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A jar begins with one bacteria. Every minute, every bacteria turns into 0, 1, 2, or 3 bacteria with a probability of 25% for each case (dies, does nothing, splits into 2, or splits into 3). What is the probability that the bacteria population eventually dies out?

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marked as duplicate by hardmath, rschwieb, Zev Chonoles Feb 22 '13 at 14:36

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Just to have a comparison with a later answer, a quick computational simulation gave me $p\approx 0.5865$ for staying alive. – cmmndy Feb 22 '13 at 12:21
It might be useful (??) to observe that the rate of growth is expected to be $1.5$: in fact, the expected contribution to the population of each bacterium every minute is $$ -1\cdot\frac 14 + 0\cdot\frac 14 + 1\cdot\frac 14 + 2\cdot\frac 14 = + 0.5 $$ Conditioning on the fact that the population has not extinguished up to time $n$, the expected number $b_n$ of bacteria at time $n$ (starting from time $0$) is $$ \mathbb{E}\big(~b_n \,\big|\, b_1,\,\ldots\,,b_{n-1}\neq 0~\big) ~=~ (1.5)^n $$ – AndreasT Feb 22 '13 at 12:41

1 Answer 1

Let $p$ be the probability of the bacteria dying out(it also means tht none of its offspring remains alive),

So the probability of this bacteria dying out = either itself dies out or the bacterias generated by it also dies out

So we have,$p=\frac{1}{4}+\frac{1}{4}p+\frac{1}{4}p^2+\frac{1}{4}p^3$

(Reason: Cases may be either the bacteria itself dies out then there is no question of its offspring, or it remains as it is(with the probaility ($1/4$ ) and then it itself dies out (along with its next generation with the probability $p$ ,or it turns into 2 (with the probaility ($1/4$ )and then both of these diesout with its next generation with prob. $p^2$ and similarly for the case when it turns into 3 bacteria.)

$\Rightarrow p^3+p^2-3p+1=0$

$\Rightarrow (p-1)(p^2+2p-1)=0$

$\Rightarrow (p^2+2p-1)=0$(As $p\ne1$ otherwise 'everything falls apart'.Truly, I cant see why it cant be 1, just my mind says so)

$\Rightarrow p=-1+\sqrt{2}$or $p=-1-\sqrt{2}$

As p cant be -ve so the answer is $p=\sqrt{2}-1$

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It's not immediately clear to me why it couldn't be the case that $p=1$. How are you ruling that out? – Zev Chonoles Feb 22 '13 at 12:53
@AbhraAbirKundu you mean smaller than 1? – cmmndy Feb 22 '13 at 13:12
Very sorry it will be smaller than 1 – Abhra Abir Kundu Feb 22 '13 at 13:14
@ZevChonoles: I'd have a non rigorous, rather intuitive, answer to the fact that the probability is not $1$. If $b_n$ denotes the population at time $n$, then the probability that eventually it dies is $$ \sum_{n=1}^\infty \mathbb P(b_n=0\,|\,b_{n-1}\neq 0) $$ Now, if we know $b_{n-1}$, then $$ \mathbb P(b_n=0\,|\,b_{n-1}\neq 0) = \frac{1}{4^{b_{n-1}}} $$ If we replace $b_{n-1}$ by its expected value $(1.5)^n$, then the probability to extinguish is approx. $$ ~\approx~ \sum_{n=0}^\infty \frac{1}{4^{(1.5)^n}} ~\approx~ 0.42941 $$ Then one should discuss what we mean by $\approx$... – AndreasT Feb 22 '13 at 13:41
Truly, I cant see why it cant be 1... Let $q_n$ denote the probability that the $n$th generation is empty, then $q_0=0$, $q_{n+1}=f(q_n)$ for the function $f$ one knows, and the probability of extinction is $q=\lim\limits_{n\to\infty}q_n$. Obviously $q$ is a fixed point of $f$, that is, $q=f(q)$. For every fixed point $x\geqslant0$ of $f$, $q_0\leqslant x$ and $f$ is increasing hence $q_n\leqslant x$ for every $n$ hence $q\leqslant x$. Thus $q$ must be the smallest nonnegative fixed point. – Did Feb 22 '13 at 14:49

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