# Calculating growth rate of a population of Minecraft chickens

I have a rather strange question (for this Stack Exchange anyway). It felt too mathematical to ask elsewhere. If this is out of place here, please let me know.

A chicken in Minecraft lays eggs; the time between layings is uniformly distributed between 5 and 10 minutes in intervals of 0.05 seconds. An egg, when thrown, produces one chick with probability $\frac3{32}$ or four chicks with probability $\frac1{32}$, and is destroyed afterwards regardless. A chick matures into an egg-laying chicken in 20 minutes.

Assuming eggs are immediately thrown upon laying, how can I estimate the number of chickens after $X$ minutes starting with 1 chicken (that is at the start of its egg-laying cycle)? Chickens are immortal.

I can figure most of it out myself, but the thing that's giving me the most trouble is the last bit. I don't know how to take into account the time delay between an egg hatching and the chick growing up.

• The notion of branching process (Galton-Watson process, equivalently) is obviously relevant but, to give quantitative estimates, the assumption "every 5-10 minutes" should be made more precise. – Did Sep 10 '16 at 6:24
• You are describing a cycle with an expected time of $27.5$ minutes that produces $(3/32)*1+(1/32)4 = 7/32$ chicken per cycle. Now factor in a little slop between eggs being laid and harvested, and whatever other work you have to do... lets call it 0.1 chickens / hour. And as time increases it starts to become an exponential distribution. – Doug M Sep 10 '16 at 6:50
• The unconventional features of the model are the inter-laying times, uniform on the interval $(u,v)=(5,10)$, and the delay to maturity, deterministic and equal to $m=20$ (we measure everything in minutes). To simplify the computations, assume that the inter-laying times are actually exponentially distributed with the same mean inter-laying time, that is, exponential with parameter $\lambda=\frac2{u+v}$, and that the delays to maturity are exponentially distributed with mean the (true) delay to maturity, that is, exponential with parameter $\mu=1/m$. Note also that each laying ... – Did Sep 10 '16 at 12:07
• ... produces $\beta=1\cdot\frac{3}{32}+4\cdot\frac{1}{32}+0\cdot\frac{28}{32}=\frac{7}{32}$ (immature) chicken in the mean. Then, the numbers of mature chicken and of immature chicken at time $t$ are described by a bivariate branching process and the mean numbers $x(t)$ of mature chicken and $y(t)$ of immature chicken at time $t$ are such that $$x(t+dt)=x(t)+y(t)\mu dt\qquad y(t+dt)=(1-\mu dt)y(t)+\beta x(t)\lambda dt$$ that is, $$x'(t)=\mu y(t)\qquad y'(t)=\beta\lambda x(t)-\mu y(t)$$ The positive eigenvalue $\alpha$ of the matrix $$\begin{pmatrix}0&\mu\\\beta\lambda&-\mu\end{pmatrix}$$ – Did Sep 10 '16 at 12:08
• @did very interesting! You might want to make that an answer. – Daffy Sep 10 '16 at 15:05