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变形虫繁殖最终灭绝概率

Amoeba Population

专题
Probability / 概率
难度
L4

题目详情

池塘里最初有 1 个变形虫。每分钟它会以概率 1/41/4 发生以下之一：

  • 死亡（变为 0 个）；
  • 保持 1 个；
  • 分裂成 2 个；
  • 分裂成 3 个。

后代彼此独立并遵循同样规则。

问：最终种群灭绝（变为 0 个并保持为 0）的概率是多少？

One amoeba is in a pond. After every minute it may (each with probability 14\frac{1}{4}):

  • Die (becomes 0),
  • Stay 1,
  • Split into 2,
  • Split into 3.
    All offspring behave identically and independently. What is the probability that the amoeba population eventually goes extinct?
解析

设灭绝概率为 pp。则下一分钟后：

p=14⋅1+14⋅p+14⋅p2+14⋅p3.p=\tfrac14\cdot 1+\tfrac14\cdot p+\tfrac14\cdot p^2+\tfrac14\cdot p^3.

在 (0,1)(0,1) 中解该方程得

p=2−1≈0.4142.p=\sqrt2-1\approx 0.4142.

Original Explanation

Let P(E)P(E) be the extinction probability. Then P(E)=14×1+14×P(E)+14×[P(E)]2+14×[P(E)]3.P(E) = \tfrac{1}{4}\times 1 + \tfrac{1}{4}\times P(E) + \tfrac{1}{4}\times [P(E)]^2 + \tfrac{1}{4}\times [P(E)]^3. Solving for P(E)P(E) in (0,1)(0,1) yields P(E)=2−1≈0.4142.P(E) = \sqrt{2} - 1 \approx 0.4142.