# probability of dying of a bacteria

Suppose that there are some bacterias. In any minute, each living dies with probability 1/4, stands still with probability 1/4, splits into 2 with probability 1/4, and splits into 3 with probability 1/4.

What is the probability of this species dying out finally, when initially there is only one bacteria?

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Do you know about probability generating functions? – Chris Taylor Feb 2 '12 at 12:02
@Chris Taylor: I am learning about it, thanks – neticin Feb 2 '12 at 14:17
Just a point of English: One bacterium, two or more bacteria. – John Bentin Feb 2 '12 at 21:19

The probability of eventual extinction of a branching process is the smallest root in $[0,1]$ of $\phi(t)=t$, where $\phi$ is the probability generating function. In your case, $\phi(t)=(1+t+t^2+t^3)/4$, and the probability is $\sqrt{2}-1=.41421$.
When I sum up over all timesteps (in my answer), I get $\sum_{t=0}^\infty (1/4)^{(3/2)^t}\approx 0.429407$, which is a little too large. Could it be used as upper bound? – draks ... Feb 2 '12 at 14:44
@draks No, it's not that simple. I can find a different branching process with $p_0=1/4$ and $\mu=3/2$, but with arbitrarily large extinction probability. – Byron Schmuland Feb 2 '12 at 15:18
In each step, each bacteria becomes $(0+1+2+3)/4=1,5$ bacteria in the mean. So after $t$ steps, we have $1,5^t$ of them and the chance that all of them die at once is $\left(\frac{1}{4}\right)^{1,5^t}$, but I'm not a bacteria expert to say how they tend to mass extinction.