Question

一枚公平骰子掷四次。最后三次中的每一次点数都至少与前一次一样大的概率可写成 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数。求 $m + n$ 。

A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：79

知识点：基本概率隔板法双射

难度评级：2230

解答：

四次掷骰结果必须形成非递减序列。从 $\{1, \ldots, 6\}$ 中选出的任意四个数的多重集合，都恰好有一种非递减排列，所以成功结果数等于这类多重集合的数量。由隔板法（四个星和五个隔板），数量为 $\binom{9}{4} = 126$ 。

概率为 $\frac{126}{6^4} = \frac{126}{1296} = \frac{7}{72}$ ，所以 $m + n = 7 + 72 = 79$ 。

The rolls must form a non-decreasing sequence. Every multiset of four values from $\{1, \ldots, 6\}$ can be arranged in non-decreasing order in exactly one way, so the number of successful outcomes equals the number of such multisets. By stars and bars (4 stars and 5 dividers), that count is $\binom{9}{4} = 126.$

The probability is $\frac{126}{6^4} = \frac{126}{1296} = \frac{7}{72},$ so $m + n = 7 + 72 = 79.$

2001 AIME I 第 6 题

答案：79

难度评级：2230

解答：

其他年份的第 6 题