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You live in Killville. Can you stay alive?

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You live in Killville - a town populated by 10 killers and 10 pacifists.

When a pacifist meets a pacifist, nothing happens.

When a pacifist meets a killer, the pacifist is killed.

When two killers meet, both die.

Assume meetings always occur between exactly two persons

and the pairs involved are completely random.

Are your odds of survival better if you are a killer? or a pacifist?

Or does it matter?

Regardless of whether you are a pacifist or a killer,

you may disregard all events in which a pacifist other than yourself is involved

and consider only events in which you are killed

or a pair of killers other than yourself is killed.

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Posted · Report post

odds of survival are better if u are a killer coz then there are 9 killers and 10 pacifists

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Posted · Report post

be a pacifist and let the killers kill themselves

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Assuming that all meetings are truly random, it is most likely that everyone dies (including you) with an even number of both groups. While it would be possible for the killers to all kill themselves off and leave 1+ pacifists behind, it's not very likely. Out of the 4 meeting types (PP, PK, KP, KK), two kill pacifists only one kills killers. Mathematically, it looks like a Pacifist should be better off in even numbers, but I have run through a couple of random scenarios that take the population decreases into account, and every time the score ends 0-0. So it would seem to me that moving is your best option to survive if the numbers are even. I don't think that probabilites/statistics type math tells the whole story due to the population decreases over time. Of course, an odd number of killers is a horse of a different color, it's better to be a killer in that scenario, because killer deaths always occur evenly, so there will be a murderous survivor with the town to himself.

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Posted · Report post

Of course, an odd number of killers is a horse of a different color, it's better to be a killer in that scenario, because killer deaths always occur evenly, so there will be a murderous survivor with the town to himself.

you're right up until that sentence. the odds will fluctuate with the decrease in population to favor either side, but will eventually end up at 50% each again. as there will never be an odd number of killers, there will not be a murderous survivor. they'll kill each other

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Posted (edited) · Report post

you're right up until that sentence. the odds will fluctuate with the decrease in population to favor either side, but will eventually end up at 50% each again. as there will never be an odd number of killers, there will not be a murderous survivor. they'll kill each other

The only way for a killer to die is if he meets another killer, in which case they both die. This means that if we start with ten killers they will eventually all kill each other off (regardless of the ratio to pacifists); however, were we to start with 11 killers and 11 pacifists (as posed by Bonanova) there would be one killer left. No matter how the population decreases affect the odds, killers can only be killed when they meet another killer. So if we have an odd number of killers in the initial population we will always be left with 1. After we're down to 1 killer it really doesn't matter how many pacifists are left (0, 3, 11), because they will eventually meet the lone killer until he is all that's left. I stand by my theory of if there is an even number of both groups (i.e. 10 to 10) the odds are everyone dies (but it would be possible for one or more Pacifists to survive if the killers killed themselves off first), but if there is an odd number of both groups (i.e. 11 to 11) then a lone killer will survive.

Edited by Aunt Minta
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Posted · Report post

you're right up until that sentence. the odds will fluctuate with the decrease in population to favor either side, but will eventually end up at 50% each again. as there will never be an odd number of killers, there will not be a murderous survivor. they'll kill each other

I think he was suggesting that Killville STARTED with an odd number of Killers. Since Killers die in pairs, if you start with an odd number of killers you will eventually have 1 left over and no one to kill him. Thus, in a Killville that starts with an odd number of Killers, it's better to be a killer and hope you're the Lucky "There Can Be Only" One.

In the original even scenario, in a even distribution of encounters, (PP, PK, KP, KK), half the Killers and half the Pacifists die, but that doesn't really matter since the Killers always die in pairs there will eventually be 0 Killers... thus it's better to be a Pacifist.

odds: The odds for the "odd" scenario is easy: 1/#of Killers

I don't know the odds for the "even" scenario

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Posted · Report post

i dont think it matters

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Pacifist

I'd have to say I'd be a pacifist and here's the deal. Killers are only killed in pairs (when 2 killers meet) so eventually, maybe after or BEFORE all pacifists are killed, 5 pairs of killers will meet and anhialate the killer race. We can't say that all pacifists will die, but I assure all killers will!!

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Posted (edited) · Report post

Pacifist

I'd have to say I'd be a pacifist and here's the deal. Killers are only killed in pairs (when 2 killers meet) so eventually, maybe after or BEFORE all pacifists are killed, 5 pairs of killers will meet and anhialate the killer race. We can't say that all pacifists will die, but I assure all killers will!!

Oh, nevermind that last post.

Edited by itachi-san
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Posted · Report post

If killers always die in pairs then the rate of decline for killers would be greater as more meetings take place. Therefore, the further in the game you are, the more chances you have that you'd bump into a pacifist rather then a killer.

Of course, this also means that the killers are also more likely to meet with a pacifist and kill them, at which point the rate of decline is even again. So, maybe it doesn't matter after all.

Wether you are a pacifist or a killer, the same applies. If you meet with a killer, you're dead. If you meet with a pacifist, you go home and get to have another meeting tomorrow. And since the rate of death is even for both parties it would mean that you are just as safe on either side.

So ultimately it is up to you to to be careful not to speak to strangers and leave the rest up to good old fashion luck. (or prayer for the religious ones)

In the case where the town is populated by 11 members of each party, then the situation is different. Since meetings only take place in pairs (an even number), and killers always die in pairs (also an even number), then the last remaining killer will have no chance of meeting another killer. This guarantees him survival. He's here to stay, and meetings will keep going until every last pacifist is dead.

I don't know what the mathematical formula for this would be. But, I am guessing that every meeting that is potentially possible before every killer is dead gives the the last killer one more chance to go home. If "x" is the amount of meetings possible between killers before they're all dead, then the chances are 1/x in the killers favor.

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If the number of killers is odd...

If there is an odd number of killers then I deffinately want to be a killer. Even if the original number of pacifists is outstanding say 1,000 and only 11 killers eventually 10 killers will be killed off leaving only one and that one will eventually kill off the remaining pacifists.

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Posted · Report post

there would be a equal chance you can stay alive or dead

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I've just taken a look at this thread for the first time (thanks Google Gadget!), and really feel no one has solved it mathematically correctly. This is months and months after the fact, and I'm sure most of you could care less about the math, but it was really bothering me, so here we go.

The concept of even killers means you should want to a pacifist, and odd killers means you should want to be a killer is correct but the chances of staying alive has not been calculated correctly.

Here's the question: what's the chances of staying alive through an indefinite period of time if you are a pacifist in a 10/10 community? Obviously, your chances of staying alive if you are a killer is zero, since even if you kill off all the pacifists, you'll still kill each other off too.

Well, starting 10K/10P, after one meeting, we could still have 10/10 if two pacifists meet, and the chances of that are 45/190. We would have 8K/10P if two killers meet, and the chances of that are 45/190. We would have 10KL/9P if a killer meets a pacifist, with a chance of 100/190.

(By the way, if the question was, for ONE iteration, would you rather be a killer or a pacifist, the answer can be solved now. If you are a pacifist, you have a 100/190 chance of being killed in the first scenario, if you have the 1/10 chance of being the one pacifist. A random pacifist has a 0.0526 chance of being killed. If you are a killer, you have a 45/190 chance of being killed in the first scenario, if you are either one of the killers who get killed. Every killer is in 9 of the 45 scenarios (90 killers killed divided by 10), so 9/190 is 0.0474. For just one meeting, you'd rather be a killer .... barely.)

Getting back to the indefinite time probability of a pacifist staying alive ... those numbers that get calculated are based on combinatorics. Let k = number of killers, p = number of pacifists, and n = total alive = k+p. The probability that two pacifists meet is always pC2/nC2; the probability that two killers meet is always kC2/nC2; and the probability that a pacifist meets a killer is k*p/nC2 [more accurately, kC1*pC1/nC2].

However, when a pacifists meets another pacifists, and they leave together, nothing has really changed, and we're still at the same spot. So, we really only care when at least one of the people are killers. Doing easy algebra division, we find the ratio of times when the population decreases by two killers to the times when the population decreases by one pacifist is kC2/k*p. Doing some combinatorics expansion and basic algebra division cancellation, this is equivalent to (k-1)/(2p). The probability that from the current population goes down two killers is (k-1)/(k+2p-1) and the probability that from the current population goes down one pacifist is (2p)/(k+2p-1)

I made an excel spreasheet to do the math and came up with this:

---All 10 pacifists survive: 0.0001

---9 pacifists survive: 0.0005

---8 pacifists survive: 0.0017

---7 pacifists survive: 0.0042

---6 pacifists survive: 0.0090

---5 pacifists survive: 0.0176

---4 pacifists survive: 0.0326

---3 pacifists survive: 0.0585

---2 pacifists survive: 0.1053

---1 pacifist survives: 0.2028

---0 pacifists survive: 0.5678

All numbers above are rounded, and would add to 1 if rounding error is taken into account.

What this means is that about 57% of the time, the killers kill off all the pacifists before killing themselves (in other words, in this town of 20 people, everyone dies more than half of the time).

What's even MORE interesting, is that when you go through all of the math, the probability that if you are a random pacifist in this town, that you'll survive the carnage is....

....

....

1 out of 11. Exactly 1/11.

Now, if anyone wants to mathematically explain THAT, I'd love to see it. I'm probably going to be bugged by this enough that I'll look through some more scenarios and see if there's a pattern (is it 1/(pacifists + 1)? 1/(killers + 1)? Something more complicated?).

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Posted · Report post

What's even MORE interesting, is that when you go through all of the math, the probability that if you are a random pacifist in this town, that you'll survive the carnage is....1 out of 11. Exactly 1/11.

Now, if anyone wants to mathematically explain THAT, I'd love to see it. I'm probably going to be bugged by this enough that I'll look through some more scenarios and see if there's a pattern (is it 1/(pacifists + 1)? 1/(killers + 1)? Something more complicated?).

In case anyone is interested, I'm about 95% sure about the following:

---A) the chance that a pacifist survives the carnage is independent of the number of pacifists. It solely depends on the number of killers. With 10 killers, the chance a pacifists survives is 1/11 regardless of whether there is one pacifist in the town, or a billion.

---B) the chance that a pacifist survives the carnage is 1/(killers + 1)

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Posted · Report post

i would rather be a pacifist if there was an even or odd # of killers cause i would try to get out of the village

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Given the origanal situation, I would NOT want to be a killer. Because, killers would kill each other in pairs until there are no killers left. I would try and advoid them until they are all dead, which would mean I would survive.

However, in the situation of there being 11 killers, I would WANT to be a killer. This is because after 5 meetings that have both sides as killers, there would only be one killer left, and no-one that can kill that killer.

Therefore, if I was in Killville, I would want to be a Killer if there are an odd number of Killers, and not one if there are an even number of killers. And if I was to move in, I wouldn't move in while there are an odd number of killers at large.

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Hint 1: What does it mean to survive?

Hint 2: Would the result be different if Killville were populated by 11 killers and 11 pacifists?

Someone up there ^ must like you.

:lol:

Just kidding....idk

Being a killer would have a better survival chance, because if a killer meets a killer then they both die. that means, in order to kill all the killers you need an even number

so basically if there is an even number of people, the pacifists would have a better chance

and if there is an odd number only one person (a killer) would survive

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If you assume the meetings are random, i would like to be a pacifist. If it is random there should be an equal number of each of the following outcomes:

Killer vs Killer

Killer vs Pacifist

Pacifist vs Pacifist

so after one of each outcome the population would be 9 pacifists and 8 killers, (two killers meet, 2 killers die, two pacifists meat noone dies, killer meets pacifist one pacifist dies)

so by the time only half the pacifists have died, all the killers will be dead, meaning any pacifist will have a 50/50 chance of surviving, wilst the chances of a killer surviving is 0.

if there were an odd number of killers, then you should be a killer, because killers only die in pairs, so one will survive to kill all the pacifists

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The riddle has already been awnsered. Unless you have anything important or a different awnser that hasn't already been posted, then you shouldn't post awnsers in an old thread. I am quite sure that a mod will probally lock this topic soon because it is already done with.

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Posted (edited) · Report post

Simple answer is:

Assuming that meetings random and that to begin there are the same amount of each type then survival is essentially equal. Both groups have a 50% chance of interacting with the other kind and therefore both groups have a survival chance of 50% for each encounter.

However, I think the long term survival chances for every individual is 5%.

Killer populations will drop in spikes and the pacifist population will drop on a gradual slope.

Edited by rpupkin77
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Kudos to writersblock [2nd post basically does it]

Only killers can kill killers, and they die doing it; so killers die in pairs.

Starting with 10 killers, all the killers will eventually die, two at a time.

[killer chance of survival = 0.]

[pacifist chance of survival depends on ability to hide until last killer is dead; risky proposition, but not 0.]

So you want to be a pacifist.

Starting with 11 killers, eventually 10 killers will annihilate themselves pairwise, and the surviving 11th will eventually kill all the pacifists.

[killer chance of survival = 1/11]

[pacifist chance of survival = 0.]

So you want to be a killer.

But do all the pacifists go first? If all the pacifists go first then nobody survives. You have an overall chance of surviving longer if you are a killer. So basically there is a possible outcome where there are no survivors if we are talking about the long run here.

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You live in Killville - a town populated by 10 killers and 10 pacifists.

When a pacifist meets a pacifist, nothing happens.

When a pacifist meets a killer, the pacifist is killed.

When two killers meet, both die.

Assume meetings always occur between exactly two persons

and the pairs involved are completely random.

Are your odds of survival better if you are a killer? or a pacifist?

Or does it matter?

Regardless of whether you are a pacifist or a killer,

you may disregard all events in which a pacifist other than yourself is involved

and consider only events in which you are killed

or a pair of killers other than yourself is killed.

Actually I think the solution is simple. Don't have to take in account probabilities and math.

When you have an odd number of people in each group, after all meetings take place, a killer will be left standing. That means if you are a killer, you have a chance of never meeting with another killer and staying alive.

In the case of an even number of people in each group, it is certain that at some point, all killers will meet each other. The "game" doesn't stop when all pacifists die. So if you are a killer it is certain that you will die, while being a pacifist you have chances of surviving if all killers die before you meet any.

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So I know this is an old topic, and I don't know if anyone will read this, but I found this site today through iGoogle, and found this puzzle.

After generating a formula for deriving possibilities and adding them up, and comparing vectors and on and on, and generally overworking excel, I went to total the odds of being the sole surviving Killer, and realized, there was no column in my table for that, and of couse then it hit me that there wasn't going to be one.

Then I came and read all the posts, and investigated myself, and the formula that the odds of surviving as a pacifist in an even Killer scenario are in fact 1 / (K+1) where K is the number of killers. This can actually be demonstrated through induction, but I'll save it.

I posed the question to a friend, and he asked if I meant that he was one of the 20, or if he was moving into town and had to choose a team, because in that instance, he found it strange, his odds were even no matter which he picks!

So the coolest result of all, is that if you have to pick a behavior pattern to adopt, then knowing the number of total residents, and the make-up of the Killer to Pacifist balance are totally irrelevant. Beyond that, you can't affect your odds by choosing either lifestyle. If the number of existing Killers is even, then you get your 1 / (K+1) shot either way, and if the number is odd, you're doomed no matter which way you go.

Cool.

Great site, btw for anyone who is listening.

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So the coolest result of all, is that if you have to pick a behavior pattern to adopt, then knowing the number of total residents, and the make-up of the Killer to Pacifist balance are totally irrelevant. Beyond that, you can't affect your odds by choosing either lifestyle. If the number of existing Killers is even, then you get your 1 / (K+1) shot either way, and if the number is odd, you're doomed no matter which way you go.

That is a very cool result. I spent a long time trying to prove you wrong, but I've convinced myself you are right.

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You live in Killville - a town populated by 10 killers and 10 pacifists.

When a pacifist meets a pacifist, nothing happens.

When a pacifist meets a killer, the pacifist is killed.

When two killers meet, both die.

Assume meetings always occur between exactly two persons

and the pairs involved are completely random.

Are your odds of survival better if you are a killer? or a pacifist?

Or does it matter?

Regardless of whether you are a pacifist or a killer,

you may disregard all events in which a pacifist other than yourself is involved

and consider only events in which you are killed

or a pair of killers other than yourself is killed.

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