2006 AMC 10A 第 13 题

Question

玩家支付

\$5

玩一个游戏。掷一个骰子。如果掷出的数是奇数，游戏失败。如果掷出的数是偶数，则再掷一次骰子。此时若第二次的数与第一次相同，玩家获胜，否则失败。若游戏公平，玩家获胜时应赢得多少钱？（在公平游戏中，获胜概率乘以奖金等于玩家应支付的费用。）

A player pays

\$5

to play a game. A die is rolled. If the number on the die is odd, the game is lost. If the number on the die is even, the die is rolled again. In this case the player wins if the second number matches the first and loses otherwise. How much should the player win if the game is fair? (In a fair game the probability of winning times the amount won is what the player should pay.)

\$12

\$30

\$50

\$60

\$100

答案：D

知识点：期望值骰子（概率）

难度评级：1390

小提示：

求获胜概率：第一次掷到偶数，然后第二次匹配。

Find the probability of winning: an even roll, then a matching roll

大提示：

获胜概率为

\frac12 \cdot \frac16 = \frac{1}{12}

；令“概率乘奖金”等于

5

。

The win probability is

\frac12 \cdot \frac16 = \frac{1}{12};

set (probability)(prize) equal to

5

解答：

玩家只有在第一次掷到偶数（概率

\frac12

）且第二次与第一次相同（概率

\frac16

）时获胜，所以获胜概率为

\frac12 \cdot \frac16 = \frac{1}{12}

。

公平游戏满足

\frac{1}{12} x = 5

，所以

x = 60

。

所以正确答案是 D。

The player wins only if the first roll is even (probability

\frac12

) and the second roll matches it (probability

\frac16

), so the win probability is

\frac12 \cdot \frac16 = \frac{1}{12}.

For a fair game,

\frac{1}{12} x = 5,

so

x = 60.

Thus, the correct answer is D.

2006 AMC 10A 第 13 题

答案：D

难度评级：1390

小提示：

大提示：

解答：

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