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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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Since you are new, I want to give some advice about the site: To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people tend to be more willing to help you if you show that you've tried the problem yourself. – Zev Chonoles Feb 22 '13 at 12:07
Just to have a comparison with a later answer, a quick computational simulation gave me $p\approx 0.5865$ for staying alive. – k1next 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

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? – k1next 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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