# When killing rabbits doesn't stop them: a mathematical horror story

Backstory

Fibonacci's famous model (on January 1 there is one baby rabbit. Every rabbit of age > 1 month will have one baby on the first day of every month) predicts an incredibly fast growing populations of rabbits with (still) 1 rabbit on February 1, 2 rabbits on March 1, 144 rabbits on December 1 and 233 rabbits on January 1 of the next year.

I want to have a more 'realistic' model, where growth is somewhat less extreme. The 'obvious' weak point to attack in Fibonacci's model is the fact that rabbits never die. So I propose the following improved model:

• At January 1, there is one rabbit.
• At the 15th day of each month, each rabbit that was alive already at the 1st of that month will have one child.
• At the 28th of each month, every rabbit that was alive on the 1st day of the previous month will die. (So from April 29 onwards, no living rabbit will remember February anymore, from May 29 onwards, no rabbit will remember March, etc.)
• It just so happens that the rabbit we started with at January 1 was already alive at December 1.

Sounds a lot more realistic, right? So let's calculate the number of rabbits throughout the year.

January 1: 1 rabbit

February 1: 1 rabbit: our old rabbit has died 4 days ago, but we have a 17 day old replacement

March 1: 2 rabbits: our young rabbit is still alive and has a a child

April 1: 3 rabbits: both rabbits have given birth, but one died, $4-1 = 3$

May 1: 5 rabbits: same reasoning, $2*3 - 1 = 5$

June 1: 8 rabbits: same reasoning, but now 2 die, $2*5-2 = 8$

July 1: 13 rabbits

August 1: 21 rabbits

September 1: 34 rabbits

October 1: 55 rabbits

November 1: 89 rabbits

December 1: 144 rabbits

January 1: 233 rabbits

Sounds familiar, no? What is going on here? Are these rabbits completely impervious to death? Why do they keep coming in exactly the same hordes as in the previous model?

More precise question

Well, comparing the two models there it is obvious what happens:

Allowing the rabbits to start having children 2 weeks earlier than in the previous model completely cancels out the effect of the introduction of death.

Checking the numbers, this is correct. But superficially speaking it sounds exactly like

'Doubling the the value of one number in a multiplication completely cancels out the effect of the other number in the multiplication being set to zero'

which is wrong. I always thought that being mortal is quite a big deal, not something you can easily 'cure' by halving your waiting time to have children, but apparently my intuition is being mislead somewhere.

So my question is:

I find the fact that reducing the number of children a given rabbit will have by an infinite amount and reducing the time it spends without children by a tiny finite amount can cancel each other out exactly on a population level, highly counter-intuitive. I can prove it, but not understand it. How would you explain it?

• The world’s human population continues to grow despite hordes of people dying each day. – amd Jul 12 '18 at 20:48
• If, in your model, you do away with death, then the population will double on the 15th of every month, beginning January 15. So death does slow the rate of population increase. What you've described is nonetheless curious: death after two procreations has the same effect as immortality with a two-month delay in maturity. – Barry Cipra Jul 12 '18 at 21:05
• @BarryCipra In a finite sequence that grows exponentially like the Fibonacci sequence, the last term will be comparable in magnitude to the sum of all the previous ones, so it's not particularly remarkable to my mind that trading mortality for older, smaller generations couldn't cancel immediate maturity for the most recent, large generation. – Connor Harris Jul 12 '18 at 21:10
• @BarryCipra No, only humans – Vincent Jul 12 '18 at 21:12
• @ConnorHarris, I prefer to think (metem)psychotically: Each rabbit's first procreation adds a new soul, but each rabbit passes its own soul onto its second procreation. – Barry Cipra Jul 12 '18 at 21:20

Your realistic model boils down to doubling a Fibonacci number and subtracting the number $2$ numbers prior which also gives the next number as the original Fibonacci series. It is easy to see why this is the same as adding the previous number.
$3, 5, 8, 13, 21$
Lets take $(21 \cdot 2) - 8 = 34$
$2\cdot 21 = 21 + 21$ which essentially comes from $8+13+8+13$. So if we subtract the number $2$ numbers prior it becomes $13+8+13 = 34$ which is identical to $13 + 21$.