Amoeba Population

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Problem Statement

There is one amoeba in a pond. After every minute the amoeba may die, stay the same, split into two or split into three with equal probability. All its offspring, if it has any, will behave the same (and independent of other amoebas). What is the probability the amoeba population will die out?

Solution

This is a classic branching process problem that can be solved using conditional probability and the Law of Total Probability.

Step 1: Recognize the Problem Structure

This is a branching process (also called a Galton-Watson process), where:

  • Each amoeba acts independently

  • Each amoeba has the same probabilistic behavior

  • We want to find the extinction probability

Step 2: Set Up the Conditional Probability Framework

Let P(E) be the probability that the amoeba population will die out (extinction probability).

We apply the Law of Total Probability conditioned on what happens to the amoeba in the first minute:

P(E) = \sum_{i=1}^{n} P(E \mid F_i) P(F_i)

where F_i are mutually exclusive events covering all possibilities.

Step 3: Identify the Possible Events

After one minute, there are four mutually exclusive events, each with probability \frac{1}{4}:

  1. F_1: The amoeba dies

  2. F_2: The amoeba stays the same (one amoeba remains)

  3. F_3: The amoeba splits into two

  4. F_4: The amoeba splits into three

Step 4: Calculate Conditional Probabilities

Event F_1: The amoeba dies

If the amoeba dies, there are no amoebas left, so extinction is certain:

P(E \mid F_1) = 1

Event F_2: The amoeba stays the same

If the amoeba stays the same, we’re back to the initial state (one amoeba). Therefore:

P(E \mid F_2) = P(E)

Event F_3: The amoeba splits into two

If the amoeba splits into two, we now have two independent amoebas. For the population to die out, both amoebas must die out. Since they behave independently:

P(E \mid F_3) = P(E) \cdot P(E) = P(E)^2

Event F_4: The amoeba splits into three

If the amoeba splits into three, we now have three independent amoebas. For the population to die out, all three amoebas must die out. Since they behave independently:

P(E \mid F_4) = P(E) \cdot P(E) \cdot P(E) = P(E)^3

Step 5: Set Up the Equation

Using the Law of Total Probability:

P(E) = P(E \mid F_1)P(F_1) + P(E \mid F_2)P(F_2) + P(E \mid F_3)P(F_3) + P(E \mid F_4)P(F_4)

Substituting the values:

P(E) = 1 \cdot \frac{1}{4} + P(E) \cdot \frac{1}{4} + P(E)^2 \cdot \frac{1}{4} + P(E)^3 \cdot \frac{1}{4}
P(E) = \frac{1}{4} + \frac{1}{4}P(E) + \frac{1}{4}P(E)^2 + \frac{1}{4}P(E)^3

Multiplying both sides by 4:

4P(E) = 1 + P(E) + P(E)^2 + P(E)^3

Rearranging:

4P(E) - P(E) = 1 + P(E)^2 + P(E)^3
3P(E) = 1 + P(E)^2 + P(E)^3

Bringing all terms to one side:

P(E)^3 + P(E)^2 - 3P(E) + 1 = 0

Step 6: Solve the Cubic Equation

We need to solve:

P(E)^3 + P(E)^2 - 3P(E) + 1 = 0

with the constraint 0 < P(E) < 1 (extinction probability must be between 0 and 1, and we assume it’s not 0 or 1).

Let’s check if P(E) = 1 is a root:

1^3 + 1^2 - 3(1) + 1 = 1 + 1 - 3 + 1 = 0

So P(E) = 1 is a root. We can factor:

(P(E) - 1)(P(E)^2 + 2P(E) - 1) = 0

Solving the quadratic:

P(E)^2 + 2P(E) - 1 = 0

Using the quadratic formula:

P(E) = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2}

Since P(E) > 0, we take:

P(E) = -1 + \sqrt{2} = \sqrt{2} - 1

The other root is -1 - \sqrt{2} < 0, which we discard.

Step 7: Verify the Solution

We have three roots:

  1. P(E) = 1

  2. P(E) = \sqrt{2} - 1 \approx 0.414

  3. P(E) = -1 - \sqrt{2} < 0 (discarded)

For a branching process, we typically want the smallest positive root that is less than 1, which is \sqrt{2} - 1.

Let’s verify: \sqrt{2} - 1 \approx 1.414 - 1 = 0.414

Step 8: Final Answer

P(E) = \sqrt{2} - 1 \approx 0.414

Interpretation

The extinction probability is approximately 41.4%. This means:

  • There’s about a 41.4% chance the population will eventually die out

  • There’s about a 58.6% chance the population will survive indefinitely (grow without bound)

Key Insights

  1. Law of Total Probability: This problem is solved by conditioning on the first step and using the Law of Total Probability.

  2. Recursive Structure: The extinction probability has a recursive structure - the probability of extinction from the current state depends on extinction probabilities from future states.

  3. Independence: When multiple amoebas exist, their fates are independent, so we multiply probabilities.

  4. Branching Processes: This is a classic example of a branching process, which models population dynamics, nuclear chain reactions, and other phenomena.

  5. Cubic Equations: The problem reduces to solving a cubic equation, which is typical for branching processes with up to 3 offspring.

  6. Multiple Solutions: The cubic equation has multiple roots, but we select the appropriate one based on the problem context (probability between 0 and 1).

Generalization

For a branching process where each individual can produce 0, 1, 2, \ldots, k offspring with probabilities p_0, p_1, p_2, \ldots, p_k, the extinction probability P(E) satisfies:

P(E) = \sum_{i=0}^{k} p_i [P(E)]^i

This is because:

  • With probability p_i, an individual produces i offspring

  • For extinction, all i offspring (and their descendants) must die out

  • Since they’re independent: P(\text{all } i \text{ die out}) = [P(E)]^i

The solution involves finding the smallest positive root of this polynomial equation in the interval (0, 1).