Pirates

[Genady]

Five aging pirates decided to split their treasure and to retire. Each one will be satisfied if in his estimate his share is not smaller than one fifth of the total. Their estimates vary. How can they do it peacefully?

[Genady]

If there were two pirates rather than five, a solution could be that one of them divides the treasure into two parts, and the other one chooses. 

Edited May 28, 2023 by Genady

[TheVat]

Spoiler

Deal out five equal piles of small objects like doubloons or jewels.

If there's a remainder of gold objects, melt them, and divide into five equal ingots.

If there's a remainder of jewels, donate to Pirate's Widows Charitable Fund.

JK

Spoiler

Looks like a Nash equilibrium.

A series of votes is held, based on the oldest pirate counting out piles.  That pirate counts out five piles of treasure, then all vote on whether to accept that division or not.   If a division of treasure is accepted by majority then the pirate who counted out the piles lives and everyone takes their portion.  If majority rejects, then counter is killed, and process starts again.  Since the counter wants to live, he will count scrupulously.  Since no other pirate wants to run the risk of being counter (except the youngest, who has no such risk), the majority will accept the first count.

 

[Genady]

29 minutes ago, TheVat said:

  Hide contents

Deal out five equal piles of small objects like doubloons or jewels.

If there's a remainder of gold objects, melt them, and divide into five equal ingots.

If there's a remainder of jewels, donate to Pirate's Widows Charitable Fund.

JK

  Hide contents

Looks like a Nash equilibrium.

A series of votes is held, based on the oldest pirate counting out piles.  That pirate counts out five piles of treasure, then all vote on whether to accept that division or not.   If a division of treasure is accepted by majority then the pirate who counted out the piles lives and everyone takes their portion.  If majority rejects, then counter is killed, and process starts again.  Since the counter wants to live, he will count scrupulously.  Since no other pirate wants to run the risk of being counter (except the youngest, who has no such risk), the majority will accept the first count.

 

Not peaceful enough

We don't want to introduce any other incentives, such as a desire not to be killed. The only criterion should be that each one is satisfied with his own portion of the treasure.

They should not be forced to accept. They should be individually satisfied.

Not by majority either. Literally, each pirate should leave satisfied.

A solution should be as good for each one of the five pirates as in the example of two pirates above.

[md65536]

6 hours ago, Genady said:

If there were two pirates rather than five, a solution could be that one of them divides the treasure into two parts, and the other one chooses. 

There are a lot of extra assumptions here. 1. the treasure can be divided to any precision. This wouldn't work if the treasure was 3 large diamonds. 2. The one dividing believes they can do so exactly. 3. There is no conflict in choosing who divides and/or who picks.

(I guess pirates are more peaceful than children.)

Spoiler

A simple extension is just have one pirate divide, and let another pirate pick. Then repeat the process with the remaining 4 pirates.

This adds an assumption that there's no conflict in who divides and (but not or) who picks. I don't think this is a good solution, but without knowing what the actual assumptions are, I don't know what needs fixing. I don't think it would work in practice.

 

[Genady]

25 minutes ago, md65536 said:

There are a lot of extra assumptions here. 1. the treasure can be divided to any precision. This wouldn't work if the treasure was 3 large diamonds. 2. The one dividing believes they can do so exactly. 3. There is no conflict in choosing who divides and/or who picks.

(I guess pirates are more peaceful than children.)

  Hide contents

A simple extension is just have one pirate divide, and let another pirate pick. Then repeat the process with the remaining 4 pirates.

This adds an assumption that there's no conflict in who divides and (but not or) who picks. I don't think this is a good solution, but without knowing what the actual assumptions are, I don't know what needs fixing. I don't think it would work in practice.

 

Yes, the assumption 1 is necessary. We assume that the treasure is as close to a continuum as needed. The assumed precision is the precision of their evaluation of the pieces.

I'm not sure why the other assumptions are required. For example, why there would be a conflict between who divides and/or who picks in the case of two, given the assumption 1.

[Genady]

1 hour ago, md65536 said:

(I guess pirates are more peaceful than children.)

They are not more peaceful, but more mature than children. Specifically, they don't care if another pirate got a larger share than them. Each one cares only that his share is not less than one fifth of the total, in his evaluation.

[md65536]

Then my previous answer is no good. I guess any pirate is satisfied if they get to divide the treasure, and any pirate is satisfied if they can pick, but obviously the others wouldn't be happy letting someone else pick.

Spoiler

One splits the treasure. Everyone else picks. The first picks from whatever wasn't picked. Any piles of loot that were picked by exactly one pirate are taken, and the remaining pirates repeat the process with the remaining loot. Each time, at least the pirate that divided the treasure is satisfied.

 

Edit: This doesn't work either, given your last post. I could divide them as 50%, 47%, 1%, 1%, 1%, hoping everyone would pick the 50% and I'd get 47%. Then the pirate who never gets to split won't be satisfied. Or someone will pick the the 47% and I won't be satisfied.

 

Edit2: I suppose they can make it work. The dividing pirate has to divide fairly if they want to be satisfied. Every other pirate, if they think the one who divided will end up with a pile that's more than a fifth, making subsequent rounds potentially unsatisfiable, they can just choose that one and be satisfied. All it would take is that they purposefully choose a satisfactory option rather than one that could cause dissatisfaction later.

 

Edited May 28, 2023 by md65536

[Genady]

10 minutes ago, md65536 said:

Then my previous answer is no good. I guess any pirate is satisfied if they get to divide the treasure, and any pirate is satisfied if they can pick, but obviously the others wouldn't be happy letting someone else pick.

  Hide contents

One splits the treasure. Everyone else picks. The first picks from whatever wasn't picked. Any piles of loot that were picked by exactly one pirate are taken, and the remaining pirates repeat the process with the remaining loot. Each time, at least the pirate that divided the treasure is satisfied.

 

This is good. I have a different procedure in mind, but yours works as well. +1

[Phi for All]

Spoiler

I would start with one pirate who divides the treasure in equal fifths so that he'll be happy with any share, then have the second pirate pick which share the first gets. Then the second divides the remains into fourths, and the third picks his share. Third divides into equal thirds and the fourth picks the third's share. The fourth splits the remains in two, and the fifth gets to pick which one the fourth gets, which means he also gets to pick which is his share, which is only fair since he had to wait.

 

[TheVat]

Is a double pan scale involved?  

[Genady]

5 minutes ago, Phi for All said:

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I would start with one pirate who divides the treasure in equal fifths so that he'll be happy with any share, then have the second pirate pick which share the first gets. Then the second divides the remains into fourths, and the third picks his share. Third divides into equal thirds and the fourth picks the third's share. The fourth splits the remains in two, and the fifth gets to pick which one the fourth gets, which means he also gets to pick which is his share, which is only fair since he had to wait.

 

Spoiler

I see a problem here. The first step will make the first and the second pirates happy, but the third or other pirate's evaluation might be that the first got more than 1/5 of the total leaving less than 4/5 to the rest of them. Somebody will not be satisfied in such a case, I think.

 

9 minutes ago, TheVat said:

Is a double pan scale involved?  

No scales involved. All evaluations are subjective.

[md65536]

My answer before isn't good enough.

Consider a worst-case scenario:

One pirate divides the loot, and the others evaluate. Say each of the other 4 think there is only one share that is less than equal, and that is the only share that they'd be satisfied with someone taking, but also each of the 4 think that a different share is the smallest. Then there is no share that can be kept by anyone, where they all agree on.

Therefore I think that no solution that involves the piles being split up, and then allowing one of them to be kept, can work. I think any solution that involves one person splitting up the loot, would require that the shares be adjusted before guaranteeing that they all agree.

[Genady]

7 minutes ago, md65536 said:

My answer before isn't good enough.

Consider a worst-case scenario:

One pirate divides the loot, and the others evaluate. Say each of the other 4 think there is only one share that is less than equal, and that is the only share that they'd be satisfied with someone taking, but also each of the 4 think that a different share is the smallest. Then there is no share that can be kept by anyone, where they all agree on.

Therefore I think that no solution that involves the piles being split up, and then allowing one of them to be kept, can work. I think any solution that involves one person splitting up the loot, would require that the shares be adjusted before guaranteeing that they all agree.

Spoiler

You are certainly right. I've missed that.

You are right in your conclusion. That's what happening in the solution I know of. I think you're very close.

 

[md65536]

2 hours ago, md65536 said:

Therefore I think that no solution that involves the piles being split up, and then allowing one of them to be kept, can work. I think any solution that involves one person splitting up the loot, would require that the shares be adjusted before guaranteeing that they all agree.

Nope, I'm wrong again! This is bad reasoning.

Lol, this puzzle has broken my brain. Every time you say I'm right I disagree. Every time I realize something I just said was wrong, I forget some other detail and say something else wrong, all the time flipping back and forth between 2 incorrect positions.

What I forgot is that is that it doesn't matter if someone is not satisfied with letting some other share be taken, IF they're also taking a share that they're satisfied with at the same time. So it might (should? but I don't want to give any more answers unless I figure it out) be possible to have someone split up the treasure, and have either everyone satisfied with the shares everyone else is taking, OR they're taking a share they're satisfied with. (However, my previous solution still doesn't work, sometimes not fulfilling either condition.)

[Genady]

Spoiler

I still think that the conclusion,

5 hours ago, md65536 said:

any solution that involves one person splitting up the loot, would require that the shares be adjusted before guaranteeing that they all agree

is correct.

 

[Intoscience]

 

Spoiler

They take it in turns to choose one item each until all the loot is gone.

They draw straws on who chooses first, second, third, fourth, last. The person drawing last goes first on the next go followed by fourth place and so on until each gets a turn of first pick per round. This process continues so each pirate picks in each position over and over until the loot is gone. If at the final round there is not enough loot to go around for everyone to have one final pick then the person who drew the short straw originally gets the final first pick followed by the fourth and so on.

 

Edited May 29, 2023 by Intoscience

[Genady]

2 hours ago, Intoscience said:

 

  Hide contents

They take it in turns to choose one item each until all the loot is gone.

They draw straws on who chooses first, second, third, fourth, last. The person drawing last goes first on the next go followed by fourth place and so on until each gets a turn of first pick per round. This process continues so each pirate picks in each position over and over until the loot is gone. If at the final round there is not enough loot to go around for everyone to have one final pick then the person who drew the short straw originally gets the final first pick followed by the fourth and so on.

 

Spoiler

I don't think this system would make them all satisfied. For example, assume the values of the items, in somebody's evaluation, are 1000, 999, 998, 997, 996, ... The first one to pick gets 1000+995+990+985+..., while the fifth one to pick gets 996+991+986+981+... At every round, the first gets higher values than the second etc. and the differences accumulate. The final totals will be very different and not satisfactory to some of them.

They could draw straws at every step, but then it is a lottery. We want a system that gives everyone a share according to their values of the items by their choice, not by chance. The example above (https://www.scienceforums.net/topic/131753-pirates/?do=findComment&comment=1240976) achieves this perfectly in case of two pirates.

 

 

[md65536]

11 hours ago, Genady said:

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I still think that the conclusion,

is correct.

What I wrote is ambiguous...

Spoiler

What I meant was, "There's no solution where one person can divide the loot, and then one or more people can take a share, to everyone's satisfaction." I meant that before any shares could be taken out, they'd have to be readjusted. I think this is wrong, because in all the cases I've looked at, if anyone's dissatisfied with one of the shares being taken, there's another share that they can take and be satisfied, or there's someone else who can take a share. I'm unable to find a case that won't work (unless I allow unreasonable pirates, eg. one who thinks a share is < 1/5 but thinks the remainder adds up to < 4/5).

It could also mean, "There's no solution where one person divides the loot, and everyone gets a share in some way that everyone is satisfied taking that share." I agree this is true, eg. if 2 pirates both only want share #1, there is no way to satisfy them. I didn't mean this. The (wrong) solution I gave involved stages, where each stage the remaining loot is divided up once by one person, and then at least one share (and pirate) is taken away, without the shares needing readjustment. I still think such a solution is possible.

It's also impossible to have a solution where one person divides the loot, and exactly one person takes a share, to everyone's agreement.

The solution I'm working on seems to work, but I'm having trouble proving it:

Spoiler

One pirate splits the loot into 5 shares, and they're willing to take any share, or let any share go. The others decide for each of the other shares, if they'd be willing to take it or if they'd be willing to let someone else take it (must be at least one, if they're reasonable pirates). Then there must be a subset of n pirates for which n shares can be divided to each of their satisfaction, with the remaining pirates satisfied to let all of those n shares go. (Proving this has been elusive.) Let the n pirates take their satisfactory share and then repeat the process with the remaining loot and pirates.

 

It seems that if you make enough pirates refuse to let shares go, it means you're making more shares acceptable for them to take, and there's always either enough shares that can be let go, or enough pirates able to find a share they're satisfied with, but I haven't proven it.

 

 

Edited May 29, 2023 by md65536

[Genady]

38 minutes ago, md65536 said:

The solution I'm working on seems to work, but I'm having trouble proving it:

Spoiler

Let's try the following scenario. The pirates are A, B, C, D, E. A divides the loot into 5 shares: U, W, X, Y, Z.

A is willing to take any of them or let go any of them. B is willing to let only U go. C is willing to let only W go. D is willing to let only X go. E is willing to let only Y go.

Seems to me that there is no "a subset of n pirates for which n shares can be divided to each of their satisfaction, with the remaining pirates satisfied to let all of those n shares go." Do I miss something?

 

[TheVat]

On 5/27/2023 at 12:11 PM, Genady said:

Each one will be satisfied if in his estimate his share is not smaller than one fifth of the total. Their estimates vary.

Seems like none of the solutions really satisfy the situation as stated.  To make an estimate requires getting some basic metric of the loot, like total mass or number of doubloons or whatever.  Seems like some prior metric of value must be agreed upon, for estimates to be generally satisfactory and to have peace.  Without the objectivity of math, how would the pirates ever avoid endless argument?  

[Genady]

2 minutes ago, TheVat said:

Seems like none of the solutions really satisfy the situation as stated.  To make an estimate requires getting some basic metric of the loot, like total mass or number of doubloons or whatever.  Seems like some prior metric of value must be agreed upon, for estimates to be generally satisfactory and to have peace.  Without the objectivity of math, how would the pirates ever avoid endless argument?  

I believe I have such a no-agreed-upon-metric solution. Do you want a little hint?

[md65536]

13 minutes ago, Genady said:

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Let's try the following scenario. The pirates are A, B, C, D, E. A divides the loot into 5 shares: U, W, X, Y, Z.

A is willing to take any of them or let go any of them. B is willing to let only U go. C is willing to let only W go. D is willing to let only X go. E is willing to let only Y go.

Seems to me that there is no "a subset of n pirates for which n shares can be divided to each of their satisfaction, with the remaining pirates satisfied to let all of those n shares go." Do I miss something?

 

Spoiler

B takes W

C X

D Y

E Z

A U

n=5. Here, if B thought that W was less than 1/5th, they'd reasonably be willing to let it go. Since they don't, they should be satisfied to take it. etc.

 

[Phi for All]

Spoiler

Since the treasure has elements that make it difficult to decide equal shares (can't go by weight, nor by value judged by the pirates), what about finding a buyer for the whole treasure willing to pay in bank notes that are easily divisible by five? Can I add an extra element or does this have to be done strictly by the pirates?

 

[Genady]

7 minutes ago, Phi for All said:

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Since the treasure has elements that make it difficult to decide equal shares (can't go by weight, nor by value judged by the pirates), what about finding a buyer for the whole treasure willing to pay in bank notes that are easily divisible by five? Can I add an extra element or does this have to be done strictly by the pirates?

 

Strictly by the pirates.

You can consider items to be small enough that they don't need to be cut or broken.

The only way to go is by the subjective value judgement of the pirates.