在超市的抽奖活动中，参与者将首先转动转盘，然后掷一枚均匀的六面骰子。 转盘有 9 个区域和一根指针，指针停在 9 个区域中任意一个区域的概率均等。转盘上有一个区域标有 “$100”，三个区域标有 “$4”，五个区域标有 “$1$”。 均匀的六面骰子有两个面标有 “0.02”，三个面标有 “0.5”，一个面标有 “2”。 参与者赢得的奖金金额是通过将转盘和骰子的结果相乘来计算的。

1. Identify all the combinations of wheel and die outcomes that result in a $2 win.

The possible (wheel, die) combinations that multiply to $2 are:

• $(\char36{}100, 0.02)$ since $100 \times 0.02 = 2$
• $(\char36{}4, 0.5)$ since $4 \times 0.5 = 2$
• $(\char36{}1, 2)$ since $1 \times 2 = 2$

2. Calculate the probability for each combination and sum them to find the total probability.

$P(\text{Win } \char36{}2) = P(\char36{}100 \text{ and } 0.02) + P(\char36{}4 \text{ and } 0.5) + P(\char36{}1 \text{ and } 2)$
$P(\text{Win } \char36{}2) = \left(\frac{1}{9} \times \frac{2}{6}\right) + \left(\frac{3}{9} \times \frac{3}{6}\right) + \left(\frac{5}{9} \times \frac{1}{6}\right)$
$P(\text{Win } \char36{}2) = \frac{2}{54} + \frac{9}{54} + \frac{5}{54}$
$P(\text{Win } \char36{}2) = \frac{16}{54} = \frac{8}{27}$

1. Retrieve or calculate the total probability of winning $2 from part (a).

From part (a), the probability of winning $2 is the sum of probabilities of the mutually exclusive winning combinations:
$P(\text{Win } \char36{}2) = P(\char36{}100 \text{ and } 0.02) + P(\char36{}4 \text{ and } 0.5) + P(\char36{}1 \text{ and } 2)$
$= \left(\frac{1}{9}\right)\left(\frac{2}{6}\right) + \left(\frac{3}{9}\right)\left(\frac{3}{6}\right) + \left(\frac{5}{9}\right)\left(\frac{1}{6}\right)$
$= \frac{2}{54} + \frac{9}{54} + \frac{5}{54}$
$= \frac{16}{54} = \frac{8}{27}$

2. Calculate the probability that the participant lands on $100 AND wins $2.

The only way to win $2 when the wheel lands on $100 is for the die to show 0.02.
$P(\text{Wheel } \char36{}100 \text{ and Win } \char36{}2) = \left(\frac{1}{9}\right)\left(\frac{2}{6}\right) = \frac{2}{54} = \frac{1}{27}$

3. Apply the conditional probability formula to find the required probability.

Using the concept of conditional probability:
$P(\text{Wheel } \char36{}100 \mid \text{Win } \char36{}2) = \frac{P(\text{Wheel } \char36{}100 \text{ and Win } \char36{}2)}{P(\text{Win } \char36{}2)}$
$= \frac{1/27}{8/27}$
$= \frac{1}{8}$

1. Deduce the exact sequence of winnings required to win $6 with three different die face values.

Let the three different die face values rolled be $0.02$, $0.5$, and $2$.
Let the corresponding wheel values landed on be $W_1, W_2, W_3 \in \{1, 4, 100\}$.
Total winnings $= 0.02 W_1 + 0.5 W_2 + 2 W_3 = 6$
Since $W_3 \ge 1$, if $W_3 \ge 4$, the total will exceed $6$. Thus, we must have $W_3 = 1$.
This leaves $0.02 W_1 + 0.5 W_2 = 6 - 2(1) = 4$.
If $W_2 = 100$, the sum exceeds $4$. If $W_2 = 1$, we get $0.02 W_1 = 3.5 \implies W_1 = 175$ (impossible since the maximum wheel value is $100$).
Thus, $W_2$ must be $4$, giving $0.5(4) = 2$. This implies $0.02 W_1 = 2 \implies W_1 = 100$.
Therefore, the only way to satisfy the conditions is to win exactly $\char36{}2$ in each of the three draws via these specific combinations:

• Wheel $\char36{}100$ and Die $0.02$
• Wheel $\char36{}4$ and Die $0.5$
• Wheel $\char36{}1$ and Die $2$

2. Calculate the total probability by considering the permutations of the three specific events.

Calculate the probability of each specific combination occurring in a single draw:
$P(\char36{}100 \text{ and } 0.02) = \frac{1}{9} \times \frac{2}{6} = \frac{1}{27}$
$P(\char36{}4 \text{ and } 0.5) = \frac{3}{9} \times \frac{3}{6} = \frac{1}{6}$
$P(\char36{}1 \text{ and } 2) = \frac{5}{9} \times \frac{1}{6} = \frac{5}{54}$

Since the participant plays three consecutive draws, these 3 distinct events can occur in any order. The number of possible arrangements is $3! = 6$.

$\text{Total Probability} = 6 \times \left( \frac{1}{27} \times \frac{1}{6} \times \frac{5}{54} \right)$
$= 6 \times \frac{5}{8748}$
$= \frac{5}{1458}$

1. Calculate the expected value (average) of the multiplier from throwing the die.

$\text{Expected multiplier} = \frac{2}{6}(0.02) + \frac{3}{6}(0.5) + \frac{1}{6}(2)$

$= \frac{0.04 + 1.5 + 2}{6}$

$= \frac{3.54}{6} = 0.59$

2. Compare the expected multiplier with 1 to justify the decision.

Since the expected multiplier $0.59$ is less than $1$, throwing the die will, on average, decrease the final prize money regardless of what section the wheel came to rest over.

Therefore, he should choose to end the game and take the amount based on the wheel.