Posted 17 April 2008 - 07:33 PM

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?).