Pirates

[Genady]

  On 5/29/2023 at 7:49 PM, md65536 said:

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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.

 

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Aha, got it. Still looking for a counterexample. Let's try this scenario:

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B wants U or W.

C, D, E each want U or X.

Seems OK for A and B to take U and W between them. C, D, and E now need to split X, Y, and Z. 

But X+Y+Z might be < 3/5, and somebody will not be satisfied.

Am I right?

 

[TheVat]

Think I see it.

 

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It is pure game theory.  Everyone decides what is a fair one fifth of the loot.  They write this down, and place in a blank envelope.  Five envelopes go in a barrel, which is rolled around the camp, then everyone draws out an envelope, not knowing whose estimate they will get.  

Everyone sees the optimal choice is to be as scrupulous as possible.  Short someone else, and you can be shorted.  Make a greedy estimate, you have only 1/5 chance of getting it.  

 

Edited May 29, 2023 by TheVat
addendum

[Genady]

  On 5/29/2023 at 11:12 PM, TheVat said:

Think I see it.

 

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It is pure game theory.  Everyone decides what is a fair one fifth of the loot.  They write this down, and place in a blank envelope.  Five envelopes go in a barrel, which is rolled around the camp, then everyone draws out an envelope, not knowing whose estimate they will get.  

Everyone sees the optimal choice is to be as scrupulous as possible.  Short someone else, and you can be shorted.  Make a greedy estimate, you have only 1/5 chance of getting it.  

 

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I see three problems:

1. The lists will overlap. One pirate gets a list of items A, B, C, D, ... and another gets a list B, D, J, K, ... What will they do?

2. Some items will be not listed.

2. Not everyone will be satisfied, because a list which is an honest 1/5 for one might be worth, say only 1/10 for another.

 

[TheVat]

  On 5/30/2023 at 12:03 AM, Genady said:

Not everyone will be satisfied, because a list which is an honest 1/5 for one might be worth, say only 1/10 for another.

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Well that is a kind of treasure where you have heterogeneous collection of objects.  If, as you indicate...

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

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...then many puzzle solvers assume it is items that are to be traded as money, like gold coins.  Hence, pirates.  My solution is workable when the treasure is fairly homogeneous and subjective estimates are simply about amount.  These are experienced pirates who know the current exchange rate for doubloons.  Estimates, in my system, will tend to converge on accurate monetary appraisals.  

With heterogeneous or sui generis items, it would seem highly improbable that any accord could be reached, and game theory goes out the window.  Time to call Sotheby's.  I know I am missing something (but that's the fun,  yes?)

[Genady]

  On 5/30/2023 at 1:21 AM, TheVat said:

heterogeneous or sui generis items

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Yes. Many of those.

 

  On 5/30/2023 at 1:21 AM, TheVat said:

game theory goes out the window

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Yes. No probabilities.

 

  On 5/30/2023 at 1:21 AM, TheVat said:

Time to call Sotheby's.

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No. All is done between the pirates.

They follow a definite procedure, and after a finite number of steps each pirate keeps a part of the loot which is at least one fifth of the total according to his own subjective evaluation.

Edited May 30, 2023 by Genady

[md65536]

  On 5/29/2023 at 9:30 PM, Genady said:

Aha, got it. Still looking for a counterexample. Let's try this scenario:

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B wants U or W.

C, D, E each want U or X.

Seems OK for A and B to take U and W between them. C, D, and E now need to split X, Y, and Z. 

But X+Y+Z might be < 3/5, and somebody will not be satisfied.

Am I right?

 

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Yes, that's right. I'm assuming that if someone wants a share, and someone else takes it, and the former doesn't also get a share that they're satisfied with from that same division of shares, then they're dissatisfied. Like a child saying they wanted the purple-flavour yogurt that their sibling took and the parent opening a new purple-flavour yogurt and trying to argue that it's the same amount... it's hopeless! Only their own judgment can determine if they're satisfied.

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But in this case, with so few shares wanted, there are a lot of shares people are willing to part with. No one wants Y or Z, so everyone is satisfied if A takes one of them, and they could go on to the next round. However, only B wants W, so they can also take that. C, D, E can't be satisfied with this division of the treasure, but they'd be okay to have U, X, Z merged and redivided again.

I considered the case where C,D,E "especially hate" Z so much while thinking that U and X are only slightly better than equal, and won't think this is fair. But if they're reasonable, they'll know that if only unwanted shares are taken away, less than 1/5th per taken share goes (by their judgment), and there must be at least 1/5th times the shares left remaining. If their only criteria for satisfaction is they want at least 1/5th by their judgment, this still works.

Oh, I just realized that if they're inconsistent with their judgment between rounds, then it doesn't work. Like, if they think the 3 options left have 61% of the total treasure, and then after they're divided again think each of the options is < 20%, they likely can't be satisfied. So I'm assuming they're reasonable and consistent.

 

Edited May 30, 2023 by md65536

[Intoscience]

  On 5/29/2023 at 11:56 AM, Genady said:

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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

 

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My thinking order of choice is something like as follows:

1.  1,5,4,3,2,1,5,4,3,2,1
2.  2,4,3,2,1,5,4,3,2,1
3.  3,3,2,1,5,4,3,2,1,5
4.  4,2,1,5,4,3,2,1,5,1
5.  5,1,5,4,3,2,1,5,1,5

Not sure if this would even out over the rounds, I was thinking about a number of different systems similar to this that might work better. 

 

[Genady]

  On 5/30/2023 at 6:15 AM, Intoscience said:

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My thinking order of choice is something like as follows:

1.  1,5,4,3,2,1,5,4,3,2,1
2.  2,4,3,2,1,5,4,3,2,1
3.  3,3,2,1,5,4,3,2,1,5
4.  4,2,1,5,4,3,2,1,5,1
5.  5,1,5,4,3,2,1,5,1,5

Not sure if this would even out over the rounds, I was thinking about a number of different systems similar to this that might work better. 

 

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I think that any order of choosing allows for a possibility that some pirate gets less than what he perceives one fifth of the loot. This is unacceptable. They have to get definitely a satisfactory share each.

  On 5/30/2023 at 5:10 AM, md65536 said:

Yes, that's right. I'm assuming that if someone wants a share, and someone else takes it, and the former doesn't also get a share that they're satisfied with from that same division of shares, then they're dissatisfied. Like a child saying they wanted the purple-flavour yogurt that their sibling took and the parent opening a new purple-flavour yogurt and trying to argue that it's the same amount... it's hopeless! Only their own judgment can determine if they're satisfied.

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But in this case, with so few shares wanted, there are a lot of shares people are willing to part with. No one wants Y or Z, so everyone is satisfied if A takes one of them, and they could go on to the next round. However, only B wants W, so they can also take that. C, D, E can't be satisfied with this division of the treasure, but they'd be okay to have U, X, Z merged and redivided again.

I considered the case where C,D,E "especially hate" Z so much while thinking that U and X are only slightly better than equal, and won't think this is fair. But if they're reasonable, they'll know that if only unwanted shares are taken away, less than 1/5th per taken share goes (by their judgment), and there must be at least 1/5th times the shares left remaining. If their only criteria for satisfaction is they want at least 1/5th by their judgment, this still works.

Oh, I just realized that if they're inconsistent with their judgment between rounds, then it doesn't work. Like, if they think the 3 options left have 61% of the total treasure, and then after they're divided again think each of the options is < 20%, they likely can't be satisfied. So I'm assuming they're reasonable and consistent.

 

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So, is there a definite procedure? 

[Intoscience]

  On 5/30/2023 at 9:23 AM, Genady said:

So, is there a definite procedure? 

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I'm not sure if there is one.

  On 5/30/2023 at 9:23 AM, Genady said:

I think that any order of choosing allows for a possibility that some pirate gets less than what he perceives one fifth of the loot. This is unacceptable. They have to get definitely a satisfactory share each

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I think there are 2 issues, first issue is, is there a system that produces definitive equal shares? The second issue is, does that system, even that which is definitive in quantity satisfy the perception of each and every pirate? Each pirate has to agree on the value of each item and that the value should match on each and every pick. Then each round should produce an equal quantity of items.

Can this be done?    

[Genady]

  On 5/30/2023 at 10:27 AM, Intoscience said:

I'm not sure if there is one.

I think there are 2 issues, first issue is, is there a system that produces definitive equal shares? The second issue is, does that system, even that which is definitive in quantity satisfy the perception of each and every pirate? Each pirate has to agree on the value of each item and that the value should match on each and every pick. Then each round should produce an equal quantity of items.

Can this be done?    

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Yes, there is a definite procedure. But you ask too much from it. They don't need to agree on the value of each item, for example, and they don't need to agree that the shares are equal. All they need to get is that each one is satisfied with his own share.

[Intoscience]

  On 5/30/2023 at 10:40 AM, Genady said:

Yes, there is a definite procedure. But you ask too much from it. They don't need to agree on the value of each item, for example, and they don't need to agree that the shares are equal. All they need to get is that each one is satisfied with his own share.

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Ah, a more simple approach needed then.

A better trick would be a way in which each perceives they themselves have the best share. 

[Genady]

  On 5/30/2023 at 11:09 AM, Intoscience said:

Ah, a more simple approach needed then.

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Yes.

 

  On 5/30/2023 at 11:09 AM, Intoscience said:

A better trick would be a way in which each perceives they themselves have the best share.

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Am not sure this is possible. For example, A can be happy with his share but also think that B screwed up and let C have too much. 

[Intoscience]

  On 5/30/2023 at 11:17 AM, Genady said:

Am not sure this is possible. For example, A can be happy with his share but also think that B screwed up and let C have too much

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So each pirate has to be happy and content with their own lot, but also happy that each other member has been treated equally fairly? 

[Genady]

  On 5/30/2023 at 12:32 PM, Intoscience said:

So each pirate has to be happy and content with their own lot 

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That's it. No other requirements.

[TheVat]

Seems to be a problem if all pirates happen to think one of the sui generis items is by far the most valuable part of the treasure.  Let's call it the Holy Grail.  They obtained it by bravely confronting the Gorge of Eternal Peril, the deadly Rabbit of Caerbannog, the Knights Who Say "Ni!", etc.  The Holy Grail cannot be divided up, therefore four pirates are dissatisfied.  Which leads me to speculative answer number three (actually I've lost track of the number, perhaps it is four)

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The OP does not indicate that pirates will do anything with the treasure but hoard it.  If simply having treasure is satisfactory, then the pirates can incorporate, and issue 5 bearer bonds to each other, each worth 1/5 of corporate assets.  When anyone dies, then the bonds are destroyed, and a new round of bonds is issued at 1/4 value.  And so on.  No outsider is required, as this is a black market corporation.  

 

[Genady]

  On 5/30/2023 at 1:45 PM, TheVat said:

Seems to be a problem if all pirates happen to think one of the sui generis items is by far the most valuable part of the treasure.

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Assume that there is no such item in the treasure. I thought such problem was eliminated by, "You can consider items to be small enough that they don't need to be cut or broken." I've meant, "small enough" by value, as size of the items doesn't seem to be a factor in this puzzle.

They want to take their shares, leave, and never see each other again  

[Genady]

Here is the first step of the procedure, if you're interested:

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A pirate takes from the treasure chest and put out a pile that he is willing to take as his share.

(I.e., he does not split the treasure into five parts.) 

 

[TheVat]

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Seems like a voting system must be found.  Other four inspect A's pile, them vote on whether it seems fair.  If any no votes, then they remove objects from A pile and put back in chest to satisfy them, then vote is taken again, until all accept A pile.  Then B takes a share....etc.

Each pirate agrees to this vote/cull system, each knowing that what is left for them depends on what they accept, that they must allow a cull of their own pile that leaves enough for those who come after, and that sufficient must be left for E, the last one, so that the final round voting will be unanimous.  If E does not accept the final pile, then the whole process starts over.

 

 

[Genady]

  On 5/31/2023 at 1:03 PM, TheVat said:

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Seems like a voting system must be found.  Other four inspect A's pile, them vote on whether it seems fair.  If any no votes, then they remove objects from A pile and put back in chest to satisfy them, then vote is taken again, until all accept A pile.  Then B takes a share....etc.

Each pirate agrees to this vote/cull system, each knowing that what is left for them depends on what they accept, that they must allow a cull of their own pile that leaves enough for those who come after, and that sufficient must be left for E, the last one, so that the final round voting will be unanimous.  If E does not accept the final pile, then the whole process starts over.

 

 

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Putting some items back to the chest is right, but the voting would not work. 

Other pirates can vote the pile down to what is unacceptable to A, or B, etc. Nobody has to take what is unacceptable to him. The procedure is not guaranteed to end in a finite number of steps.

The following is a kind of hint. When you have mentioned,

  On 5/30/2023 at 1:21 AM, TheVat said:

Sotheby's

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I thought for a moment that you're on the right track, but not in the way you've meant it.

 

[md65536]

  On 5/31/2023 at 11:49 AM, Genady said:

Here is the first step of the procedure, if you're interested:

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A pirate takes from the treasure chest and put out a pile that he is willing to take as his share.

(I.e., he does not split the treasure into five parts.) 

 

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Well that's much simpler!

The first pirate makes one share of the treasure that they're willing to take AND willing to give away.

If any of the others think it's too big a share, they can decrease it until they're willing to take it or give it away. After all pirates have had a chance to do that, the last pirate who adjusted the pile is willing to take it, and everyone else was already satisfied with letting go of the share when it was larger, so they're still satisfied letting them take it.

Repeat the process, removing one share at a time, until there are 2 left, at which point it's clear the last pirate would be satisfied with the remaining treasure (but if that's not clear, they could always fall back on "one divides, the other picks". Actually, if you can remove 0 treasure and call that an adjustment, then it's the same as "one divides, the other picks").

 

Any pirate who adjusts the pile to their satisfaction must also be willing to let someone else take it, or it's possible that someone else removes treasure and the former pirate is still unwilling to let it go.

 

Edited May 31, 2023 by md65536

[Genady]

  On 5/31/2023 at 4:31 PM, md65536 said:

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Well that's much simpler!

The first pirate makes one share of the treasure that they're willing to take AND willing to give away.

If any of the others think it's too big a share, they can decrease it until they're willing to take it or give it away. After all pirates have had a chance to do that, the last pirate who adjusted the pile is willing to take it, and everyone else was already satisfied with letting go of the share when it was larger, so they're still satisfied letting them take it.

Repeat the process, removing one share at a time, until there are 2 left, at which point it's clear the last pirate would be satisfied with the remaining treasure (but if that's not clear, they could always fall back on "one divides, the other picks". Actually, if you can remove 0 treasure and call that an adjustment, then it's the same as "one divides, the other picks").

 

Any pirate who adjusts the pile to their satisfaction must also be willing to let someone else take it, or it's possible that someone else removes treasure and the former pirate is still unwilling to let it go.

 

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Right! +1

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It is actually just a form of auction.

 

[TheVat]

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That's great.  Oddly enough, I was thinking of how some auction system might work (with bidding on each item, which reflected how much each pirate valued it) but got bogged down in how it would be refereed.  Overthinking is deadly!  😀

 

[Jez]

I would say it is impossible to be sure a solution can be found, as one pirate may always be unreasonable. I mean, it's in the job description, right?

So I think there would have to be some component of compulsion that forces a pirate to agree what they are presented with is a fair share, even if they don't really think it is.

Like the 'Here is $100 for you two people, you can give the other person $60 and you keep the $40 and agree that's fair, or neither of you gets anything'. Like that.

There has to be something like a democratic acceptance of a given rule, and then to blindly agree that it is fair even if it isn't really.

In fact, maybe I should say just like democracy! 😄

Here is what I'd say;

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Five pieces of random treasure are placed in a row. Each takes a piece of treasure in turn (we are assuming it is a random set of dissimilar objects) and places it in the pile 'they' think is the least valuable pile.

Once all the treasure is allocated into 5 piles, then they all have to agree that each pile is exactly 1/5th of the total to be allowed to get to draw a straw which allocates a random pile to each of the pirates. If any refuse to accept all 5 piles are an exact division of 1/5th then they forfeit their share and get nothing.

 

[Genady]

  On 6/3/2023 at 7:29 PM, Jez said:

I would say it is impossible to be sure a solution can be found, as one pirate may always be unreasonable. I mean, it's in the job description, right?

So I think there would have to be some component of compulsion that forces a pirate to agree what they are presented with is a fair share, even if they don't really think it is.

Like the 'Here is $100 for you two people, you can give the other person $60 and you keep the $40 and agree that's fair, or neither of you gets anything'. Like that.

There has to be something like a democratic acceptance of a given rule, and then to blindly agree that it is fair even if it isn't really.

In fact, maybe I should say just like democracy! 😄

Here is what I'd say;

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Five pieces of random treasure are placed in a row. Each takes a piece of treasure in turn (we are assuming it is a random set of dissimilar objects) and places it in the pile 'they' think is the least valuable pile.

Once all the treasure is allocated into 5 piles, then they all have to agree that each pile is exactly 1/5th of the total to be allowed to get to draw a straw which allocates a random pile to each of the pirates. If any refuse to accept all 5 piles are an exact division of 1/5th then they forfeit their share and get nothing.

 

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However, there IS a more satisfactory procedure. The pirates are reasonable to the extent that each wants to get a share that he thinks is at least one fifth of the total value.

Think about it this way: if such a procedure exists in the case of two pirates, what would prevent it to exist in the case of five?

[Jez]

  On 6/3/2023 at 8:00 PM, Genady said:

However, there IS a more satisfactory procedure. The pirates are reasonable to the extent that each wants to get a share that he thinks is at least one fifth of the total value.

Think about it this way: if such a procedure exists in the case of two pirates, what would prevent it to exist in the case of five?

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You can run the 'take my pile if you think it's bigger' method for powers of 2, but not 5.

OK, so the first step is to get the 1st pirate out of the way. 

1. Rough out 1/5 of the loot, and ask if anyone wants it, if no-one wants it then keep adding to the pile until someone takes it, or if everyone wants it, keep removing pieces until there is only one pirate asking for it, then they get that. A pirate considering whether to take it has to decide whether they are more likely to get more at this stage than if they go to the next, so there is a motivation to 'lock in the deal' early.

Then you can do this with any power of 2;

2. Roughly halve the remaining pile and the two pirates standing next to a given pile will spilt it like for two people. If there are 3 or more pirates standing next to a pile, move the treasure piece by piece to the other pile until each pile has two pirates who want to share that particular pile.