Suppose there is a bacterium in a bottle, it has $\frac{1}{3}$ chance to die and it has $\frac{2}{3}$ chance to split into 2 individuals, and the new individuals will follow this rule and so on. So here is the question, what is the probability that all bacteria are dead in the bottle?
Denote by p the probability that all the bacteria are dead. $$ p =\frac{1}{3}+\frac{2}{3}p^2$$ and it gives that $p = 0.5$ or $1$, so what is the next step? Which one is the answer? thanks.
Another way to write this is as $$ p_{k}=\frac{1}{3}+\frac{2}{3}p_{k-1}^2, $$ or $$ p_{k}-p_{k-1}=\frac{2}{3}p_{k-1}^2-p_{k-1}+\frac{1}{3}=\frac{2}{3}\left(p_{k-1}-1\right)\left(p_{k}-\frac{1}{2}\right), $$ where $p_{k}$ is the probability that a bacteria and all its descendants are dead after $k$ generations. So if $p_{k}\in[0,1/2)$, $p_{k+1}>p_{k}$ (that is, the probability increases with each additional generation), while if $p_{k}\in(1/2,1)$, $p_{k+1}< p_{k}$ (the probability decreases). From this you can see that there are exactly two fixed points, at $p=1/2$ and $p=1$, and that the fixed point at $p=1/2$ is the attractive one.
This is a supercritical Galton-Watson process (supercritical meaning each individual has an average of more than one offspring). It's a classical result that such a process survives with positive probability, so your p cannot be 1 and must be 1/2. You can probably find a proof in many places; I know it's in Durrett's PTE.
So we know that if there is a valid probability $p$, it must satisfy the equation $$ p = \frac 13 + \frac 23 p^2 $$ That is: for the first cell that splits, we note that in order for all trace of the cell to be gone after a certain time, either the first cell has to die, or the first cell splits and the trace of the offspring is gone after a certain time.
For any cell, $q$ is the probability that all trace of the cell is gone after a long enough time.
So, we conclude that $p$ satisfies the above equation. We have $$ 2 p^2 - 3p + 1 = 0\\ (2p - 1)(p - 1) = 0\\ p = \frac 12, \quad p = 1 $$ So why can't the probability be $p=1$? The answer really lies in the "limit".
What is the probability that all cells are gone after one generation? That's just $$ p_1 = 1/3 $$ Two generations? $$ p_2 = 1/3 + 2/3\cdot (1/3)^2 $$ Three generations? $$ p_3 = 1/3 + 2/3 \cdot (1/3 + 2/3\cdot (1/3)^2)^2 $$ And so on $$ p_4 = 1/3 + 2/3 \cdot (1/3 + 2/3\cdot (1/3 + 2/3\cdot (1/3)^2)^2)^2 $$
$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ Let's $p_{-}$ and $p_{+}$ the probability a bacteria dies or split in two bacteria, respectively. The probability of a bacteria generates $n$ bacteria will be $$P_{n} = p_{-}\delta_{n,0}\ +\ p_{+}\delta_{n,2}\tag{1}$$ Given a population of $N$ bacterias we'll calculate the probability ${\cal P}_{N \to N'}$ the population will be $N'$. ${\cal P}_{N \to N'}$ is given by:
\begin{align} {\cal P}_{N \to N'}&=\sum_{n_{1}=0}^{\infty}P_{n_{1}}\ldots \sum_{n_{N}=0}^{\infty}P_{n_{N}}\,\delta_{n_{1} + \cdots + n_{N},N'} \\[3mm]&=\sum_{n_{1}=0}^{\infty}P_{n_{1}}\ldots\sum_{n_{N}=0}^{\infty}P_{n_{N}}\ \overbrace{\int_{\verts{z} = 1}{1 \over z^{-n_{1} - \cdots - n_{N} + N' + 1}} \,{\dd z \over 2\pi\ic}}^{\ds{\delta_{n_{1} + \cdots + n_{N},N'}}} \\[3mm]&=\int_{\verts{z} =1}\pars{\sum_{n = 0}^{\infty}P_{n}z^{n}}^{N}\, {1 \over z^{N' + 1}}\,{\dd z \over 2\pi\ic} =\int_{\verts{z} =1}\pars{p_{-} + p_{+}z^{2}}^{N}\,{1 \over z^{N' + 1}} \,{\dd z \over 2\pi\ic} \end{align} where we used expression $\pars{1}$.
\begin{align} {\cal P}_{N \to N'}&= \int_{\verts{z} = 1}\sum_{\ell = 0}^{N} {N \choose \ell}p_{-}^{N - \ell}\pars{p_{+}z^{2}}^{\ell}\,{1 \over z^{N' + 1}} \,{\dd z \over 2\pi\ic} \\[3mm]&=\sum_{\ell = 0}^{N}{N \choose \ell}p_{-}^{N -\ell}p_{+}^{\ell}\ \overbrace{\int_{\verts{z} = 1}{1 \over z^{N' + 1 - 2\ell}}\,{\dd z \over 2\pi\ic}} ^{\ds{\delta_{2\ell,N'}}}\tag{2} \end{align}
From expression $\pars{2}$ we conclude: $$ {\cal P}_{N \to N'} =\left\lbrace% \begin{array}{lcl} {N \choose N'/2}p_{-}^{N - N'/2}\ p_{+}^{N'/2} & \mbox{if} & N'\ \mbox{is even and}\ 0 \leq N' \leq 2N \\[2mm] 0, &&\mbox{otherwise} \end{array}\right. $$
For the present question $$ p_{-}^{N - N'/2}\ p_{+}^{N'/2} = \pars{1 \over 3}^{N - N'/2}\pars{2 \over 3}^{N'/2} = {2^{N'/2} \over 3^{N}} $$ $$\color{#00f}{\large% {\cal P}_{N \to N'} =\left\lbrace% \begin{array}{lcl} {1 \over 3^{N}}\,{N \choose N'/2}2^{N'/2} & \mbox{if} & N'\ \mbox{is even and}\ 0 \leq N' \leq 2N \\[2mm] 0, &&\mbox{otherwise} \end{array}\right.} $$
With this expression for ${\cal P}_{N \to N'}$ we can answer many questions about the bacteria population.
