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You live in Killville. Can you stay alive?


Best Answer Writersblock , 10 October 2007 - 11:14 PM

Hmmm. This is an ambiguous question I think. I read it originally as surviving one encounter, but it can also be understood to be the "last man standing."

Spoiler for solution
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#31 Aunt Minta

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Posted 26 March 2008 - 06:10 PM

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, 26 March 2008 - 06:11 PM.

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#32 redshift

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Posted 26 March 2008 - 06:28 PM

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

Spoiler for solution


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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#33 hopstop

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Posted 30 March 2008 - 05:51 PM

i dont think it matters
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#34 Felipe

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Posted 30 March 2008 - 07:33 PM

Spoiler for Solution

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#35 itachi-san

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Posted 30 March 2008 - 07:49 PM

Spoiler for Solution



Oh, nevermind that last post.

Edited by itachi-san, 30 March 2008 - 07:50 PM.

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#36 David Deutsch

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Posted 02 April 2008 - 05:21 PM

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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#37 finance_it

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Posted 02 April 2008 - 05:54 PM

If the number of killers is odd...
Spoiler for Odd number of killers (11)

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#38 puzzlemaster

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Posted 03 April 2008 - 12:55 PM

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

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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?).
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#40 rhapsodize

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Posted 17 April 2008 - 08:38 PM

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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