$1/2$ or $1$? probability that all bacteria will die 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.
 A: 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.
A: 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.
A: 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
$$
p_4 = 1/3 + 2/3 \cdot (1/3 + 2/3\cdot (1/3 + 2/3\cdot (1/3)^2)^2)^2
$$
A: $\newcommand{\+}{^{\dagger}}%
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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.

