23 replies to this topic

[bonanova]

Subtle variation on the original problem.

You are transferred to Killville, by your employer, for a one-year temporary assignment.
Its population comprises 10 Killers and 10 Pacifists.
A Killer immediately guns down any other person he meets.
A pacifist never harms a soul.

When you arrive at Killville, you are required to register, either as a Killer or as a Pacifist.
If you register as a Killer, you are issued a gun and required to act as a Killer.
If you register as a Pacifist, you are given the town's best wishes for good luck and advised to careful.

You desire to live long and prosper. Which way do you register?

Assume all encounters are 1-on-1 and random.

Use spoilers please.

[mmiguel]

Subtle variation on the original problem.

You are transferred to Killville, by your employer, for a one-year temporary assignment.
Its population comprises 10 Killers and 10 Pacifists.
A Killer immediately guns down any other person he meets.
A pacifist never harms a soul.

When you arrive at Killville, you are required to register, either as a Killer or as a Pacifist.
If you register as a Killer, you are issued a gun and required to act as a Killer.
If you register as a Pacifist, you are given the town's best wishes for good luck and advised to careful.

You desire to live long and prosper. Which way do you register?

Assume all encounters are 1-on-1 and random.

Use spoilers please.

Spoiler for

I think the main difference here is that you are the 11th person for whatever group you decide to be in.
I notice that when you register as a pacifist, you are not required to act like a pacifist (the promise they make registered killers make), but I think that this problem likely has more depth in its true solution than just this - not sure if this can even offer you much of an advantage any way.

I'll assume after you register, the game proceeds like the last problem, except there are now 11 people in your group and 10 in the other.
The only real way to survive is if all the killers kill each other (other than you) without any of them meeting you.

How to model the randomness of these meetings?
Assume each valid configuration of pairings is equally likely in each round.
What if there are an odd number of people?
Then I guess one person doesn't meet with anyone.
First assume n is even.
Number of possible pairings is given by multinomial formula (divided by (n/2)! to make ordering of pairings not matter):
n!/((n/2)!2!*2!*2!...)
where there are n/2 many 2!s on the bottom, and (n/2)! i.e.
n!/(2^(n/2)*(n/2)!)

If there are k killers (excluding you) of those n, then the chance of you meeting with one of them is....
well, how many possible ways can you meet with a killer?
need to choose 1 of the k killers
k possibilities for that
The other pairings need to happen now. The rest of the folks now number n-2 and there should be n/2-1 pairings now.
The probability of you dying is therefore
k*(n-2)!/((n/2-1)!2^(n/2-1))/(n!/((n/2)!2^(n/2)))

This simplifies to:
k/(n-1)

This is the chance of you meeting a killer.
The chance of you surviving is one minus this.
1 - k/(n-1)

If n is odd, the you have a 1/n chance of being selected as the odd man out.
The probability becomes:
1/n + (1-1/n)*(1-k/(n-1))

In the beginning, n = 21 and k=10, which gives you a 11/21 ~ 0.52 chance of surviving the first round.

The way you register does not affect your survival chances on the first round (k excludes you, so numerically your choice changes nothing).
Also, if you are a killer, it won't really help you, since you only get the opportunity to kill when you are dying.
In either case, your only hope is to never touch a killer.

If you are a killer and that happens, then each round you were either an odd man out or you killed some pacifists each round.

Are you better off with more pacifists or less pacifists?
Well, if we believe that at each round, if there are n people, the probability of you surviving is
given above, then it is better to have more pacifists around, since those equations increase with n (positive derivative).

Therefore, you don't want to kill pacifists, because they are protecting you by taking bullets for you.
You therefore want to be a pacifist to maximize the number of people who can take bullets for you in the hopes that the killers all kill eachother and other pacifists before ever meeting you.

Answer: Pacifist

Edited by mmiguel, 06 September 2012 - 08:54 AM.

[mmiguel]

Spoiler for

I think the main difference here is that you are the 11th person for whatever group you decide to be in.
I notice that when you register as a pacifist, you are not required to act like a pacifist (the promise they make registered killers make), but I think that this problem likely has more depth in its true solution than just this - not sure if this can even offer you much of an advantage any way.

I'll assume after you register, the game proceeds like the last problem, except there are now 11 people in your group and 10 in the other.
The only real way to survive is if all the killers kill each other (other than you) without any of them meeting you.

How to model the randomness of these meetings?
Assume each valid configuration of pairings is equally likely in each round.
What if there are an odd number of people?
Then I guess one person doesn't meet with anyone.
First assume n is even.
Number of possible pairings is given by multinomial formula (divided by (n/2)! to make ordering of pairings not matter):
n!/((n/2)!2!*2!*2!...)
where there are n/2 many 2!s on the bottom, and (n/2)! i.e.
n!/(2^(n/2)*(n/2)!)

If there are k killers (excluding you) of those n, then the chance of you meeting with one of them is....
well, how many possible ways can you meet with a killer?
need to choose 1 of the k killers
k possibilities for that
The other pairings need to happen now. The rest of the folks now number n-2 and there should be n/2-1 pairings now.
The probability of you dying is therefore
k*(n-2)!/((n/2-1)!2^(n/2-1))/(n!/((n/2)!2^(n/2)))

This simplifies to:
k/(n-1)

This is the chance of you meeting a killer.
The chance of you surviving is one minus this.
1 - k/(n-1)

If n is odd, the you have a 1/n chance of being selected as the odd man out.
The probability becomes:
1/n + (1-1/n)*(1-k/(n-1))

In the beginning, n = 21 and k=10, which gives you a 11/21 ~ 0.52 chance of surviving the first round.

The way you register does not affect your survival chances on the first round (k excludes you, so numerically your choice changes nothing).
Also, if you are a killer, it won't really help you, since you only get the opportunity to kill when you are dying.
In either case, your only hope is to never touch a killer.

If you are a killer and that happens, then each round you were either an odd man out or you killed some pacifists each round.

Are you better off with more pacifists or less pacifists?
Well, if we believe that at each round, if there are n people, the probability of you surviving is
given above, then it is better to have more pacifists around, since those equations increase with n (positive derivative).

Therefore, you don't want to kill pacifists, because they are protecting you by taking bullets for you.
You therefore want to be a pacifist to maximize the number of people who can take bullets for you in the hopes that the killers all kill eachother and other pacifists before ever meeting you.

Answer: Pacifist

Spoiler for mathematical correction

"
If n is odd, the you have a 1/n chance of being selected as the odd man out.
The probability becomes:
1/n + (1-1/n)*(1-k/(n-1))
"
should be
1/n + (k/n)*(1-(k-1)/(n-2)) + (n-k-1)/n*(1-k/(n-2))

need to account for odd man out being killer.
also denominator should be n-2, which is what i originally had before, but changed due to nonsensical answers for k=0 and/or k=20, without figuring out exactly what was wrong.

this simplifies to
(n-k)/n = 1 - k/n
amazingly


So survival if n is even is:
1 - k/(n-1)
survival if n is odd is:
1 - k/n

Conclusion from before remains the same.
Chance of surviving first round is still 11/21
haha, intuition could have given those probabilities from the beginning, but i tend to be paranoid with probability, and not try to go by intuition, unless problems are very simple.

Edited by mmiguel, 06 September 2012 - 09:27 AM.

[ujjagrawal]

Spoiler for Logic

Register as KILLER

Then you will have atleast 1/11 chance of surviving and prosper killing the other Pacifists.

Otherwise if you register yourself as Pacifist then surely all killers will kill themselves and your chances of surviving will depends solely on your luck.

[emeraldfyr3]

Spoiler for

I would just quit my job.

[phil1882]

personally i think

Spoiler for

either way works.
if you're a pacifist, that makes 11 pacifists, and 10 killers, making it possible, all-be-it unlikely that all the killers will kill
themselves off, while still leaving some pacifists. if you're a killer, that makes 11 killers and 10 pacifists. meaning 1 killer is guaranteed to survive.

[jim]

Spoiler for .

As long as the process reaches completition within the year--or as long as the probability of A meeting B in a short interval of time does not depend on how many other people there are (a kind of Ideal Gas Law) your chances of survival are the same whether you chose to be a killer or a pacifist. That is because the only encounters that matter for you are when two killers meet, or when a killer meets you. Whether a pacifist dies when meeting you are not is irrelevant to your survival.

[Saxguy01]

Spoiler for My take

The only way to survive is to be a killer. As the prompt says, killers immediately shoot anyone they meet. The only way two killers would end up simultaneously killing each other is if they simultaneously shoot each other (the old Wild West Gunslinger scene), an unlikely situation in any context. Barring that, 1 killer will survive (all the others having been killed by the other killers).

Edited by Saxguy01, 06 September 2012 - 08:56 PM.

[jim]

Saxguy....Realistically you are correct--but in the original version of the problem it was specifically stated that both killers die when they meet each other.

[mmiguel]

Spoiler for

I think the main difference here is that you are the 11th person for whatever group you decide to be in.
I notice that when you register as a pacifist, you are not required to act like a pacifist (the promise they make registered killers make), but I think that this problem likely has more depth in its true solution than just this - not sure if this can even offer you much of an advantage any way.

I'll assume after you register, the game proceeds like the last problem, except there are now 11 people in your group and 10 in the other.
The only real way to survive is if all the killers kill each other (other than you) without any of them meeting you.

How to model the randomness of these meetings?
Assume each valid configuration of pairings is equally likely in each round.
What if there are an odd number of people?
Then I guess one person doesn't meet with anyone.
First assume n is even.
Number of possible pairings is given by multinomial formula (divided by (n/2)! to make ordering of pairings not matter):
n!/((n/2)!2!*2!*2!...)
where there are n/2 many 2!s on the bottom, and (n/2)! i.e.
n!/(2^(n/2)*(n/2)!)

If there are k killers (excluding you) of those n, then the chance of you meeting with one of them is....
well, how many possible ways can you meet with a killer?
need to choose 1 of the k killers
k possibilities for that
The other pairings need to happen now. The rest of the folks now number n-2 and there should be n/2-1 pairings now.
The probability of you dying is therefore
k*(n-2)!/((n/2-1)!2^(n/2-1))/(n!/((n/2)!2^(n/2)))

This simplifies to:
k/(n-1)

This is the chance of you meeting a killer.
The chance of you surviving is one minus this.
1 - k/(n-1)

If n is odd, the you have a 1/n chance of being selected as the odd man out.
The probability becomes:
1/n + (1-1/n)*(1-k/(n-1))

In the beginning, n = 21 and k=10, which gives you a 11/21 ~ 0.52 chance of surviving the first round.

The way you register does not affect your survival chances on the first round (k excludes you, so numerically your choice changes nothing).
Also, if you are a killer, it won't really help you, since you only get the opportunity to kill when you are dying.
In either case, your only hope is to never touch a killer.

If you are a killer and that happens, then each round you were either an odd man out or you killed some pacifists each round.

Are you better off with more pacifists or less pacifists?
Well, if we believe that at each round, if there are n people, the probability of you surviving is
given above, then it is better to have more pacifists around, since those equations increase with n (positive derivative).

Therefore, you don't want to kill pacifists, because they are protecting you by taking bullets for you.
You therefore want to be a pacifist to maximize the number of people who can take bullets for you in the hopes that the killers all kill eachother and other pacifists before ever meeting you.

Answer: Pacifist

man... nobody agrees with me lol