How Do You Share an Inheritance Fairly?

Each sibling counts by their own prices. Maya’s add up to 60 lakh, so her third is 20; Ramesh’s add up to 54, his third 18; Sita’s 57, her third 19. A share is fair to someone if it is worth at least their own third. Nobody needs to agree with anybody else’s prices.

Two ways of being fair will matter, the same two that matter for sharing a cake (the Grades 6–8 topic How Do You Cut a Cake Fairly for Three?):

• Proportional: each of the n people gets at least 1/n of everything, by their own prices.
• Envy-free: nobody would rather have someone else’s share, by their own prices.

A cake can be cut as finely as you like. An inheritance often can’t: a house, a buffalo, a motorbike, a shop are indivisible. That is why the widget had no answer. Something must be added that can be cut.

Money is the cake in this problem: it can be cut into any amount, and a lakh is a lakh to everybody. So let the person who gets something valuable pay the others.

• a.Give the land to Maya, the house to Sita and the buffalo to Ramesh. Find payments that give everyone at least a third, by their own prices.
• b.Now make the payments so that nobody would swap, too.
• c.Is there only one answer?

Here is a rule that needs no arguing over prices. The Polish mathematician Bronisław Knaster devised it, and Hugo Steinhaus published it in 1948 (the same Steinhaus who asked how to cut a cake for three). It is called the method of sealed bids, or Knaster’s procedure.

1. Each heir writes down, in secret, what each thing is worth to them: their bids. The bids are opened together.
2. Each thing goes to whoever bid the most for it.
3. Each heir’s fair share is 1/n of their own bids added up (for three heirs, a third).
4. Each heir who got things worth more than their fair share (by their own bids) pays the difference into a pot. Each heir who got less takes from the pot what they are short.
5. Money is left in the pot. It is shared out equally.

With honest bids: Maya gets the land (42) and her fair share is 20, so she pays 22 into the pot. Sita gets the house (27), her fair share is 19, and she pays 8. Ramesh gets the buffalo (6), his fair share is 18, and he takes out 12. That leaves 22 + 8 − 12 = 18 lakh in the pot, 6 each. In the end Maya pays 16, Sita pays 2, Ramesh gets 18: the last section’s fairest answer, found by a rule. By their own prices Maya has 42 − 16 = 26, Ramesh 6 + 18 = 24 and Sita 27 − 2 = 25: each exactly 6 more than their third.

Why everyone gets more than a third

That 6 each is no coincidence. Follow any heir. Say their things are worth v to them and their fair share is f. They pay v − f into the pot (or take out f − v if that is negative, which is the same thing), and then get back a third of the pot, P/3. So in the end their share is worth, by their own prices,

$$v - (v - f) + \tfrac{P}{3} = f + \tfrac{P}{3}.$$

What they got, v, cancels out. Everyone ends with exactly their fair share plus an equal slice of the pot, however many things they got. So everyone gets at least a third if the pot is never negative. Is it?

Look at one thing at a time. The money that goes into the pot for the land is the top bid, 42, which Maya “pays” by taking the land; the money that comes out for it is a third of everyone’s bid for it, (42 + 30 + 27)/3 = 33, the average bid. So the land adds 42 − 33 = 9 to the pot, the house 27 − 20 = 7 and the buffalo 6 − 4 = 2: 18 in all (Figure 40.1).

A top bid can never be below the average of the bids (the average is between the smallest and the largest). So the pot is never negative, and the method is always proportional: with any honest bids, any number of heirs and any number of things, everyone gets at least 1/n of everything by their own prices.

The pot is zero only if everyone bids the same for every thing: then each ends with exactly a third. The more the heirs’ tastes differ, the bigger the pot, and the more each of them gains. Different tastes are not the problem in sharing; they are what makes good sharing possible.

There is a second good thing. Each thing goes to the person who values it most, so the total of what the heirs get, each counting their own things at their own price, is as big as it can be: 42 + 27 + 6 = 75 lakh. Our script tried every other way of giving out the three things: the best of them makes only 72. So the division is efficient: nobody could be made better off without someone else being made worse off, because that would raise the total, and the money always adds up to zero.

Shading a bid

Bids are supposed to be honest. But suppose Sita has guessed that Maya and Ramesh value the house at no more than 18. Choose “Sita bids 22.5” in the widget: she writes 22.5 lakh for the house instead of 27.

She still gets the house, but her fair share drops and in the end she pays nothing. By her real prices her share is now worth 27 lakh instead of 25; Maya and Ramesh each lose 1.

For any bid h above 18 and up to 27, the script’s numbers fit one line: Sita’s share is worth $25 + \tfrac{4}{9}(27 - h)$ to her, more the lower she goes. So why not bid 15? Choose “Sita bids 15”: the house goes to Ramesh, and Sita is left with 19.33 lakh, well below the 25 an honest bid gave her. Now she would rather have Ramesh’s share.

So the method is not strategy-proof: an heir who knows the others’ bids can gain by lying, and one who guesses wrong loses. Honest bidding is safe: whatever the others do, you are sure of at least 1/n by your real prices. Clever bidding is a gamble.

Envy

With two heirs, sealed bids never leave anyone envious. With three, they can. Choose “Sita likes the land less”: Sita now honestly values the land at only 9 lakh, and everything else is the same. Everyone still gets more than a third, but Sita’s own share (the house and paying 6) is worth 21 to her, while Ramesh’s share (the buffalo and 20 lakh) is worth 3 + 20 = 23 to her. She would rather have his.

Ramesh’s fair share comes from his prices, and he valued the land at 30, so he is owed a lot of money; to Sita, who doesn’t care for the land, that looks like cash for nothing. This isn’t rare: in our script’s 20,000 random sets of bids for three heirs and three things, about 2 in 5 left someone envious.

Cash, and what money means

Two more problems lie outside the mathematics. Maya must find 16 lakh to keep the land, and she may not have it. And the method assumes a lakh means the same to everyone, when to a sister with no savings, 16 lakh to pay may weigh far more than 16 lakh of land.

Sometimes nobody has the cash. Steven Brams and Alan Taylor published a method for two people in 1996 that needs none: the adjusted winner procedure. Each person shares out 100 points among the things, in secret. Points only say how much you want each thing compared with the others, so nobody needs any money at all.

Two sisters, Anju and Binita, share their mother’s gold jewelry, sewing machine, shop, a plot of land and a car.

The rules:

1. Each thing goes to whoever gave it more points. (Here Anju has 80 of her points, Binita 50 of hers.)
2. The one who is ahead must give something up. Divide her points for each of her things by the other person’s points for it, and start with the smallest ratio.
3. Move that thing across, all of it if that keeps her ahead or makes them equal, then the next, and so on. The thing that would make the leader fall behind is split instead, in just the share that makes the two equal.

Anju is ahead, and her smallest ratio is the shop’s, 30 ÷ 20 = 1.5. Moving all of it would leave her with 50 and give Binita 70, too much. So the shop is split: if a fraction p of it goes to Binita,

$$80 - 30p = 50 + 20p, \qquad p = \tfrac{30}{50} = \tfrac{3}{5}.$$

Binita gets three fifths of the shop, Anju two fifths, and each sister ends with 62 of her own 100 points. In practice, a shared shop is one they own together, or one that one sister buys from the other.

The method promises four things at once: nobody envies the other (envy-free), each has the same number of their own points (equitable), nobody could be made better off without making the other worse off (efficient: giving away the smallest ratio first loses the leader the fewest points for each point the other gains), and at most one thing is shared. No other method for two people does all four. With three or more people, there may be no division that manages the first three together.

A linear program in disguise

Grade 10 Optional Maths solves linear programming problems: the best plan when the limits are straight lines. The sisters’ problem is one. Let x₁, …, x₅ be the fractions of the five things that go to Anju, each between 0 and 1. Then Anju’s points are

$$15x_1 + 10x_2 + 30x_3 + 40x_4 + 5x_5$$

and Binita’s are 25(1 − x₁) + 5(1 − x₂) + 20(1 − x₃) + 25(1 − x₄) + 25(1 − x₅). Ask for the two to be equal, and make Anju’s points as large as possible. Every limit (each fraction between 0 and 1, the two totals equal) is linear, and so is what is made largest: a linear program, with five unknowns instead of Grade 10’s two. Our script gave it to a linear-programming solver, which returned exactly the adjusted winner’s answer: 62 each, the shop shared 2/5 to 3/5.

Grade 10’s rule that the best plan is at a corner explains why at most one thing is split. At a corner of the region where the points are equal, all but one of the fractions sit at 0 or 1: only one thing is shared. With whole things only and no sharing at all, the best the smaller share can be is 55 points.

• How Do You Cut a Cake Fairly for Three? (Grades 6–8) Fair division when everything can be cut: cut and choose, the moving knife, and a share for three that nobody envies.
• Which Voting System Is Fair? When a group must choose one thing together, “fair” turns out to be harder still.
• Should You Cooperate or Cheat? Game theory: what happens when each person does what is best for themselves, as Sita did with her bid.
• Can Anyone Beat the Stock Market? Money, risk and what investors can and can’t know.
• How Can Students Be Matched to Colleges So No One Wants to Swap? (Computing, Going Further 9–12) Another kind of fairness without money, where nobody would rather trade. (Coming soon.)
• Grade 10 Optional Math: Linear Programming The best plan at a corner, the idea behind sharing at most one thing.