3 张 A 与 3 张 J：最优猜牌的期望正确次数

最优策略：每一步都猜“剩余数量更多”的那种牌（若数量相同则随便猜）。

令 $E(a,b)$ 表示剩余 $a$ 张 A、$b$ 张 J 时的最优期望正确次数，则可递推计算得到

$$E(3,3)=\frac{41}{10}=4.1.$$

因此期望赚 4.1 元。

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Original Explanation

Let $X_1, X_2, \ldots, X_6$ denote the amount of money won on turn $1, 2, \ldots, 6$, respectively. By linearity of expectation, the total amount of money won is simply $\mathbb{E}[X_1] + \mathbb{E}[X_2] + \ldots + \mathbb{E}[X_6]$. Clearly,

$\mathbb{E}[X_1] = \frac{1}{2}$, so let's begin our discussion with the second turn.

After Jamie makes a guess from round 1, regardless of the guess and the true card, Jamie will have information for the correct 3 card-2 card split of the remaining 5 cards. Under the optimal strategy, Jamie should guess the value of the card that appears 3 times in the remaining 5 cards. Hence,

$\mathbb{E}[X_2] = \frac{3}{5}$.

The expected value of round 3 depends on the previous two guesses; specifically, there is either a 3-1 split or a 2-2 split. Once again, Jamie should know the true split and guess accordingly. It is not difficult to conclude that the probability of a 3-1 split is $\frac{2}{5}$, since there are $20$ total orderings of 3 aces and 3 jacks, and $\frac{4}{1} \cdot 2$ of those orderings begin with ace-ace or jack-jack. That means that there is a $\frac{3}{5}$ chance that there is a 2-2 split. By the law of total expectation, we have

$\mathbb{E}[X_3] = \frac{3}{5} \cdot \frac{1}{2} + \frac{2}{5} \cdot \frac{3}{4} = \frac{3}{5}$.

The expected value of round 4 can be computed similarly. There is either a 2-1 split or a 3-0 split. The 3-0 split occurs with probability $\frac{1}{10}$, while the 2-1 split occurs with probability $\frac{9}{10}$. By linearity of expectation, we find

$\mathbb{E}[X_4] = \frac{1}{10} \cdot 1 + \frac{9}{10} \cdot \frac{2}{3} = \frac{7}{10}$.

The expected value of round 5 can be conditioned on a 2-0 split which occurs with probability $\frac{2}{5}$ or a 1-1 split which occurs with probability $\frac{3}{5}$. By linearity of expectation, we find

$\mathbb{E}[X_5] = \frac{2}{5} \cdot 1 + \frac{3}{5} \cdot \frac{1}{2} = \frac{7}{10}$.

By round 6, there is only one card remaining, so

$\mathbb{E}[X_6] = 1$.

Putting it all together, our expected value is

$\frac{1}{2} + \frac{3}{5} + \frac{3}{5} + \frac{7}{10} + \frac{7}{10} + 1 = \frac{5+6+6+7+7+10}{10} = \frac{41}{10}$.