My question is about the given solution to problem 4 in Newman's book 'A Problem Seminar'. Note that the book is available online at Springer.
Problem 4
A microbe either splits into two perfect copies of itself or else disintegrates. If the probability of splitting is p, what is the probability that one microbe will produce an everlasting colony?
Solution: We can (almost) solve this problem by expressing the unknown probability, $x$, in terms of itself. For, after all, how can the one microbe multiply forever? Only in one way. It must at the first moment split, and then at least one of its two daughters must succeed in the task of the mother (i.e., lead to an eternal progeny). In short $x= p\cdot(1-(1-x)^ 2)$. We can solve for x and get the answer! ...
From my understanding, $(1-(1-p)^2)$ means the probability of at least one microbe splitting. However, I have no idea why $x= p\cdot(1-(1-x)^ 2)$ instead of $x= p\cdot(1-(1-p)^ 2)$. How does $x$ get into the $(1-x)^2$? What does $(1-(1-x)^ 2)$ intuitively mean?
Since $x$ is the probability that a microbe produces an everlasting colony, $1-x$ is the probability that a microbe’s line will die out. Now we start with one microbe; with probability $p$ it splits. The daughter lines survive or die out independently of each other, so the probability that both daughter lines die out is $(1-x)^2$. The probability that at least one daughter line is everlasting is therefore the complementary probability $1-(1-x)^2$.
Thus, the probability that the original microbe splits and has at least one everlasting daughter line is $p\big(1-(1-x)^2\big)$. But this is just the probability that the original microbe’s line is everlasting, so $x=p\big(1-(1-x)^2\big)$.