), so on average they're at k/53 for k=1,2,3,…,52. The four aces are on average at 1/5, 2/5, 3/5, 4/5. So 0.2=k/53 implies k=10.6, done!