# Silly Malthusian probability question **URL:** \ **Category:** Factual Questions\ **Created:** [March 31, 2016, 10:17pm UTC](https://boards.straightdope.com/t/silly-malthusian-probability-question/750787 "2016-03-31T22:17:43Z")\ **Posts on this page:** 7\ **Page:** 4
**Author:** ![OldGuy](https://avatars.discourse-cdn.com/v4/letter/o/3bc359/32.png) [@OldGuy](https://boards.straightdope.com/u/OldGuy)\ **Post date:** [April 3, 2016, 5:26am UTC](https://boards.straightdope.com/t/silly-malthusian-probability-question/750787/61 "2016-04-03T05:26:41Z")
> [@septimus](#): > > **Indistinguishable** agreed with my #47. Yours now makes the vote tally 1-to-1. Care to wager? I would, but I stay anonymous here. But to make sure we’re communicating, this is the problem I’m solving: We start with N people. One person is chosen at random. That person kills someone else chosen at random, and that person dies with 100% certainty before he can do anything. At each point after that there are k “killers” alive who have killed one other person and i “innocents” alive who have killed no one else. One of the i innocents is chosen at random and he kills one of the other k + i -1 people with 100% certainty. No corpses do any shooting or are shot a second time. The procedure continues until there are no innocents left alive only killers. What is the ex ante expected number of people remaining alive. My answer, which I was not first to propose, is approximately N/e. Your answer N/2. If that’s the problem we’re agreeing on, then N/2 cannot be correct. Clearly there can never be more than N/2 people left. If there are more than N/2 people left at any time than fewer than N/2 people have shot someone else so there must be fewer than N/2 killers alive or dead. That means there must be more than N/2 innocents alive or dead so there must be some innocents still alive, so the process continues. As the process can’t stop with more than N/2 people alive, the only way N/2 could be the average number to stay alive is if exactly N/2 people always live; otherwise the average would have to be less than N/2. But the only way exactly N/2 people survive is if each person killed is an innocent so that we’re left with N/2 killers. That means when there are N/2 + 1 people left alive (N/2 - 1 killers and 2 innocents) one of the two innocents must be picked at random and he must pick the other innocent as his target. The probability of that is 2/(N/2 + 1) \* 1/(N/2). That’s a quite low probability, but you must have it happen with certainty. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [April 3, 2016, 5:35am UTC](https://boards.straightdope.com/t/silly-malthusian-probability-question/750787/62 "2016-04-03T05:35:00Z")
> [@septimus](#): > > And FWIW, I did jet over to Monte Carlo, observe a billion roulette spins or so, and confirm the 50.00% estimate. Yes, I figured you did a simulation, which is why I questioned my recurrence calculation. And now I in fact see the error in said calculation: > [@Indistinguishable](#): > > 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). This only counts deaths of people who have not yet killed. Whoops! Let me think about how to rigorize your argument. I agree there’s something there. ---
**Author:** ![septimus](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/septimus/32/410_2.png) [@septimus](https://boards.straightdope.com/u/septimus)\ **Post date:** [April 3, 2016, 5:35am UTC](https://boards.straightdope.com/t/silly-malthusian-probability-question/750787/63 "2016-04-03T05:35:23Z")
> [@OldGuy](#): > > But to make sure we’re communicating, this is the problem I’m solving: > … > No corpses do any shooting or are shot a second time. No. We’re solving FOUR different problems. (Was it so very hard to scroll up to post #47 ? 😕 ) You’ve solved one of the easier variations whose solution has already been posted several times. The least trivial is the one we’re discussing now: corpses don’t shoot but can be shot repeatedly. > [@septimus](#): > > 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? ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [April 3, 2016, 6:26am UTC](https://boards.straightdope.com/t/silly-malthusian-probability-question/750787/64 "2016-04-03T06:26:39Z")
Thoughts on the **septimus** argument (for the asymptotic survival rate if corpses cannot shoot but can be shot): Take T people and assign them a shooting order; I’ll speak of Person #1, #2, …, through #T. Let g(p, n) = the probability Person #p is still alive after the first n players have shot. Of course, g(p, 0) = 1, while g(p, n) = g(p, n - 1) \* (1 - (the probability that Person #n is alive after the first n - 1 players have shot, given that Person #p is)/(T - 1)), for positive n distinct from p. Finally, we have g(p, p) = g(p, p - 1). [This accounts for people not shooting themselves; if we wanted people able to shoot themselves, it would be a minor correction] Ideally, “the probability that Person #n is alive after the first n - 1 players have shot, given that Person #p is” would just be the same as g(p, n - 1) itself; however, even apart from the minor symmetry-breakings introduced by people not shooting themselves, there’s also the fact that people’s chances of being alive at any given moment aren’t quite independent; knowing that one particular person is alive gives us slightly more reason to suspect another particular person has been shot instead. However… that all does seem intuitively negligible, and if we did indeed ignore it, g(p, n) should be approximately 1/(1 + n/T) for large T, so that the asymptotic survival rate would be 50%. [All I’m doing is rewriting things as explicitly discrete, and noting the few minor wrinkles I see preventing this from being completely rigorous yet] ---
**Author:** ![eschereal](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/eschereal/32/18939_2.png) [@eschereal](https://boards.straightdope.com/u/eschereal)\ **Post date:** [April 3, 2016, 6:35am UTC](https://boards.straightdope.com/t/silly-malthusian-probability-question/750787/65 "2016-04-03T06:35:55Z")
The requirement is that each person _kill_ one other person. The method of killing was not dictated. Shooting a corpse is all well and good, whatever floats your boat, but it does not fulfill the requirement of killing someone else. If you use up your bullets plugging corpses, you have to, say, break out the old wakashi and get all midæval on someone – or join the i group. ---
**Author:** ![Chessic\_Sense](https://avatars.discourse-cdn.com/v4/letter/c/7c8e57/32.png) [@Chessic\_Sense](https://boards.straightdope.com/u/Chessic_Sense)\ **Post date:** [April 3, 2016, 1:57pm UTC](https://boards.straightdope.com/t/silly-malthusian-probability-question/750787/66 "2016-04-03T13:57:43Z")
> [@SpyOne](#): > > 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. Kills = Deaths. Every time someone kills, someone dies. For Group 2, they both kill and die, so they cancel themselves out. All that’s left is people who kill once and people who die once, and since kills=deaths, their populations must be equal. Put another way, ask all the group 1s who killed them. Either their murderer is in group 3, or that killer is in group 2 and himself has a killer in group 3, or the killer’s killer is a 2 who was killed by a 3…and so on. No matter what, there has to be exactly one 3 that ends any murder chain started by a 1, for each and every 1. The flaw in your assertion is the “group 3 smaller” part. Yes, that Group 2 person takes over the _direct_ link from 3 to 1, but he himself necessitates the existence of another group 3er…after all, if he’s in group 2, who killed him? > [@t-bonham@scc.net](#): > > infants born that day… are too weak & uncoordinated to kill another person that day (or even for many days, months, or years). That’s what they _want_ you to think! > [@Lemur866](#): > > Even if you die, you must still kill another person in some way–simultaneous bullets, time bombs, and so forth… > Babies can’t kill someone… 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. Not true at all. You guys are thinking of complicated ways like time bombs to kill people after one’s own death, but there’s an easier way that actually happens frequently - plague. Just yesterday, my baby shoved his hand in my mouth unexpectedly, cutting my gum. Suppose that got infected w/ a horrendous germ that killed me. Babies are nasty carriers, capable of spreading all kinds of illnesses. Ask any preschool teacher. They cough, sneeze, touch everything, lick everything, and other people handle their poop. They can definitely kill. ---
**Author:** ![LSLGuy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lslguy/32/5813_2.png) [@LSLGuy](https://boards.straightdope.com/u/LSLGuy)\ **Post date:** [April 3, 2016, 2:30pm UTC](https://boards.straightdope.com/t/silly-malthusian-probability-question/750787/67 "2016-04-03T14:30:15Z")
Plus … In pre-20th Century conditions pre-birth and trans-birth infants commit an awful lot of matricide too. The darn things look cute, but they’re surprisingly lethal. Rather like the [Rabbit of Caerbannog](https://en.wikipedia.org/wiki/Rabbit_of_Caerbannog). [Previous page](https://boards.straightdope.com/t/silly-malthusian-probability-question/750787.md?page=3)