A colony begins with a cell, which can die, do nothing, transform to two or three cells, with probability 1/4 for each case at next time point. Children cells share the same property described above. What's the probability of this colony's extinction?

I got two solutions, $1$ and $\sqrt{2}-1$, from a simple recursive equation, $p=\frac{1}{4}(1+p+p^2+p^3)$. But I've no idea which one is correct. Thanks for your help!

  • $\begingroup$ It's certainly not 1, as it guarantees extinction, which is possible iff the probability to die is greater than $\frac{1}{2}$ $\endgroup$
    – Alex
    Jul 20, 2013 at 15:06
  • $\begingroup$ I forget how to prove it, but the probability of extinction is $<1$ if the expected number of offspring is $>1$. $\endgroup$ Jul 20, 2013 at 18:49

3 Answers 3


You've done most of the work; what remains is to decide whether $1$ or $\sqrt2-1$ is the probability of eventual extinction. For that purpose, let $x_n$ be the probability of the event that the population goes extinct at or before the $n$-th time step. Notice that these events form an increasing sequence with respect to $\subseteq$, so the probability of their union, the event of eventual extinction, is the supremum (and also the limit) of the increasing sequence of numbers $x_n$. The issue is therefore whether this sequence ever gets above $\sqrt2-1$.

We have $x_0=0$ (since the population is initially a single cell, not extinct), and $$ x_{n+1}=\frac14(1+x_n+{x_n}^2+{x_n}^3) $$ (because of the rules for how the cells multiply or die or do nothing). Notice that this equation says $x_{n+1}=f(x_n)$, where $f(x)=\frac14(1+x+x^2+x^3)$ is the function whose fixed points you already calculated. In particular, you know that $f(\sqrt2-1)=\sqrt2-1$. But $f(x)$ is clearly an increasing function of $x$ as long as $x\geq0$. So, since $x_0<\sqrt2-1$, we get, by induction on $n$, that $x_n<\sqrt2-1$ for all $n$. Therefore, the probability of eventual extinction is not $1$ but $\sqrt2-1$.

  • $\begingroup$ can you explain why the sign of $x_n$ is a plus and not a minus? $\endgroup$ Jul 20, 2013 at 20:26
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    $\begingroup$ @al-Hwarizmi If you mean the linear term $x_n$ in the formula for $x_{n+1}$, I can't imagine why you'd think there should be a minus sign. The probability that a cell does nothing (neither dies nor multiplies) is $1/4$, not $-1/4$. $\endgroup$ Jul 20, 2013 at 21:08
  • $\begingroup$ Isnt this the death term that cant be from the same sign as the birth terms? I just cannot follow how you derived this equation? $\endgroup$ Jul 21, 2013 at 8:56
  • $\begingroup$ @al-Hwarizmi No, the death term is the constant term 1. Each of the four possibiliites (death, remaining one cell, becoming 2 cells, and becoming 3 cells) has probability $1/4$. In the first of these 4 cases, the cell is dead, so the probability of extinction at or before $n+1$ is $1$. In the second case, it remains one cell, so the probability of extinction by $n+1$ equals the probability that the one cell existing after the first stage becomes extinct in at most $n$ additional stages, $x_n$. [continued in next comment] $\endgroup$ Jul 21, 2013 at 15:25
  • $\begingroup$ In the 3rd case, it's the probability that both of the cells existing after $1$ stage proceed to extinction in at most $n$ more stages, ${x_n}^2$. And in the 4th case it's the probability that all $3$ of the cells after the first stage proceed to extinction in at most $n$ further stages, ${x_n}^3$. All four of these cases contribute to the overall probability $x_{n+1}$ with coefficients $1/4$. (As I said before, I can't imagine why you expect a negative coefficient.) $\endgroup$ Jul 21, 2013 at 15:28

I think I can shed some light on the recursion identity.

Let's call $p$ the probability that the population starting from one cell dies out. We have 4 scenarios:

1) The cell dies with probability $1/4$, hence extinction (term $1/4$).

2) It does nothing with the same probability, hence postponing the question (term $p/4$).

3) It spawns another cell, hence we study the extinction of two populations (term $p^2/4$)

4) It spawns two cells, hence we study the extinction of three populations (term $p^3/4$)

Thus, $$p=\frac 14 (1+p+p^2+p^3),$$ which gives us the roots $1$, $-1\pm \sqrt 2$.

Now we need to chose the correct root between $1$ and $\sqrt 2-1$. My intuition suggests that thanks to the expectation to generate half a cell on each step rules out the root $p=1$ (i.e. extinction almost surely), but I might be wrong. So my money is on $p=\sqrt 2-1$, and I would be glad if someone could give a formal proof of this result (or prove me wrong, of course).

  • $\begingroup$ I am afraid, you have a time dependent process and not a stationary and will need to set up the master equations and regard the transition probabilities so the birth/death process. Then build the expectation values for the living cells a.s.f. $\endgroup$ Jul 20, 2013 at 19:23
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    $\begingroup$ @al-Hwarizmi I don't see a time-dependent process here, I study only the probabilistic events (not even Markov's chains). If I want an estimation on the number of cells at a given time, then yes, I would need some other instruments. $\endgroup$ Jul 20, 2013 at 19:28
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    $\begingroup$ @al-Hwarizmi There're no steps, there're no functions of time. I have a random event with four outcomes, each of them is characterised by a certain quantative property. I calculate the expectation of that property given the probabilities of outcomes. $\endgroup$ Jul 20, 2013 at 19:54
  • $\begingroup$ @al-Hwarizmi I don't have generations/colonies/etc. I have one event with 4 outcomes. $\endgroup$ Jul 20, 2013 at 20:02
  • $\begingroup$ Yes. Your calculation is absolute fine. I am just considering how we can take it foward to an answer to the question. $\endgroup$ Jul 20, 2013 at 20:15

I have translated this into a birt/death-process. Often one can set the master equations and derive the macroscopic relations from the microscopic. As there is no non-linearity (no liason of the cells) the rarction rates correspond to your probability of $1/4$ and we get

$$C \xrightarrow{k} D$$ $$C \xrightarrow{k} 2C$$ $$C \xrightarrow{k} 3C$$ $$C \xrightarrow{k} C$$

$$\frac{d C}{d t}=-k\,C+kC+k\,C+k\,C$$

$$\frac{d C}{d t}=2k\,C$$

Not sure if there is any threat to the colony to ever die out. $C$ in the equations represents the Expectation value (first moment) of the living cells over the population.

The above calculation holds if you start with a large population. If you start with one cell you will need to set up the master equations with the transition probailities and follow the progress of how your probability distribution changes over each time step. At the start it is $1/4$ then less and less. Larger the population gets less the probability that it dies out. Once large enough (law of large numbers) it will follow the above equations.

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    $\begingroup$ The probability of extinction is very obviously at least $1/4$. $\endgroup$ Jul 20, 2013 at 18:50
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    $\begingroup$ I have no idea what you're using the words "microscopic" and "macroscopic" to mean. $\endgroup$ Jul 20, 2013 at 18:53
  • $\begingroup$ But this is false: there is probability $1/4$ that the initial cell dies immediately. $\endgroup$ Jul 20, 2013 at 19:12
  • $\begingroup$ Have you read the question? We begin with ONE CELL. $\endgroup$ Jul 20, 2013 at 19:22
  • $\begingroup$ No matter how cells you have at a given step, there's a non-zero probability that they all die, so your word "never" is out of place here. $\endgroup$ Jul 20, 2013 at 19:41

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