骰子游戏期望值

为了确定我们是否应该玩这个游戏，我们可以计算玩游戏的预期值并确定该值是否为正。获得 0 个匹配的概率可以通过我们掷 4 个唯一数字的方式数除以我们可以掷 4 个骰子的所有可能方式来计算。这相当于： $$\begin{equation}P(\textrm{0 matches}) = \frac{\textrm{Number of ways to roll 4 unique dice}}{\textrm{Total number of ways to roll 4 dice}} = \frac{6 * 5 * 4 * 3}{6^4} = \frac{5}{18}\end{equation}$$ 为了计算预期收益，我们可以计算： $$\begin{equation} \textrm{Expected Payoff} = P(\textrm{At least 1 match}) * 1 - P(\textrm{0 matches}) * 1 = \frac{13}{18} - \frac{5}{18} = \boxed{\frac{8}{18}} \end{equation}$$ 由于预期回报是积极的，我会玩这个游戏。

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Original Explanation

To determine whether we should play this game we can calculate the expected value of playing and determine whether that value is positive. The probability of getting 0 matches can be calculated as the number of ways we roll 4 unique numbers divided by all possible ways we can roll 4 dice. This would equate to:

$$\begin{equation}P(\textrm{0 matches}) = \frac{\textrm{Number of ways to roll 4 unique dice}}{\textrm{Total number of ways to roll 4 dice}} = \frac{6 * 5 * 4 * 3}{6^4} = \frac{5}{18}\end{equation}$$

To calculate the expected payoff we can compute:

$$\begin{equation} \textrm{Expected Payoff} = P(\textrm{At least 1 match}) * 1 - P(\textrm{0 matches}) * 1 = \frac{13}{18} - \frac{5}{18} = \boxed{\frac{8}{18}} \end{equation}$$

Since the expected payoff is positive, I would play this game.

考虑两种情况：保留第一次结果，或重掷一次。设游戏期望收益为 $EP$，则

$$EP=EP(\text{只掷一次并保留})\cdot P(\text{只掷一次})+EP(\text{重掷})\cdot P(\text{重掷}).$$

如果只能掷一次，奇数点数收益分别为 1、3、5 美元，偶数点数收益分别为 4、8、12 美元，因此一次掷骰的期望收益为

$$\frac{1}{6}(1+3+5+4+8+12)=\frac{33}{6}=5.5.$$

既然有一次重掷机会，只有当第一次掷出 4 或 6 时，保留第一次结果才优于重掷；对应收益分别为 8 美元和 12 美元。因此

$$EP(\text{只掷一次并保留})=\frac{8+12}{2}=10.$$

若选择重掷，第二次必须接受结果，因此期望收益仍为

$$EP(\text{重掷})=5.5.$$

第一次只在掷出 4 或 6 时保留，所以

$$P(\text{只掷一次})=\frac{2}{6}=\frac{1}{3},\qquad P(\text{重掷})=\frac{2}{3}.$$

代入得到

$$EP=10\cdot\frac{1}{3}+5.5\cdot\frac{2}{3}=\boxed{7}.$$

所以这个游戏的期望收益为 7 美元。

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Original Explanation

For this question, we have to consider two primary cases: the player rolling once and the player rolling twice. We can calculate the expected payoff ($EP$) of the game as:

$$\begin{equation} \textrm{EP} = EP(\textrm{game | play once}) \times P(\textrm{play once}) + EP(\textrm{game | play twice}) \times P(\textrm{play twice}) \end{equation}$$

Let's break down the equation above into parts. First we'll find $EP(\textrm{game | play once})$. Since we are rolling a fair dice, each outcome of the dice roll is equally likely (1/6) and the payoffs for rolling an odd number are 1, 3, and 5, while the payoffs for rolling an even number are 4, 8, and 12 (double of 2, 4, and 6). Considering the fact that we are given the ability to reroll the dice, it would only make sense to keep the outcome of the first dice roll if we rolled a 4 or 6 (payoffs of $8 and $12 respectively).

The rationale for this stems from calculating the expected payoff if we were to play a game in which we could only roll the dice once. In this case the expected payoff is simply the probability of each dice roll outcome times the corresponding value of each dice roll.

$$\begin{equation} EP(\textrm{game where you only roll once}) = \frac{1}{6} \times (1 + 3 + 5 + 4 + 8 + 12) = \frac{33}{6} = \$5.5 \end{equation}$$

Since we know that we have the ability to reroll the dice, we would only want to keep the outcome of our first dice roll if the outcome was greater than $EP(\textrm{game where you only roll once})$ which only occurs if we roll a 4 or 6.

Therefore, we calculate $EP(\textrm{game | play once})$ as:

$$\begin{equation} EP(\textrm{game | play once}) = \frac{1}{2} \times (8 + 12) = \frac{20}{2} = \$10 \end{equation}$$

Now we need to calculate $EP(\textrm{game | play twice})$. Recall that if we are on our second attempt of playing the game, this is our last chance to roll the dice. Therefore, the expected payoff is the same as that of the hypothetical game we mentioned earlier where you can only roll the dice once.

$$\begin{equation} EP(\textrm{game | play twice}) = \frac{1}{6} \times (1 + 3 + 5 + 4 + 8 + 12) = \frac{33}{6} = \$5.5 \end{equation}$$

The last part involves figuring out $P(\textrm{play once})$ and $P(\textrm{play twice})$. The $P(\textrm{play once})$ is $1/3$ because we will only play once if we roll a 4 or a 6 which represents two of the six possible outcomes of rolling a fair dice.

Since there are only two choices in the game (play once or play twice), we know that $P(\textrm{play twice})$ must be the complement of the probability of playing once which would be $1 - P(\textrm{play once}) = 1 - 1/3 = 2/3$.

Putting this all together we get:

$$\begin{equation} \textrm{EP} = 10 \times 1/3 + 5.5 \times 2/3 = \boxed{7} \end{equation}$$