23 replies to this topic

[Rob_Gandy]

Spoiler for My choice is...

completely irrelelvant.
(Assuming that when I leave the town the only possible meetings left involve no deaths and all meetings occur one at a time.)

No matter which I choose to be I have about a 9% chance of surviving. That is without counting any meeting that didn't involve me or a killer dying. Because none of the other meetings affect my chance of surviving at the end.

Basically it comes down to me vs. killers. Whether I'm a killer or a pacifist I cannot meet a killer or else I die.

So I need there to be 5 meetings of killers for me to survive. For each set of possible meetings, that count, my chance of survival is (n-1)/(n+1) where n is the number of killers left that aren't me. Which gives us...

9/11 * 7/9 * 5/7 * 3/5 * 1/3 = 1/11 or .0909... or 9.0909...%
And that is my take on it.

[mmiguel]

Spoiler for My choice is...

completely irrelelvant.
(Assuming that when I leave the town the only possible meetings left involve no deaths and all meetings occur one at a time.)

No matter which I choose to be I have about a 9% chance of surviving. That is without counting any meeting that didn't involve me or a killer dying. Because none of the other meetings affect my chance of surviving at the end.

Basically it comes down to me vs. killers. Whether I'm a killer or a pacifist I cannot meet a killer or else I die.

So I need there to be 5 meetings of killers for me to survive. For each set of possible meetings, that count, my chance of survival is (n-1)/(n+1) where n is the number of killers left that aren't me. Which gives us...

9/11 * 7/9 * 5/7 * 3/5 * 1/3 = 1/11 or .0909... or 9.0909...%
And that is my take on it.

Spoiler for but

Let's say there were 2 killers left, and 100,000 pacifists (I know, can't be more than 10 -- have patience).

Do you really think your chances of dying here are
(2-1)/(2+1) = 1/3?

Of all the other people out there they could kill, why would the killers be so likely to kill you?

Answer:
Because your chance of dying is not (n-1)/(n+1), but depends on number of non-killers out there too.

[bonanova]

Spoiler for but

Let's say there were 2 killers left, and 100,000 pacifists (I know, can't be more than 10 -- have patience).

Do you really think your chances of dying here are
(2-1)/(2+1) = 1/3?

Of all the other people out there they could kill, why would the killers be so likely to kill you?

Answer:
Because your chance of dying is not (n-1)/(n+1), but depends on number of non-killers out there too.

Spoiler for but

How many initial pacifists would there have to have been for there to be only 100,000 pacifists remaining when the killers numbered but two? What are your chances of being among those 100,000?

Actually, what if the puzzle had said Killville has K (an even number) killers and P pacifists when you arrive. You may join the killers, and act like one, or you may join the pacifists and remain unarmed. Which party would get your vote?

BTW, RG is correct to observe that your survival can be defined as follows: among the six events, KK, KK, KK, KK, KK, K-you, the K-you event must happen last. That is, be impossible.

How do the other events (K-other pacifists) affect you or thin outthe killers?

[Rob_Gandy]

Spoiler for but

Let's say there were 2 killers left, and 100,000 pacifists (I know, can't be more than 10 -- have patience).

Do you really think your chances of dying here are
(2-1)/(2+1) = 1/3?

Of all the other people out there they could kill, why would the killers be so likely to kill you?

Answer:
Because your chance of dying is not (n-1)/(n+1), but depends on number of non-killers out there too.

Spoiler for I guess I should explain my logic some more.

And yes I do still think that my chances of survival are still 1/3 in that situation.

My chance of dying in any given encounter does depend on the total number of people, but, if we are thinking end game, it doesn't matter to my final survival what happens to those 100,000 pacifists. My survival depends on the two killers meeting each other before one of them meets me. So the only meetings that affect my survival are (K1,K2), (K1,Me) and (K2,Me). So based on those meetings I have a 1/3 chance or survival.

I worked under the assumption that there comes a point where either there is a single killer left or no killers and some number of pacifists and thus no more meetings are important. And that meetings occur one at a time.
Also, I only considered meetings that affected my survival as important and thats how I came up with (n-1)/(n+1).

Am I on the right track or way off?

Edit:In red.

Edited by Rob_Gandy, 07 September 2012 - 05:18 AM.

[mmiguel]

Spoiler for but

How many initial pacifists would there have to have been for there to be only 100,000 pacifists remaining when the killers numbered but two? What are your chances of being among those 100,000?

In short, what if the puzzle had said Killville has K (an even number) killers and P pacifists when you arrive. You may join the killers, and act like one, or you may join the pacifists and remain unarmed. Which party would get your vote?

BTW, RG is correct to observe that your survival can be defined as follows: among the six events, KK, KK, KK, KK, KK, K-you, the K-you event must happen last.

Other events (K-other pacifists) can transpire at any time. They don't harm you, but they don't decrease the killer population either.

Spoiler for

chances of survival are probably not very high either way.
but, i showed mathematically in a prior post (unless someone finds a flaw in my reasoning, which is of course very possible), that the chance of surviving a single encounter (assuming everyone is paired together in such an encounter, with the exception of maybe one odd man out), increases as the total number of pacifists increase while the number of killers remain the same.

it is thus in your best interest to make any possible decision you can to result in as many pacifists being alive in each round as possible.
it's like the evolutionary strategy of a school of fish - band together so that although the chances of a shark getting a meal are large, within the school itself, the likelihood of any particular fish getting eaten in a single shark encounter decreases.

being a killer erodes your "protection" faster.

you may likely die either way, but you have a slightly better chance of surviving if you don't kill others, letting them live in order to take a bullet for you later in the future.

Pacifist

[bonanova]

Spoiler for I guess I should explain my logic some more.

And yes I do still think that my chances of survival are still 1/3 in that situation.

My chance of dying in any given encounter does depend on the total number of people, but, if we are thinking end game, it doesn't matter to my final survival what happens to those 100,000 pacifists. My survival depends on the two killers meeting each other before one of them meets me. So the only meetings that affect my survival are (K1,K2), (K1,Me) and (K2,Me). So based on those meetings I have a 1/3 chance or survival.

I worked under the assumption that there comes a point where either there is a single killer left or no killers and some number of pacifists and thus no more meetings are important.
Also, I only considered meetings that affected my survival as important and thats how I came up with (n-1)/(n+1).

Am I on the right track or way off?

Yes. See my spoiler above.

Spoiler for another formula

In the general case I alluded to above, with K (even) killers and P pacifists, P/ (K+1) pacifists survive on average. If K is odd, no pacifists survive.

When K=P=10 initially, by joining the pacifists you make P=11, and on average 1 pacifist can attend the killers' funerals. You have 1/11 chance of being that survivor.

If you join the killers, you make K=11 (odd) so it is certain that exactly1 killer is standing at the end. He wipes out any pacifists still breathing when killers 9 and 10 shoot each other. Again, you have 1/11 chance of being the surviving person.

[mmiguel]

Spoiler for I guess I should explain my logic some more.

And yes I do still think that my chances of survival are still 1/3 in that situation.

My chance of dying in any given encounter does depend on the total number of people, but, if we are thinking end game, it doesn't matter to my final survival what happens to those 100,000 pacifists. My survival depends on the two killers meeting each other before one of them meets me. So the only meetings that affect my survival are (K1,K2), (K1,Me) and (K2,Me). So based on those meetings I have a 1/3 chance or survival.

I worked under the assumption that there comes a point where either there is a single killer left or no killers and some number of pacifists and thus no more meetings are important.
Also, I only considered meetings that affected my survival as important and thats how I came up with (n-1)/(n+1).

Am I on the right track or way off?

Spoiler for

well in your last post, i had thought you said that (n-1)/(n+1) were the chances of surviving any round, but maybe i mis-interpreted.

well, i need to think about your response...

i agree that with 2 killers and 100000 pacifists, the last important turn has 3 possibilities.
but have we established that those 3 possibilities are equally likely?


what if there are 100 killers and you are the only pacifist?
your formula says you have 99/101 chance of surviving (pretty good chance), which doesn't seem believable to me.

what if there are 3 killers and you are the only pacifist?
in this case, you basically have to die, no matter what.
in my formula, the chance of survival would be
1 - k/(n-1) = 1-3/3 = 0

in yours, it's 1/2

Edited by mmiguel, 07 September 2012 - 05:30 AM.

[Rob_Gandy]

Spoiler for

chances of survival are probably not very high either way.
but, i showed mathematically in a prior post (unless someone finds a flaw in my reasoning, which is of course very possible), that the chance of surviving a single encounter (assuming everyone is paired together in such an encounter, with the exception of maybe one odd man out), increases as the total number of pacifists increase while the number of killers remain the same.

it is thus in your best interest to make any possible decision you can to result in as many pacifists being alive in each round as possible.
it's like the evolutionary strategy of a school of fish - band together so that although the chances of a shark getting a meal are large, within the school itself, the likelihood of any particular fish getting eaten in a single shark encounter decreases.

being a killer erodes your "protection" faster.

you may likely die either way, but you have a slightly better chance of surviving if you don't kill others, letting them live in order to take a bullet for you later in the future.

Pacifist

Spoiler for I see an important difference in our assumptions.

You: assuming everyone is paired together in such an encounter, with the exception of maybe one odd man out
Me: I worked under the assumption...And that meetings occur one at a time.

[mmiguel]

Yes. See my spoiler above.

Spoiler for another formula

In the general case I alluded to above, with K (even) killers and P pacifists, P/ (K+1) pacifists survive on average.

When K=P=10 initially, by joining the pacifists you make P=11, and on average 1 pacifist will stand and can attend the killers' funerals. You have 1/11 chance of being that survivor. If you join the killers, you make K=11 (odd) so it is certain that exactly1 killer is standing at the end. He wipes out any pacifists still breathing when killers 9 and 10 shoot each other. Again, you have 1/11 chance of being the surviving person.

Spoiler for

can you share the derivation of P/(K+1)?

[mmiguel]

Spoiler for I see an important difference in our assumptions.

You: assuming everyone is paired together in such an encounter, with the exception of maybe one odd man out
Me: I worked under the assumption...And that meetings occur one at a time.

ah! haha
i suppose i should spend more time reading the prompt then, i guess i was solving the wrong problem