Question

一只虫子从由边长为 $1$ 的等边三角形组成的网格的一个顶点出发。每一步，虫子都沿网格线在 $6$ 个可能方向中随机且独立地等概率移动。经过 $5$ 步后，虫子在整个过程中从未距离起点超过 $1$ 个单位的概率是多少？

A bug starts at a vertex of a grid made of equilateral triangles of side length $1.$ At each step the bug moves in one of the $6$ possible directions along the grid lines randomly and independently with equal probability. What is the probability that after $5$ moves the bug never will have been more than $1$ unit away from the starting position?

$\dfrac{13}{108}$

$\dfrac{7}{54}$

$\dfrac{29}{216}$

$\dfrac{4}{27}$

$\dfrac{1}{16}$

答案：A

知识点：随机游走递推概率

难度评级：2230

解答：

保持在距离 $1$ 以内意味着虫子始终在原点或其 $6$ 个相邻顶点上。从原点出发，所有 $6$ 个方向都可行；从相邻顶点出发，只有 $3$ 个方向可行：回到原点，或到两个相邻的邻点之一。

令 $a_k$ 和 $b_k$ 分别表示有效的 $k$ 步路径中，结束在原点和结束在某个相邻点的路径数。则 $a_{k+1} = b_k$ 、 $b_{k+1} = 6a_k + 2b_k$ ，初值为 $a_0 = 1$ 、 $b_0 = 0$ 。

迭代得 $b_1 = 6$ ， $(a_2, b_2) = (6, 12)$ 、 $(a_3, b_3) = (12, 60)$ 、 $(a_4, b_4) = (60, 192)$ 、 $(a_5, b_5) = (192, 744)$ 。有效路径总数为 $192 + 744 = 936$ 。

所求概率为 $\dfrac{936}{6^5} = \dfrac{936}{7776} = \dfrac{13}{108}$ 。

所以正确答案是 A。

Staying within distance $1$ means the bug is always at the origin or one of its $6$ neighbors. From the origin, all $6$ moves are allowed. From a neighbor, only $3$ moves keep it in range: back to the origin, or to either of the two adjacent neighbors.

Let $a_k$ and $b_k$ count valid $k$ -step paths ending at the origin and at a neighbor. Then $a_{k+1} = b_k$ and $b_{k+1} = 6a_k + 2b_k,$ starting from $a_0 = 1,$ $b_0 = 0.$

Iterating gives $b_1 = 6,$ then $(a_2, b_2) = (6, 12),$ $(a_3, b_3) = (12, 60),$ $(a_4, b_4) = (60, 192),$ $(a_5, b_5) = (192, 744).$ The total number of valid paths is $192 + 744 = 936.$

The probability is $\dfrac{936}{6^5} = \dfrac{936}{7776} = \dfrac{13}{108}.$

Thus, the correct answer is A.

2021 AMC 12B Fall 第 17 题

答案：A

难度评级：2230

解答：

其他年份的第 17 题