# Silly Malthusian probability question

If the problem is that every person on Earth must kill one and only one other person, then there are two ways we could assume this must happen.

One is that you’re not excused from killing just because you’re dead. Even if you die, you must still kill another person in some way–simultaneous bullets, time bombs, and so forth. In this case, everyone on Earth dies, except perhaps for one person if there is an odd number of people on Earth. And even then I think this person must commit suicide.

The other options is that if you’re dead you can no longer kill anyone, so no suicides and no simultaneous or posthumous kills. In this case, I think Oukile must be correct.

Babies and other incapacitated people don’t count in the second scenario. They can’t kill someone, but they can be killed. Everyone incapable of killing another person for whatever reason will end up dead. But this lowers Oukile’s number, which assumes a homogenous population who are all capable of murder and choose a murder victim at random. Hmm. Or does it? Obviously everyone incapable of murder has to be dead at the end of the day, otherwise not everyone will have killed someone. But if everyone has to kill one and only one other person, then what happens if there are large numbers of people incapable of killing, yet every person capable of killing has already killed?

An obvious example, what if there were three people in the world: an able bodied person, a newborn baby, and a coma patient. The able bodied person kills the coma patient. But the baby can’t kill the able-bodied person, and the able person has already killed and can’t kill the baby and so we have a survivor who didn’t kill anyone.

But how close to zero? I’m asking what that probability calculation would look like.

Yes, but that only works if the next-to-last murder victim slays his killer with his last breath, or via slow-acting poison, etc.

What seems more likely to me is that, in this genocidal day doubtless inspired by Inanna, each person is commanded to kill one and only one other person. Persons who die themselves are released from said obligation but still wind up in the embrace of lipless Ereshkigal. At some point, everyone left alive has fulfilled his or her obligation to Inanna. The population will be halved,. and Inanna persuades Ea that humanity will be quiet enough now that there is no need for another flood.

At some point someone will have a flower stolen by a snake.

No, Group 2’s size isn’t dependent on Group 1 or 3. Because people can kill people who may or may not have committed their own murder already.

Consider this scenario; everyone is given a gun and then lines up in single file facing east. The second person in line then shoots the first person. Then the third person shoots the second person and the fourth person shoots the third person and so on down the line. The last person in line shoots the penultimate person and there’s nobody behind him or her so that last person survives.

Group 1 consisted of just one person; the first person in line. Group 3 also consisted of one person; the last person in line. Everyone else is in Group 2.

What if everyone forms a circle, and shoots simultaneously?

The premise seems to be that each murder is temporally separate. If A kills B then B is no longer able to kill A.

Let’s assume players shoot in a fixed order, selecting uniform randomly among eligible targets. There are four cases, depending on whether corpses are allowed to shoot, and whether shooting at corpses is permitted. For the four cases, expected deaths are
100%, 63.21%, 63.21%, 50.00%

The case where corpses are permitted to be both shooters and shootees yields an ordinary Poisson distribution with parameter 1, so the 63.21% outcome is no surprise. What about the other cases?

Corpses (people who are actually dead, as in not responsive, not breathing and having no heartbeat) cannot participate in killing, that is just absurd. This is not a Zombie Apocalypse situation. However, a mortally wounded person could possibly still kill someone else. One might assume a mortally wounded person would be quite angry and might go over their quota of one, if they have the strength.

More importantly, accidental and natural deaths happen every day. Some accidents will occur in the process of killing, but it is otherwise a day like any other: people will go about their business, based on the assumption that they will be one of the survivors (the lion’s share of people will be ones that would prefer to not die). The number is very small, probably around a thousand, but on a tago de tutmonda mortigo, people might be inclined to engage in riskier behaviors – there could be some personal satisfaction in drawing one’s last breath with the knowledge that no survivor will be able to claim them as their kill.

Why can’t we all just get along?

Oh, where’s the sport in that?

If corpses are allowed to shoot and shooting at corpses is not permitted, then all n people shoot n distinct people; thus, 100% of people die, guaranteed. This was essentially the model panache45, et al, considered.

If corpses are allowed to shoot and shooting at corpses is permitted, any particular person’s probability of getting shot at least once is 1 - (1 - 1/n)^n, where n is the number of other people (or the number of people total, if one may shoot at themself). In the large n limit, this is 1 - 1/e; thus, approx. 63.21% of people die on probabilistic average. This was the model Buck Godot considered, before deleting their post.

If corpses are not allowed to shoot and shooting at corpses is not permitted, then we have the recurrence studied by Oukile, Lance Turbo, etc., which yields a 63.21% probabilistic death rate for reasons partially but not yet fully explained in this thread.

Finally, if corpses are not allowed to shoot and shooting at corpses is permitted (but not shooting at oneself), and we start with n people total, then we have the recurrence ExpectedDeaths(i living innocents) = (i - 1)/(n - 1) * [ED(i - 2) + 1] + (n - i)/(n - 1) * ED(i - 1), with ED(0) = ED(1) = 0, and the value we are interested in is ED(n). Not sure why this apparently comes out to 50% * n, but there you go.

I can’t help but think that the last situation, where corpses are valid targets but not valid shooters, is related to the problem with airplane seating where the first person takes a random seat. But I can’t quite come up with a correspondence.

Here’s how far I get: Suppose that we put all people in the order in which they will shoot. Then everyone before you goes. If in that time anyone has killed you, then you won’t kill anyone. Likewise, if in that time anyone has killed your intended target, then you also won’t kill anyone. Otherwise, you will kill someone. Now, the probability that someone has already killed you is equal to the probability that someone has already killed your intended target… but the two events are non-exclusive, and what that probability is depends on where you are in the order. Nor is there any convenient symmetry with the people early in the order canceling out those late in the order: The first person is guaranteed to get off their kill, but the last person might or might not get off their kill.

EDIT:
Oh, in the case where corpses are valid targets, we also have to assume something about the target distribution, presumably that it’s uniform. In the real world, it’s more likely that there would be a small number of unpopular people who a lot of people would want to kill, but it’s difficult to quantify that.

We have a most direct correspondence if we use the model where corpses aren’t valid targets but are the ONLY valid shooters (apart from the very first shooter), and ask “What’s the probability that the last person in the assigned shooter order survives?”. Then A shooting B is like A taking B’s seat, while A surviving is like A taking their own seat. That may not be of any use in understanding the “corpses can’t be targets but everyone can be shooters” problem, though.

Let f(x) be the probability of survival (for any player, e.g. George) after portion x of the population has taken their turn (if not then dead). Then
f(x+dx) = f(x) * (1 - f(x) dx)
where dx is the portion corresponding to a single player, e.g. dx = .001 when N = 1000. In the right side, f(x) is the probability George is alive after portion x has shot, the next f(x) is the probability the next shooter is still alive, and dx the probability he targets George.

If N is large enough that dx is infinitesimal, then the above becomes df(x)/dx = - f[sup]2/sup or f(x) = 1 / (1 + x). When all players have taken their turn x = 1 so f(x) = 1/(1+1) = .500.

This is laughably non-rigorous, but still should convince that any answer is asymptotically 0.5000.

It’s not necessary to allow corpses to be able to shoot. :rolleyes: All that’s needed is that death is not necessarily instantaneous, and someone who’s in the process of dying is still able to kill.

I adopted that fiction simply because phrases like “shooters but not shootees may be corpses” seemed to be a minimal-word approach to denoting the several mathematical cases.

Except, of course it’s not. We know that on average the fraction 1/e of the population survives. So that’s George’s probability as well.

Indistinguishable agreed with my #47. Yours now makes the vote tally 1-to-1. Care to wager?

I should note, I didn’t actually check your math for the 50% figure in #47 (which hadn’t been revealed at the time); I just took your word for it. Actually running the recurrence I noted, I see instead something like between 30.6% and 30.7% of people getting shot (thus, neither 1 - 1/e nor 50%) for large n. But it’s possible I made some error there as well.

[Similarly, I didn’t check Lance Turbo’s claim in #30/#36, if anyone assumed I did. I just took their word for it as well.]

Was my demonstration of
f(x) = 1/(1+x)
however non-rigorous, unconvincing?
And FWIW, I did jet over to Monte Carlo, observe a billion roulette spins or so, and confirm the 50.00% estimate.