This series of posts are inspired by two questions that Jarrett Byrnes asked:

*Given the extinction of E predators out of N, what is the probability that a prey species will still have at least one predator remaining?**Given E out of N, what then is the probability that***all**prey species will have at least one predator remaining?

As Jarrett and I have been discovering, these are actually quite difficult questions to answer in a general manner, i.e. for all topologies of a certain size!

Say we have a two trophic level food web with N predators, what is the probability that a prey species has **at least one predator** remaining after the extinction of E predators? The solution provided here depends on having the out-degree of prey species, and finding the probability that all predators of a prey species become extinct as a result of E. Say that the out-degree of the prey species is , then that probability is a hypergeometric solution

which reduces to

The probability then of a prey species having **at least one prey** is 1 minus the above

that is, the sum of the probabilities of having 1 predator, 2 predators, etc.

## Example

Let the adjacency matrix of a food web be

where predators are rows and prey are columns. Our prey out-degree set is therefore {3, 2}. For E=1, both prey will have at least one predator since their out-degrees both exceed 1. For E=2, the possible resulting topologies are

For s=2

This is correct since our prey species of out-degree 2 (second column of **A**) has at least one predator in two of our three post-extinction topologies. The probability should be zero for s=3 (since E<s). If we add a third prey species, with s=1, making

then for E=1, the post-extinction topologies are

The probability that this third species has at least one prey is also 2/3.

## A further example

So far so good, right? Well, Jarrett posed this example,

Notice that we now have two prey of out-degree 2. For E=1, the post-extinction topologies are

Applying the above formula yields

which is correct, since two of the three topologies maintain at least one predator for each prey. When E=2, the topologies become

Obviously, . But the formula gives

What went wrong?! The answer points to just how devilish the questions are, and how deceptive! There are two species of out-degree 2 (s=2) in the food web, hence the second term in the formula is squared (see above). BUT, the predator-prey topologies of the species are different, meaning that simple hypergeometric counting cannot work. We literally must list and examine all the post-extinction topologies, but this is prohibitively impractical for food webs and networks of even modest size (a dozen species). So there we stand. We currently have a partial solution, and I will explore the difficulty and the partial solution in the next post.

jebyrnes

said:Ha! FYI, I’m starting a series on all of this stuff here

proopnarine

said:Already spotted it dude. I’ll point my next post to your blog. I also have some ideas on reducing the Inclusion Exclusion problem that I’ll get to whenever I get to the next post.