# What is the probability of extinction?

In every second, the probability that a certain bacterium will die is p and the probability that it will divide into two identical bacteria is 1-p. (The offsprings will die or divide independently of each other.) What is the probability of extinction?

Relevance

This has already been answered correctly here, but I'll offer the summation for the exact probability:

p( ∑(n = 0 to ∞) (p(1-p))^n) ( (2n)!/(n!(n+1)!) ) ) = p/(1-p)

so that for p = 1/2, the probability is 1. For p < 1/2, the probability is < 1, so that for p = 0, the probability is 0.

We can look at this as an asymmetrical random walk, where increasing the bacteria population by 1 has probability p, and decreasing the same by 1 has probability (1-p). For every increase there has to be a decrease, hence the (p(1-p))^n term. The (2n)!/(n!(n+1)!) term counts the different ways a 2n step path can be done.

Edit: Fixed summation to show that n starts at 0, not 1, excuse me.

Edit 2: I'm still uncomfortable about this result, since my first guess was the same as First Grade Rocks!'s answer, "certain exinction, if given enough time". I keep thinking maybe I have made a mistake or overlooked something.

If the probability of the line associated with a single bacterium will die out is E, then

E = p + (1-p) probability both offspring will die

E = p + (1-p) E^2

Thus: (1-p)E^2 - E + p = 0

Given p, you can solve for E. (Note that E=1 is always one root. Use the other if it is in [0,1].)

For p = 1/2, E = 1, which is what you'd expect:

50% chance of extinction in the the first generation,

25% chance of extinction in the second generation,

etc.

In fact, whenever p>= 1/2, the probability of extinction should be 1.

This is a branching process. The way to get the extinction probability is to first find the PGF (probability generating function) as: ((1-p) + pt^2) and solve for the smallest positive root in the equation:

(p + (1-p)t^2) = t

⇒ (1-p)t^2 - t + p = 0

[1 +/- (1-2p)]/[2(1-p)]

i.e., p/(1-p) or 1

If p > 0.5, then p/(1-p) > 1 i.e., prob of extinction = 1.

• Anonymous

If at time t0 the population is N0, then after i seconds the expected population is Ni = N0*(2^i)*(1-p)^i.

After a lot of seconds i→∝, the population will be close to Ni:

1) If p<=1/2 (and N0 big enough), then the probability of extinction is a little bigger than p^N0

2) If p>1/2 then the probability of extinction is 100%

Using simplicitus's notation, where E is the probability of a bacterium's lineage dying out, then

E = min( 1, p / (1-p) ).

If you started with some initial population of bactiera A0, then

E(A0) = E^A0 = probability of all A0 lineages dying out

= min( 1, (p / (1-p))^A0 )

where E(A0) is the probability of extinction starting with an initial population of A0.