病蚂蚁传播：最终感染数

Sick ants

专题: Probability / 概率
难度: L4
来源: QuantQuestion

题目详情

A total of 25 ants are scattered on a horizontal meter stick. Simultaneously, each ant picks a random direction (left or right) independently of the others, and starts marching at $1\mathrm{cm}/$ second in the chosen direction. Whenever two ants meet, they switch the directions of their movement. When an ant reaches the end of the stick, it falls off. The ant in the middle is sick with cold, A sick ant will spread the cold to any ant it meets. By the time all ants have fallen off the stick, what is the expected number of sick ants?

解析

相遇时“交换方向”等价于“相互穿过但交换身份”，因此只需追踪“感冒”在相遇时沿着路径传递。

设中点病蚂蚁为 A。右侧有 12 只蚂蚁，其中朝左走的都会先与 A 相遇而感染，并把感冒继续传给左侧朝右走的蚂蚁。

令

$X$ ：A 右侧中一开始朝左走的只数，则 $\mathbb{E}[X]=12\cdot\tfrac12=6$
$Y$ ：A 左侧中一开始朝右走的只数，则 $\mathbb{E}[Y]=6$

若 $X=0$ （概率 $(1/2)^{12}$ ），右侧全向右，永不与 A 相遇，左侧也不会被感染；若 $X\ne 0$ ，左侧朝右者最终都会被感染。

因此最终感染总数 $N$ 满足

\mathbb{E}[N]=1+\mathbb{E}[X]+\mathbb{E}[Y]\,\mathbb{P}(X\ne 0) =1+6+6\left(1-2^{-12}\right).

即

\boxed{\mathbb{E}[N]=13-\frac{6}{2^{12}}}.