塞缪尔抛硬币

收益取决于第一次出现反面前一共抛了多少次。因为收益为 $\min(64,2^n)$，当 $n\ge 6$ 时收益封顶为 64 美元；少于 6 次时收益为 $2^n$。

因此可把期望收益拆成两部分：

1. 第一次反面出现在第 1 到第 6 次。
2. 第一次反面出现在第 7 次或之后。

前 6 次的贡献为

$$\sum_{i=1}^{6} P(\text{第 } i \text{ 次首次出现反面})\cdot 2^i =\sum_{i=1}^{6}\frac{1}{2^i}\cdot 2^i=6.$$

第 7 次及之后的贡献为

$$\sum_{i=7}^{\infty} P(\text{第 } i \text{ 次首次出现反面})\cdot 64 =\sum_{i=7}^{\infty}\frac{64}{2^i}=1.$$

所以总期望收益为

$$6+1=\boxed{7}.$$

因此，塞缪尔的期望收益是 7 美元。

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Original Explanation

Based on the problem description, we know that the expected payoff that Samuel will receive is dependent upon the number of times he flips a coin before observing a tails. Since he will receive $\min(64, 2^n)$ dollars, we know that Samuel will receive $64 as long as he flips the coin 6 or more times. Anything less than 6 flips, would result in a payoff less than $64.

With this observation, we split our expected payoff calculation by considering:

1. The payoff if Samuel flips the coin 6 or fewer times along with the probability of this event.
2. The payoff if Samuel flips the coin 7 or more times along with the probability of this event.

$$\begin{equation*} \mathbb{E}(P) = P( \leq 6 \ \textrm{flips}) \times \textrm{Payoff}(\leq 6 \ \textrm{flips}) + P(\geq 7 \ \textrm{flips}) \times \textrm{Payoff}(\geq 7 \ \textrm{flips}) \end{equation*}$$

The payoff from flipping the coin 6 or fewer times can be calculated as:

$$\begin{equation*} \sum_{i=1}^6 P(\textrm{First tails observed on ith flip}) \times \textrm{Payoff on ith flip} \end{equation*} \\ = \sum_{i=1}^6 \frac{1}{2^i} \times 2^i = \frac{1}{2} \times 2 + \frac{1}{4} \times 4 + ... + \frac{1}{6} \times 6 = 6 \times 1 = 6$$

Now we can calculate the the payoff from flipping the coin 7 or more times as:

$$\begin{equation*} \sum_{i=7}^\infty P(\textrm{First tails observed on ith flip}) \times \textrm{Payoff on ith flip} \end{equation*} \\ = \sum_{i=7}^\infty \frac{1}{2^i} \times 64 = \frac{1}{2^7} \times 2^6 + \frac{1}{2^8} \times 2^6 + ... + \frac{1}{2^\infty} \times 2^6 \\ = \frac{1}{2} + \frac{1}{4} + ... + \frac{1}{\infty} = 1 \\$$

Putting these two pieces together, we get that the expected payoff from playing this game is 6 + 1 = $\boxed{7}$ dollars.