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A wealthy person has passed away, naming three siblings as beneficiaries of the vast estate. Unfortunately, the will states that:

My estate shall be divided evenly, such that no sibling shall feel their share is smaller than any other share.

As the estate contains many illiquid assets of sentimental value, one cannot simply convert everything into monetary value and numerically divide. For simplicity however, assume that all assets can be arbitrarily divided.

As it turns out, the three siblings came together and, after some back-and-forth, arrived at a partition where no one sibling felt they received a smaller share than any other, fulfilling the will.

How did they arrive at their partition?

This is a well known puzzle, which is usually posed as sharing a cake, and how to cut it. Although the solution is simple to understand, it might not be so easy to independently come up with it.

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  • $\begingroup$ "This is a well known puzzle" - Have you tried to check whether it is already present here first? $\endgroup$ – klm123 Oct 15 '14 at 6:37
  • $\begingroup$ @klm123 yes, didn't find one. Feel free to flag as duplicate if it's already here. $\endgroup$ – congusbongus Oct 15 '14 at 6:48
  • $\begingroup$ More general question were asked here: puzzling.stackexchange.com/questions/1653/… . But from your spoiler-comment about well known puzzle and simple solution I think you confused the formulation and meant "such that no sibling shall feel their share is smaller than 1/3 of the estate". In current formulation puzzle doesn't have simple to understand solution and therefore is not well known at all. $\endgroup$ – klm123 Oct 15 '14 at 8:06
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    $\begingroup$ puzzling.stackexchange.com/questions/1653/… $\endgroup$ – d'alar'cop Oct 15 '14 at 10:13
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    $\begingroup$ "[T]he estate contains many illiquid assets of sentimental value" makes it unsolvable. If Alice and Carol both value their late mother's wedding ring above all else, even above everything else combined, then even if Bob is too struck with grief to want anything and is willing to forego his inheritance to keep the peace, the one that doesn't get the ring will feel the one that does, gets more than her fair share. $\endgroup$ – SQB Oct 21 '14 at 12:11
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  1. Monetize what can be monitized (aka cash and bank accounts). This is the pot and has a value P.

  2. Have a referee separate all of the remaining pieces into 3 (what he sees even) regions. Call these the chunks.

  3. Have all players chose which of the chunks is the biggest. If they all disagree, split any remaining pot evenly between them.

  4. If none choose one chunk, place all of the pot on that chunk. If none wants two chunks, place P evenly on the other two.

  5. Repeat step 3.

  6. If multiple people still want a part with no money, the referee failed to split well and should be fired. Repeat from step 1. (this is the hole is this plan for most simular questions. It works well for spliting apartment rent though).

  7. Take the option which multiple people want. Have the referee take money from that chunk slowly and add it to the pot. When someone wants to switch chunks, have them yell stop.

  8. repeat step 3.

  9. Have the referee move to the one no one wants, slowly remove identical amounts from the other two. Have someone yell stop when they want to switch.

  10. repeat step 3. It should work this time. Everyone is holding onto the one they think is the largest so they should not be envious.

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  • $\begingroup$ Alternatively use Sperner's Lemma: en.wikipedia.org/wiki/Sperner's_lemma $\endgroup$ – kaine Oct 15 '14 at 14:10
  • $\begingroup$ The referee should be one of the 'players', since the question states that they divide it themselves. $\endgroup$ – SQB Oct 15 '14 at 19:41
  • $\begingroup$ Ah, sorry. Doesnt really matter though. As long as long as there is a large portion of the inheretance can be monetized. The referee does not have a means to manipulate the game too much. If P exeeds the value of the maximum percieved chunk size differences ((A+B-2C) for who ever thinks C is the most unfairly small) you should be fine. $\endgroup$ – kaine Oct 15 '14 at 19:48
  • $\begingroup$ The flaw with this setup is similar to @SQB's solution: the player that left first (say from step 6) can still be unhappy with how the remaining 2/3 is split, as one of them could have been greater than what the first left with. Since this player left first, she had no say in how the remaining 2/3 were partitioned, so she could be envious. $\endgroup$ – congusbongus Oct 17 '14 at 6:39
  • $\begingroup$ @congusbongus I removed the line that gave him permission to leave though it was unlikely to begin with. That line was not important to the strategy. $\endgroup$ – kaine Oct 21 '14 at 17:34
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Another answer because this is inherently a different method:

  1. Make 3 piles. Two are actually empty. One contains everything.
  2. Have all 3 sibilings in random order pick a pile they want but a maximum of two can pick the same pile.
  3. Intentionally at #3 again: if at any point all 3 have selected different piles and no one wants to switch, everyone has their piles. We will be done then.
  4. If the rules are being followed: 2 people want 1 pile (call it A), 1 person has chosen another pile (call it B), one pile is undesired (call it C). The person at pile B takes any 1 object from A and places it in C.
  5. The player at pile B leaves pile B. The 2 players at pile A are given the option to select to any pile. The remaining player is allowed to pick his new pile. Still, a maximum of 2 is allowed on any one pile.
  6. Repeat from step 3 until the loop ends.

Please let me know how this would break apart if it would. It will be envy free.

The motivation of the player at B is to increase the net value of C. I find this interesting as a lock + a key has alot more value that either separately. Two copies of GTA5 have little more value than one. He has an economic incentive to cherry pick from A to make A, B, and C have the highest possible values from his perspective whilst leaving in A the things they want more than him. As a result this will work alot better than a random cut or refereed impartial separation.

This falls apart if one inseparatable object has more value to multiple players than half of everything else. It also fails if one party has the more motivation to extend the game than to obtain the largest pile or some party is willing to accept a smaller inheretance and whine about it later. I think, however, that any plan would fail under these conditions.

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Have Alice divide the estate into 3 fair (from her POV) parts. Offer Bob and Eve the choice of the three portions. If they choose different ones, we're done.

If they choose the same portion, split it between them. Then have them each choose one of the remaining 2 portions. If they choose the same portion, split that between them. They each now have a fair third, and Alice takes the remainder. If they choose different portions at this point, have them each share the portion they choose with Alice.

Splitting a portion between two people is, of course, done by having one person divide it and the other choose.

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  • $\begingroup$ You are missing an important point. Alice doesn't just want to have a fair share. She wants a piece that is either the largest or tied for the largest. If she does not agree with how they split it, she will not be happy. $\endgroup$ – kaine Oct 21 '14 at 17:16
  • $\begingroup$ You're right, of course. I was trying to solve "make sure each sibling believes they got at least one third", which I mistakenly thought to be equivalent. $\endgroup$ – frodoskywalker Oct 21 '14 at 18:19
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The classic problem is that you have two children fighting for a cake. Child A divides the cake in two parts such that it considers both of equal size. Child B now chooses the part it considers bigger. Note that child A might have considered only cake mass whereas child B might have considered the number of cherries, i.e. the value they attach to the parts isn't necessarily equal.

Now how to generalize this to three people? Additionally we're looking for a sharing protocol that is 1. envy-free (i.e. it suffices not to have at least a third but nobody else may have more) and 2. discrete (since the inheritance is not perfectly divisible).

Now, I could write down the protocol here, but it's easier to directly point you to the solution which is the first result on Google: http://www.mathematics-in-europe.eu/home/13-frontpage/popular-articles/576-the-cake-cutting-problem

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  • $\begingroup$ Note that this currently counts as a link only answer. $\endgroup$ – kaine Mar 13 '15 at 18:04
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Alice divides the estate in three parts, and lets Bob and Carol choose. If Bob and Carol choose different parts, Alice gets the part that's left. Since Alice divided and Bob and Carol got to choose, all should be happy. If Bob and Carol choose the same part, Alice chooses one of the other parts. The parts left are put together and now divided between Bob and Carol by the usual way of having one partition and the other one choose.

This way, each sibling got to either divide or choose and should be happy.

This is different from the cake solution, since a cake can be cut, but not joined.

To address some points raised in the comments, I propose the following adjustment:

If Bob and Carol choose different parts, Alice still gets's what left. But if Bob and Carol choose the same part, they then choose the part Alice should get. If they choose the same part for Alice, she gets that, if they choose differently, she gets to choose. Only then the remaining parts are joined and redivided.

This still doesn't address the point of Alice disagreeing with how Bob and Carol choose to redivide, should that happen.

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    $\begingroup$ -1. Alice will not be happy for sure. Alice got 1/3 exactly in her opinion, meanwhile in Alice opinion Bob and Carol did NOT get 1/3 exactly, but something different; so one of them for sure got more that 1/3, lets say it was Bob. That means Alice's part in Alice's opinion is smaller than Bob's part. $\endgroup$ – klm123 Oct 15 '14 at 10:34
  • $\begingroup$ So in the second case, where Bob and Carol redivide, Alice did get 1/3, but may be jealous of either since she may disagree with their repartition of the 2/3 left? That seems to be inline with your comment on the question, that it should've read "such that no sibling shall feel their share is smaller than 1/3 of the estate". That makes sense, but I still feel that my answer (or a variation thereof) is what OP meant. $\endgroup$ – SQB Oct 15 '14 at 11:20
  • $\begingroup$ @SQB, yes. If OP would change formulation this would not be a problem. But in current formulation your answer confuses people. (btw for bob and carol it still will not work for all cases). $\endgroup$ – klm123 Oct 15 '14 at 11:30
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    $\begingroup$ This is not a fair system. If Alice is dishonest, she can game it pretty easily. Let's assume the entire estate is worth 1,000,000€. Alice divides it into 500,000€, 499,999€ and 1€. Bob and Carol both want the largest share, so Alice picks the 499,999€. Now Bob and Carol have only 500,001€ to share – they both get about half of what Alice gets. $\endgroup$ – Moyli Oct 15 '14 at 11:30
  • $\begingroup$ Alice thinks X=Y=Z, Bob and Carol think that Z<X<Y such that X+j-k=Y-j-k=Z+2k. Alice gets Z. Bob gets X+j. Carol gets Y-j. Alice is upset as she thinks Bob got a bigger piece. $\endgroup$ – kaine Oct 15 '14 at 18:50

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