TITLE: Modeling the Decay of a Pack of Cannibalistic Hyenas
QUESTION [7 upvotes]: A population of $p_0$ hyenas has run out of food in their ecosystem, and so sadly they have resorted to eating each other.
Hyenas need to consume one meal a day, and so exactly once per day, any given hyena will kill another hyena.  The time at which this happens is random, meaning each hyena's mealtime is uniformly distributed throughout a set of $24$ hours.
Assuming that the last hyena will get hungry and die trying to eat himself, after how long will the population of hyenas become extinct?

I'm can think of two different ways to answer this (neither of which I know how to solve).  The first is simpler and less accurate (and will therefore merit less glory).
Answer 1
We can discretize time into seconds, and so the population $p(t)$ hyenas after $t$ seconds can be written as:
$$p(t) = p(t-1) - \frac{p(t-1)}{86400}.$$
(Why?   At any given second--there are $60*60*24 = 86400$ seconds in a day--each hyena has a $1$ in $86400$ chance of eating another hyena.)
Clearly this is sloppy, since it is not guaranteed that such a number of hyenas will die every second.  Also, when one hyena remains, it gets messy (although you could say it will take him a full day to die).  I think this is actually modelling the expected value of $p(t)$, but I'm not sure.
But still, I'd like to see how we can get a clean formula for $p(t)$ from this recursion, and see when it becomes $0$.
Answer 2
We think of time as continuous, and the cannibalism of the hyenas as a (Poison?) process, in which each event--the death of a hyena--occurs at some rate.  The tricky part is that this rate is dependent on the current population of the hyenas...
I assume the solution will be given by $E[t | p(t) = 0]$.

I've been thinking about this for a while, and am genuinely curious to see what you think!

REPLY [2 votes]: Here is a sketch of an approach that should show that the number of days to termination is asymptotic to $c\log (n)$ for some computable $c$. I'm a bit busy now, but I can add more details later if necessary.
Let $n$ be the number of hyenas at the beginning of a day. It suffices to show that the fraction of hyenas surviving to the next day is asymptotic to some constant.
Arrange the hyenas from left to right by their planned meal times. Now construct the directed graph on the hyenas where an edge points from hyena $A$ to hyena $B$ if $A$ eats $B$. Observe that every valid graph is equally probable.
What does this graph look like? A component consists of a path of leftward edges terminating in a single rightward edge. The number of surviving hyenas is the number of components in the graph.
Let the number of such graphs be $f(n)$. We can count the number of such graphs with an EGF. We will find that $f(n+1)/f(n)\sim \alpha n^d$ for some constants $\alpha$ and $d$.
Now we can calculate the expected number of components of size $k$, since we can count the number of ways to choose a component of size $k$ and then fill in the rest of the graph in $f(n-k)\sim \alpha^d n^{kd}f(n)$ ways. Summing over $k$ will yield the expected number of components, which should be asymptotically $cn$ for some constant $c$.
To prove the desired concentration result, we can use Chebyshev's inequality. Calculating the variance of the number of components is equivalent to counting the expected number of pairs of components.