Silly Malthusian probability question

General Questions
#1

On another thread, someone said “if on a specified day, everyone on earth kills one other person, we would cut the global population in half in one day”.

Now, I am fairly certain the comment was made in jest, but it got me to wondering. That figure seems somehow wrong, how would one calculate the actual outcome?

Presumably, random, unscheduled murders would take place throughout the day, so there is a non-zero probability that when Susan kills Edmund around tea-time, Edmund may have already killed Lucy around Lunchtime, but Lucy had slain Peter over breakfast. On the other hand, when Polly ended Eustace’s life as the sun set, Eustace had not yet had a chance to kill anyone else yet. It remains to be seen whether Polly survives her movie date with Diggory later on.

So one killing could be the apex of a series of connected killings or it could be a singular event. Is there an elaborate or simple way to determine how many people will see the following day’s sunrise?

#2

I don’t have the math for it, but I can tell you (by thought experiment) that there isn’t a single answer. If each person is only allowed to, and required to, kill one person, there’s a range of potential answers between “one person left” and “1/2 of original sample”.

One person left: A is killed by B. B is killed by C… so on, until 2nd to last person is killed by last one.

1/2 of starting: “each person kills one other person, and only one other person, all at the same time, in fixed pairs.”

#3

It was obvious to me that there was no clear answer, which is why I presented it as a probability question. I am fairly certain that that answer would be some number between one and 3.6 billion, but there should be a way to get close to what the answer would be (using spherical chickens in a vacuum).

#4

There is a clear answer: since everyone is killing one other person, the entire population of the earth is wiped out in one day.

#5

Seems to me like this would reduce the global population to zero. How else could you have each person killing another one?
Edit: Ninja’d by Northern Piper!

#6

Given the subject matter of the thread, being ninja’d seems particularly appropriate!

#7

Given typical human behavior, I think there would be a tendency to clique up and agree to kill other people and not each other. The likely size of cliques might determine how many people survive.

#8

The answer depends on how the victims are chosen. Most notably, are they chosen at the start of the day, or are they chosen at the time of the killing? If they’re chosen at the start of the day, then it’s possible that Alice and Bob both decide to kill Carol, and whichever one of them gets to her first, the other is going to have their kill negated. On the other hand, if it’s chosen at the time of the killing, then your first-choice target dying just means that you move on to your second-choice, or third, or however many it takes to get to someone that you haven’t been beaten to.

#9

I would think that the event would be ad hoc, because no one wants to know that they are scheduled to die tomorrow (either designated as a victim or by dint of not receiving a kill-chit). If people know they are marked for death tomorrow, it would greatly affect their behavior today, probably in very adverse ways.

#10

If each person must kill one, and only one, person, and no one can kill or be killed twice . . . then each person is also killed by one and only one person. This assumes that no new person is born, and nobody dies of other causes. It also assumes that there are an even number of individuals on the planet, and nobody is without access to/from others, e.g. people in solitary confinement or hermits. If all these factors are true, the probability is 100% that everybody on earth is killed.

#11

Panache45 I don’t think that you need an even number of people. Jim kills Sam, Sam kill Sarah and Sarah kills Jim. Some of the confusion about this might be coming from the issue with how can Sam kill Sarah if he has been killed by Jim.

#12

Well, the following simulation gives the following answer.
(1) Start with n person that haven’t killed anyone yet.
(2) Select one person i that hasn’t killed anyone yet.
(3) Select any other person j and remove it from the population. Mark person i as having killed someone already (note that I don’t allow anyone to kill himself).
(4) Repeat (2,3) until everyone has killed someone or only one person is left
(5) Count the number K(n) of person left.

I simulated this 1000 times with n varying from 10 to 2500 and found that

  • K appears normally distributed.
  • the average E(K(n)) fits precisely with E(K(n)) = 0.3679*N
  • the variance VAR(K(n)) fits precisely with VAR(K(n)) = 0.0375*N
    When you plot these relationship, they really form a neat line.

So, if you follow my definitions, 36.8% of the population survives, and for a large population this becomes a very accurate figure!

Edit: I just realize that 0.3679 = exp(-1)…

#13

I think the best way to fill in the details of the premise would be to say that one twenty-four hour period has been designated for the big murder spree. And on that day everyone will be trying to kill another person. But only one other person. Once they’ve killed their victim, they’ll stop. And we’ll assume nobody will fail outright.

So at the end of the day, we’ll have three groups:

  1. People who are killed before they make their kill. (A)
  2. People who make their kill but are subsequently killed themselves. (B)
  3. People who kill somebody but are not targeted by anyone. ©

These three groups will comprise the entire pre-murder population (D). And the sizes of group 1 and group 3 will be equal; everyone who survives represents a murder that didn’t happen. So the post-murder population (E) will be:

E = (D - B) / 2

So we need to figure out B - the number of people who make their kill but are subsequently killed.

#14

… and I have a recursive formula

Assume that you have k killers (people who have killed someone already) and i innocents (people who haven’t killed anyone yet). An innocent kill someone else (not himself), so he kills either one of the k killers or one of the other i-1 innocents.
So you get two alternatives:

  • kill another innocent with probability (i-1)/(i+k-1) and you will get k+1 killers and i-2 innocents.
  • or kill a killer with probability (k)/(i+k-1) and you will get k killers and i-1 innocents.

Write E(k,i) the expected number left alive when you start with k killers and i innocents.
Based on the above,

E(k,i) = (i-1)/(k+i-1)*E(k+1,i-2) + k/(k+i-1)*E(k,i-1)

With E(k,0) = k, E(k,1)=k if k>0 and E(0,1)=1 and a little dynamic programming, I can confirm the figure in the last post, i.e. E(0,n)=0.3679*n.

#15

Interesting. The appears to be a likely number which also works out to about e[sup]-1[/sup]. I wonder if there is a reason, like some kind of theorem, that it works out that way.

#16

To begin to assess the probability, we need to define things more clearly.
For example, we can’t even say the bounds for sure. The lower bound for deaths is pretty solid: if every survivor killed one person, there are at least as many dead as surviving, so at least 50% dead. But if death is not instantaneous, then it is possible for someone to kill while dying, and that means 100% is possible. (If death is instantaneous, then there must be at least one survivor.)

There are way too many variables involved that it is hard to make reasonable guesses about.
Like, is killing going to be more likely at certain times of day?
And how are we measuring this “day”? Is it midnight-to-midnight at the International Dateline?

At a guess, most of the people killed in the first hour will not have killed someone themselves, and most of the people killed in the last hour will have already killed someone. But the rate at which the population changes from “not killer” to “killer” is not going to be constant, it is going to be based on things like how dense the population is in areas where there is daylight. Most people are going to do their killing during hours when they are normally awake, and most people are normally awake during daylight.

But if we ignore all that:
Lets just pretend that the killing happens every hour on the hour (because that makes the math simple). Let’s also pretend there are 6 billion people (because that also makes the math simpler, at least at the start).
At hour 1, 250 million people will kill 250 million other people.
At hour 2, another 250 million people will kill 250 million people, but 1/23 of them (or about 10 million) will be people who already killed someone.
By my VERY crude math, after hour 15 there will be nobody left alive who hasn’t already killed someone. About 2.2 billion survivors. About 1.4 billion of the dead had killed someone.

#17

I don’t see why this would be true.

If Group 2 is zero, then groups 1 and 3 will be equal, But every person in group 2 makes group 1 bigger (or group 2 bigger) and group 3 smaller.

#18

Thanks, folks. I was never any good at probability math.

#19

Nope.
Who will kill the last person?

#20

Whoever hasn’t killed anyone . . . and who’s in the process of dying.