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A Mathematical View of Our World. 1 st ed. Parks, Musser, Trimpe, Maurer, and Maurer. Chapter 4. Fair Division. Section 4.1 Divide and Choose Methods. Goals Study fair-division problems Continuous fair division Discrete fair division Mixed fair division Study fair-division procedures I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
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1. A Mathematical View of Our World 1st ed. Parks, Musser, Trimpe, Maurer, and Maurer

2. Chapter 4 Fair Division

3. Section 4.1Divide and Choose Methods • Goals • Study fair-division problems • Continuous fair division • Discrete fair division • Mixed fair division • Study fair-division procedures • Divide-and-choose method for 2 players • Divide-and-choose method for 3 players • Last-diminisher method for 3 or more players

4. 4.1 Initial Problem • The brothers Drewvan, Oswald, and Granger are to share their family’s 3600-acre estate. • Drewvan: • Values vineyards three times as much as fields. • Values woodlands twice as much as fields. • Oswald: • Values vineyards twice as much as fields. • Values woodlands three times as much as fields.

5. 4.1 Initial Problem, cont’d • Granger: • values vineyards twice as much as fields. • Values fields three times as much as woodlands. • How can the brothers fairly divide the estate? • The solution will be given at the end of the section.

6. Fair-Division Problems • Fair-division problems involve fairly dividing something between two or more people, without the aid of an outside arbitrator. • The people who will share the object are called players. • The solution to a problem is called a fair-division procedure or a fair-division scheme.

7. Types of Fair-Division Problems • Continuous fair-division problems: • The object(s) can be divided into pieces of any size with no loss of value. • An example is dividing a cake or an amount of money among two or more people.

8. Types of Fair-Division, cont’d • Discrete fair-division problems: • The object(s) will lose value if divided. • We assume the players do not want to sell everything and divide the proceeds. • However, sometimes money must be used when no other fair division is possible • An example is dividing a car, a house, and a boat among two or more people.

9. Types of Fair-Division, cont’d • Mixed fair-division problems: • Some objects to be shared can be divided and some cannot. • This type is a combination of continuous and discrete fair division. • An example is dividing an estate consisting of money, a house, and a car among two or more people.

10. Question: Three cousins will share an inheritance. The estate includes a house, a car, and cash. What type of fair-division problem is this? a. continuous b. discrete c. mixed

11. Types of Fair-Division, cont’d • This section will consider only continuous fair-division problems. • We make the assumption that the value of a player’s share is determined by his or her values. • Different players may value the same share differently.

12. Value of a Share • In a fair-division problem with n players, a player has received a fair share if that player considers his or her share to be worth at least 1/n of the total value being shared. • A division that results in every player receiving a fair share is called proportional.

13. Value of a Share, cont’d • We assume that a player’s values in a fair-division problem cannot change based on the results of the division. • We also assume that no player has any knowledge of any other player’s values.

14. Fair Division for Two Players • The standard procedure for a continuous fair-division problem with two players is called the divide-and-choose method. • This method is described as dividing a cake, but it can be used to fairly divide any continuous object.

15. Divide-And-Choose Method • Two players, X and Y, are to divide a cake. • Player X divides the cake into 2 pieces that he or she considers to be of equal value. • Player X is called the divider. • Player Y picks the piece he or she considers to be of greater value. • Player Y is called the chooser. • Player X gets the piece that player Y did not choose.

16. Divide-And-Choose Method, cont’d • This method produces a proportional division. • The divider thinks both pieces are equal, so the divider gets a fair share. • The chooser will find at least one of the pieces to be a fair share or more than a fair share. The chooser selects that piece, and gets a fair share.

17. Example 1 • Margo and Steven will share a \$4 pizza that is half pepperoni and half Hawaiian. • Margo likes both kinds of pizza equally. • Steven likes pepperoni 4 times as much as Hawaiian.

18. Example 1, cont’d • Margo cuts the pizza into 6 pieces and arranges them as shown. • What monetary value would Margo and Steven each place on the original two halves of the pizza?

19. Example 1, cont’d • Solution: The whole pizza is worth \$4. • Margo values both kinds of pizza equally. To her each half is worth half of the total value, or \$2. • Steven values pepperoni 4 times as much as Hawaiian. To him the pepperoni half is worth 4/5 of the total value, or \$3.20. The Hawaiian half is worth 1/5 of the total, or \$0.80.

20. Example 1, cont’d • What value would each person place on each of the two plates of pizza? • What plate will Steven choose?

21. Example 1, cont’d • Solution: The whole pizza is worth \$4. • Margo values both kinds of pizza equally. To her each plate of pizza is worth half of the total value, or \$2.

22. Example 1, cont’d • Solution, cont’d: • Steven values pepperoni 4 times as much as Hawaiian. • To him each pepperoni slice is worth \$3.20/3 = \$1.067 and each Hawaiian slice is worth \$0.80/3 = \$0.267. • The first plate is worth 2(\$0.267) + 1(\$1.067) = \$1.60. • The second plate is worth 1(\$0.267) + 2(\$1.067) = \$2.40

23. Example 1, cont’d • Solution: • Steven will choose the second plate, with one slice of Hawaiian and two slices of pepperoni. • Margo gets a plate of pizza that she feels is worth half the value. • Steven gets a plate of pizza that he feels is worth more than half the value.

24. Example 2 • Caleb and Diego will drive 6 hours during the day and 4 hours at night. • Caleb prefers night to day driving 2 to 1. • Diego prefers them equally, or 1 to 1. • How should they divide the driving into 2 shifts if Caleb is the divider and Diego is the chooser?

25. Example 2, cont’d • Solution: • Caleb can assign 2 points to each hour of night driving and 1 point to each hour of day driving. • Caleb values the entire drive at 1(6) + 2(4) = 14 points. • To Caleb a fair share will be worth half the total value, or 7 points.

26. Example 2, cont’d • Solution, cont’d: • A possible fair division for Caleb is to create shifts of: • 6 hours of daytime driving and 0.5 hours of nighttime driving. • 3.5 hours of nighttime driving. • Both shifts are worth 7 points to Caleb.

27. Example 2, cont’d • Solution, cont’d: • Diego can assign 1 point to each hour of night driving and 1 point to each hour of day driving. • Diego values the entire drive at 1(6) + 1(4) = 10 points. • To Diego a fair share will be worth half the total value, or 5 points.

28. Example 2, cont’d • Solution, cont’d: • Diego values the first shift at 1(6) + 1(0.5) = 6.5 points. • Diego values the second shift at 1(3.5) = 3.5 points. • Diego will choose the first shift, because it is worth more to him.

29. Two Players, cont’d • Notice that in both of the previous examples: • The divider got a share he or she felt was equal to exactly half of the total value. • The chooser got a share he or she felt was equal to more than half of the total value. • It is often advantageous to be the chooser, so the roles should be randomly chosen.

30. Fair Division for Three Players • In a continuous fair-division problem with 3 players, it is still possible to have one player divide the object and the other players choose. • This method is also called the lone-divider method.

31. Divide-And-Choose Method • Three players, X, Y, and Z are to divide a cake. • Player X (the divider) divides the cake into 3 pieces that he/she considers to be of equal value. • Players Y and Z (the choosers) each decide which pieces are worth at least 1/3 of the total value. • These pieces are said to be acceptable. • The choosers announce their acceptable pieces.

32. Divide-And-Choose Method, cont’d • There are 2 possibilities: • If at least 1 piece is unacceptable to both Y and Z, Player X gets that piece. • If Y and Z can each choose acceptable pieces, they do so. • If Y and Z cannot each choose acceptable pieces, they put the remaining pieces back together and use the two player method to re-divide.

33. Divide-And-Choose Method, cont’d • Cont’d: • If every piece is acceptable to both Y and Z, they each take an acceptable piece. Player X gets the leftover piece. • Note: The divide-and-choose method can be extended to more than 3 players. The more players, the more complicated the process becomes.

34. Question: The divide-and-choose method for 3 players is being used to divide a pizza. Player A has cut a pizza into what she views as 3 equal shares. Player B thinks that only shares 2 and 3 are acceptable. Player C thinks that only share 2 is acceptable. What is the fair division? a. Player A gets share 3, Player B gets share 1, and Player C gets share 2. b. Player A gets share 1, Player B gets share 3, and Player C gets share 2. c. Player A gets share 2, Player B gets share 3, and Player C gets share 1. d. Player A gets share 1, Player B gets share 2, and Player C gets share 3.

35. Example 3 • Emma, Fay, and Grace will divide 24 ounces of ice cream, which is made up of equal amounts of vanilla, chocolate, and strawberry. • Emma likes the 3 flavors equally well. • Fay prefers chocolate 2 to 1 over either other flavor and prefers vanilla and strawberry equally well. • Grace prefers vanilla to chocolate to strawberry in the ratio 1 to 2 to 3. • If Emma is the divider, what are the results of the divide-and-choose method for 3 players?

36. Example 3, cont’d • Solution: Suppose Emma divides the ice cream into 3 equal parts, each consisting of one of the flavors.

37. Example 3, cont’d • Solution, cont’d: Fay is one of the choosers. • Faye finds portions 1 and 3 unacceptable.

38. Example 3, cont’d • Solution, cont’d: Grace is the other chooser. • She finds portion 1 unacceptable

39. Example 3, cont’d • Solution, cont’d: All of the players’ values are summarized in the table below.

40. Example 3, cont’d • Solution, cont’d: • Portion 1 is unacceptable to both Fay and Grace. As the divider, Emma will receive portion 1. • Only portion 2 is acceptable to Faye. • Portions 2 and 3 are acceptable to Grace. • The division is Emma: portion 1; Fay: portion 2; Grace: portion 3.

41. Last-Diminisher Method • A method for continuous fair-division problems with 3 or more players is called the last-diminisher method. • Suppose any number of players X, Y, … are dividing a cake. • Player X cuts a piece of cake that he or she considers to be a fair share.

42. Last-Diminisher Method, cont’d • Each player, in turn, judges the fairness of the piece. • If a player considers the piece fair or less than fair, it passes to the next player. • If a player considers the piece more than fair, the player trims the piece to make it fair, returning the trimming to the undivided portion and passing the trimmed piece to the next player.

43. Last-Diminisher Method, cont’d • The last player to trim the piece, gets the piece as his or her share. • If no player trimmed the piece, player X gets the piece. • After one player gets a piece of cake, the process begins again without that player and that piece. • When only 2 players remain, they use the divide-and-choose method.

44. Example 4 • Hector, Isaac, and James will divide 24 ounces of ice cream, which is equal parts vanilla, chocolate, and strawberry. • Hector values vanilla to chocolate to strawberry 1 to 2 to 3. • Isaac likes the 3 flavors equally. • James values vanilla to chocolate to strawberry 1 to 2 to 1.

45. Example 4, cont’d • Using the last-diminisher method with Hector as the first divider and Isaac as the first judge, find the results of the division. • Solution: • Hector assigns 1 point to each ounce of vanilla, 2 points to each ounce of chocolate, and 3 points to each ounce of strawberry.

46. Example 4, cont’d • Solution, cont’d: A fair share of ice cream, to Hector, is worth 48/3 = 16 points.

47. Example 4, cont’d • Solution, cont’d: • One possible fair share for Hector would be all 8 ounces of vanilla plus 4 ounces of chocolate. • This share is worth 1(8) + 2(4) = 16 points to Hector, so he would be happy with this share. • Next, Isaac must decide whether the share is fair, according to his values.

48. Example 4, cont’d • Solution, cont’d: • Isaac assigns 1 point to each ounce of vanilla, 1 point to each ounce of chocolate, and 1 point to each ounce of strawberry. • Isaac values all of the ice cream at 1(8) + 1(8) + 1(8) = 24 points. • A fair share to Isaac is 8 points.

49. Example 4, cont’d • Solution, cont’d: • Isaac’s value for Hector’s serving is 1(8) + 1(4) = 12 points. • Isaac thinks it is more than a fair share. • Isaac trims off 4 points worth of ice cream. • Suppose he trims off the 4 ounces of chocolate.

50. Example 4, cont’d • Solution, cont’d: • Next, James must judge the share. • James assigns 1 point to each ounce of vanilla, 2 points to each ounce of chocolate, and 1 point to each ounce of strawberry. • James values all of the ice cream at 1(8) + 2(8) + 1(8) = 32 points. • A fair share to James is worth 32/3 points.