A Puzzle A Day: 2026-04-13

Dr. Anya is studying a newly discovered microbial organism that behaves uniquely. She places a single organism in a nutrient solution. Each hour, the organism either perfectly divides into two identical organisms (with a 50% probability) or it dies (with a 50% probability). If it divides, both new organisms continue this same independent process, following the same probabilities. What is the probability that the entire lineage of organisms, starting from the single original organism, will eventually die out?
Correct: 100%
Let P be the probability that the entire lineage eventually dies out. We need to consider the fate of the initial single organism: 1. **With 50% probability, the initial organism dies.** In this scenario, the lineage immediately dies out. So, this contributes 0.5 * 1 to the total probability P. 2. **With 50% probability, the initial organism divides into two new, independent organisms.** For the *entire original lineage* to eventually die out in this case, *both* of these new organisms' lineages must also eventually die out. Since each new organism behaves identically and independently, the probability that one of their lineages dies out is P. Therefore, the probability that *both* new lineages die out is P multiplied by P (or P^2). Combining these two possibilities, we can form an equation for P: P = (0.5 * 1) + (0.5 * P^2) P = 0.5 + 0.5P^2 Now, we solve this quadratic equation for P: Multiply by 2 to clear the fraction: 2P = 1 + P^2 Rearrange into standard quadratic form: P^2 - 2P + 1 = 0 This is a perfect square trinomial: (P - 1)^2 = 0 Taking the square root of both sides: P - 1 = 0 P = 1 So, the probability that the entire lineage will eventually die out is 1, or 100%. This is often counter-intuitive, as it seems there's always a chance for the population to grow, but for the lineage to *never* die out, it must grow infinitely large, which requires an unbroken chain of divisions without any organism dying, which becomes infinitesimally improbable over infinite time.
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