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A single-celled organism is placed into an infinite culture. This cell has the genetic property that it has a $p = 3/4$ chance of splitting into two cells before it dies, and all its children will have the exact same property.

What is the probability that the population spawned by this single cell will last forever? How does this value vary with $p$?

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Let the probability of survival be $x$. The culture survives if the cell splits and the descendants of at least one daughter cell survive. The probability of at least one of two cultures surviving is $2x-x^2$ by inclusion-exclusion, so $x=p(2x-x^2)$. This solves to $x=0$ or $x=2-\frac{1}{p}$.

So the answer is

$2-\frac{1}{p}$ if $p>\frac{1}{2}$ and $0$ otherwise.

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  • $\begingroup$ Hmm. I remember the puzzle book I got this out of having a much more complicated explanation than that. However, the answer it gives (2/3) is correct. $\endgroup$ – Joe Z. May 30 '16 at 20:19
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    $\begingroup$ Is there room to elaborate on this answer, or am I being dense? I'm confused by the second sentence, which seems to imply that survival to the third generation guarantees population survival (did you mean "at least one daughter cell splits"?). $\endgroup$ – Roland May 31 '16 at 17:52
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To show how f" arrived at the first equation:

  • For one cell, let the probability of spawning an eternal population be $x$.

  • For a cell to spawn an eternal population, it must first split with probability $p$.

  • Then, at least one of its offspring must spawn an eternal population with probability $x$ each. The probability of neither offspring doing so is $(1-x)^2$, so the probability of at least one doing so is $1-(1-x)^2$, rewritten as $2x-x^2$.

  • Therefore, the probability of spawning an eternal population is equal to the probability of splitting and having at least one offspring spawn an eternal population; $x = p*(2x-x^2)$

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