2020 AMC 10A 第 13 题

Question

一只青蛙坐在点

(1, 2)

，开始连续跳跃。每次跳跃都平行于某条坐标轴，长度为

1

，方向为上、下、右、左之一，且每次独立随机选择。当青蛙到达以

(0, 0)

、

(0, 4)

、

(4, 4)

和

(4, 0)

为顶点的正方形的一条边时，跳跃序列结束。跳跃序列在正方形的竖直边上结束的概率是多少？

A frog sitting at the point

(1, 2)

begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length

1,

and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices

(0, 0),

(0, 4),

(4, 4),

and

(4, 0).

What is the probability that the sequence of jumps ends on a vertical side of the square?

\displaystyle \frac{1}{2}

\displaystyle \frac{5}{8}

\displaystyle \frac{2}{3}

\displaystyle \frac{3}{4}

\displaystyle \frac{7}{8}

答案：B

知识点：随机游走方程组对称性

难度评级：1950

小提示：

让每个内部格点表示先碰到竖直边的概率。

Let each interior lattice point store the probability of hitting a vertical side first

大提示：

用对称性把随机游走方程减少到四个未知数。

Use symmetry to reduce the random-walk equations to four unknowns

视频讲解：

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文字解答：

设

p(x,y)

为从

(x,y)

出发最终先碰到竖直边的概率。由对称性，令

a=p(1,1)=p(1,3)

、

b=p(2,1)=p(2,3)

、

c=p(1,2)

，并令

d=p(2,2)

。

每一步取四个相邻点概率的平均，得到

a=\dfrac{1+b+c}{4}

、

b=\dfrac{2a+d}{4}

、

c=\dfrac{1+2a+d}{4}

，以及

d=\dfrac{b+c}{2}

。解得

c=\dfrac58

。起点为

(1,2)

，所以这就是所求概率。正确答案是 B。

Let

p(x,y)

be the probability of eventually hitting a vertical side first from point

(x,y)

. By symmetry, set

a=p(1,1)=p(1,3)

,

b=p(2,1)=p(2,3)

,

c=p(1,2)

, and

d=p(2,2)

.

The averaging equations are

a=\dfrac{1+b+c}{4}

,

b=\dfrac{2a+d}{4}

,

c=\dfrac{1+2a+d}{4}

, and

d=\dfrac{b+c}{2}

. Solving gives

c=\dfrac58

, which is the desired probability from

(1,2)

. Thus, B is the correct answer.