1. A Mathematical View of Our World 1 st ed. Parks, Musser, Trimpe, Maurer, and Maurer

2. Chapter 4 Fair Division

3. Section 4.1 Divide and Choose Methods GoalsGoals Study fair-division problemsStudy fair-division problems Continuous fair divisionContinuous fair division Discrete fair divisionDiscrete fair division Mixed fair divisionMixed fair division Study fair-division proceduresStudy fair-division procedures Divide-and-choose method for 2 playersDivide-and-choose method for 2 players Divide-and-choose method for 3 playersDivide-and-choose method for 3 players Last-diminisher method for 3 or more playersLast-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.The brothers Drewvan, Oswald, and Granger are to share their family’s 3600- acre estate. Drewvan:Drewvan: Values vineyards three times as much as fields.Values vineyards three times as much as fields. Values woodlands twice as much as fields.Values woodlands twice as much as fields. Oswald:Oswald: Values vineyards twice as much as fields.Values vineyards twice as much as fields. Values woodlands three times as much as fields.Values woodlands three times as much as fields.

5. 4.1 Initial Problem, cont’d Granger:Granger: values vineyards twice as much as fields.values vineyards twice as much as fields. Values fields three times as much as woodlands.Values fields three times as much as woodlands. How can the brothers fairly divide the estate?How can the brothers fairly divide the estate? The solution will be given at the end of the section.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.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 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.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:Continuous fair-division problems: The object(s) can be divided into pieces of any size with no loss of value.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.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:Discrete fair-division problems: The object(s) will lose value if divided.The object(s) will lose value if divided. We assume the players do not want to sell everything and divide the proceeds.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 possibleHowever, 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.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:Mixed fair-division problems: Some objects to be shared can be divided and some cannot.Some objects to be shared can be divided and some cannot. This type is a combination of continuous and discrete fair division.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.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.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.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.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.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.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 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.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.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.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.Two players, X and Y, are to divide a cake. 1)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 X is called the divider. 2) Player Y picks the piece he or she considers to be of greater value. Player Y is called the chooser.Player Y is called the chooser. 3)Player X gets the piece that player Y did not choose.

16. Divide-And-Choose Method, cont’d This method produces a proportional division.This method produces a proportional division. The divider thinks both pieces are equal, so the divider gets a fair share.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.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 and Steven will share a $4 pizza that is half pepperoni and half Hawaiian. Margo likes both kinds of pizza equally.Margo likes both kinds of pizza equally. Steven likes pepperoni 4 times as much as Hawaiian.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.Margo cuts the pizza into 6 pieces and arranges them as shown. a)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.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.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.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 b)What value would each person place on each of the two plates of pizza? c)What plate will Steven choose?

21. Example 1, cont’d b)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.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 b)Solution, cont’d: Steven values pepperoni 4 times as much as Hawaiian.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.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 first plate is worth 2($0.267) + 1($1.067) = $1.60. The second plate is worth 1($0.267) + 2($1.067) = $2.40The second plate is worth 1($0.267) + 2($1.067) = $2.40

23. Example 1, cont’d c)Solution: Steven will choose the second plate, with one slice of Hawaiian and two slices of pepperoni.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.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.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 and Diego will drive 6 hours during the day and 4 hours at night. Caleb prefers night to day driving 2 to 1.Caleb prefers night to day driving 2 to 1. Diego prefers them equally, or 1 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?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:Solution: Caleb can assign 2 points to each hour of night driving and 1 point to each hour of day driving.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.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.To Caleb a fair share will be worth half the total value, or 7 points.

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

27. Example 2, cont’d Solution, 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 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.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.To Diego a fair share will be worth half the total value, or 5 points.

28. Example 2, cont’d Solution, cont’d:Solution, cont’d: Diego values the first shift at 1(6) + 1(0.5) = 6.5 points.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 values the second shift at 1(3.5) = 3.5 points. Diego will choose the first shift, because it is worth more to him.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: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 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.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.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.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.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.Three players, X, Y, and Z are to divide a cake. 1)Player X (the divider) divides the cake into 3 pieces that he/she considers to be of equal value. 2) 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.These pieces are said to be acceptable. 3)The choosers announce their acceptable pieces.

32. Divide-And-Choose Method, cont’d 3)There are 2 possibilities: a)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 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.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 3) Cont’d: b)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.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, 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.Emma likes the 3 flavors equally well. Fay prefers chocolate 2 to 1 over either other flavor and prefers vanilla and strawberry 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.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?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.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.Solution, cont’d: Fay is one of the choosers. Faye finds portions 1 and 3 unacceptable.Faye finds portions 1 and 3 unacceptable.

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

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

40. Example 3, cont’d Solution, cont’d:Solution, cont’d: Portion 1 is unacceptable to both Fay and Grace. As the divider, Emma will receive portion 1.Portion 1 is unacceptable to both Fay and Grace. As the divider, Emma will receive portion 1. Only portion 2 is acceptable to Faye.Only portion 2 is acceptable to Faye. Portions 2 and 3 are acceptable to Grace.Portions 2 and 3 are acceptable to Grace. The division is Emma: portion 1; Fay: portion 2; Grace: portion 3.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.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.Suppose any number of players X, Y, … are dividing a cake. 1)Player X cuts a piece of cake that he or she considers to be a fair share.

42. Last-Diminisher Method, cont’d 2)Each player, in turn, judges the fairness of the piece. a)If a player considers the piece fair or less than fair, it passes to the next player. b)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 3)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.If no player trimmed the piece, player X gets the piece. 4)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.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, 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.Hector values vanilla to chocolate to strawberry 1 to 2 to 3. Isaac likes the 3 flavors equally.Isaac likes the 3 flavors equally. James values vanilla to chocolate to strawberry 1 to 2 to 1.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.Using the last-diminisher method with Hector as the first divider and Isaac as the first judge, find the results of the division. Solution: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.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.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:Solution, cont’d: One possible fair share for Hector would be all 8 ounces of vanilla plus 4 ounces of chocolate.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.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.Next, Isaac must decide whether the share is fair, according to his values.

48. Example 4, cont’d Solution, 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 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.Isaac values all of the ice cream at 1(8) + 1(8) + 1(8) = 24 points. A fair share to Isaac is 8 points.A fair share to Isaac is 8 points.

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

50. Example 4, cont’d Solution, cont’d:Solution, cont’d: Next, James must judge the share.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 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.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.A fair share to James is worth 32/3 points.

51. Example 4, cont’d Solution, cont’d:Solution, cont’d: The existing share is now just 8 ounce of vanilla.The existing share is now just 8 ounce of vanilla. To James, the share is worth 1(8) = 8 points.To James, the share is worth 1(8) = 8 points. James thinks this is less than a fair share.James thinks this is less than a fair share. James will not trim the share.James will not trim the share.

52. Example 4, cont’d Solution, cont’d:Solution, cont’d: Isaac was the last-diminisher, and gets the share of ice cream.Isaac was the last-diminisher, and gets the share of ice cream. Hector and James will divide the remaining ice cream using the divide- and-choose method.Hector and James will divide the remaining ice cream using the divide- and-choose method. Note: This is only one of many different possible solutions.Note: This is only one of many different possible solutions.

53. 4.1 Initial Problem Solution The brothers Drewvan, Oswald, and Granger are to share their family’s estate, which is 1200 acres each of vineyards, woodlands, and fields.The brothers Drewvan, Oswald, and Granger are to share their family’s estate, which is 1200 acres each of vineyards, woodlands, and fields. Drewvan prefers vineyards to woodlands to fields 3 to 2 to 1.Drewvan prefers vineyards to woodlands to fields 3 to 2 to 1. Oswald prefers vineyards to woodlands to fields 2 to 3 to 1.Oswald prefers vineyards to woodlands to fields 2 to 3 to 1. Granger prefers vineyards to woodlands to fields 2 to 1 to 3.Granger prefers vineyards to woodlands to fields 2 to 1 to 3.

54. Initial Problem Solution, cont’d Use the divide-and-choose method for 3 players.Use the divide-and-choose method for 3 players. Let Drewvan be the divider.Let Drewvan be the divider.

55. Initial Problem Solution, cont’d Drewvan values the entire estate at 7200 points.Drewvan values the entire estate at 7200 points.

56. Initial Problem Solution, cont’d To Drewvan, a fair share is worth 7200/3 = 2400 points.To Drewvan, a fair share is worth 7200/3 = 2400 points. One possible fair division is shown below.One possible fair division is shown below.

57. Initial Problem Solution, cont’d Next, the two choosers will consider this division.Next, the two choosers will consider this division. Granger and Oswald both value the entire estate at 7200 points also.Granger and Oswald both value the entire estate at 7200 points also. To Oswald, a fair share is worth 2400 points.To Oswald, a fair share is worth 2400 points. To Granger, a fair share is worth 2400 points.To Granger, a fair share is worth 2400 points.

58. Initial Problem Solution, cont’d Oswald considers piece 1 to be unacceptable.Oswald considers piece 1 to be unacceptable.

59. Initial Problem Solution, cont’d Granger considers pieces 1 and 2 to be unacceptable.Granger considers pieces 1 and 2 to be unacceptable.

60. Initial Problem Solution, cont’d Both choosers think piece 1 is unacceptable, so Drewvan gets piece 1.Both choosers think piece 1 is unacceptable, so Drewvan gets piece 1. Granger thinks only piece 3 is acceptable, so he gets that piece.Granger thinks only piece 3 is acceptable, so he gets that piece. Oswald thinks pieces 2 and 3 are acceptable, so Oswald gets piece 2.Oswald thinks pieces 2 and 3 are acceptable, so Oswald gets piece 2.

61. Section 4.2 Discrete and Mixed Division Problems GoalsGoals Study discrete fair-division problemsStudy discrete fair-division problems The method of sealed bidsThe method of sealed bids The method of pointsThe method of points Study mixed fair-division problemsStudy mixed fair-division problems The adjusted winner procedureThe adjusted winner procedure

62. 4.2 Initial Problem When twins Zack and Zeke turned 16 they received:When twins Zack and Zeke turned 16 they received: A pickup truckA pickup truck A horseA horse A cowA cow How can they share these three things?How can they share these three things? The solution will be given at the end of the section.The solution will be given at the end of the section.

63. Discrete Fair Division Recall that discrete fair division problems involve sharing objects that cannot be divided without losing value.Recall that discrete fair division problems involve sharing objects that cannot be divided without losing value. Two methods for solving discrete fair- division problems are:Two methods for solving discrete fair- division problems are: The method of sealed bids.The method of sealed bids. The method of points.The method of points.

64. Method of Sealed Bids Any number of players, N, are to share any number of items.Any number of players, N, are to share any number of items. If necessary, money will be used to insure fairness.If necessary, money will be used to insure fairness. 1)All players submit sealed bids, stating monetary values for each item. 2)Each item goes to the highest bidder. The highest bidder places the dollar amount of his or her bid into a compensation fund.The highest bidder places the dollar amount of his or her bid into a compensation fund.

65. Method of Sealed Bids, cont’d 3)From the compensation fund, each player receives 1/N of his or her bid on each item. 4)Any money leftover in the fund is distributed equally to all players. Note: This method is also called the Knaster Inheritance Procedure.Note: This method is also called the Knaster Inheritance Procedure.

66. Example 1 Three sisters Maura, Nessa, and Odelia will share a house and a cottage.Three sisters Maura, Nessa, and Odelia will share a house and a cottage. Apply the method of sealed bids to divide the property, using the bids shown below.Apply the method of sealed bids to divide the property, using the bids shown below.

67. Example 1, cont’d Solution: Each piece of property goes to the highest bidder.Solution: Each piece of property goes to the highest bidder. Odelia gets the family home and places $301,000 into the compensation fund.Odelia gets the family home and places $301,000 into the compensation fund. Nessa gets the cottage and places $203,000 into the compensation fund.Nessa gets the cottage and places $203,000 into the compensation fund.

68. Example 1, cont’d Solution, cont’d: The compensation fund now contains a total of $203,000 + $301,000 = $504,000.Solution, cont’d: The compensation fund now contains a total of $203,000 + $301,000 = $504,000.

69. Example 1, cont’d Solution, cont’d: Each sister receives 1/3 of her total bids from the compensation fund.Solution, cont’d: Each sister receives 1/3 of her total bids from the compensation fund. Maura receivesMaura receives Nessa receivesNessa receives Odelia receivesOdelia receives

70. Example 1, cont’d Solution, cont’d: After the distributions, there is $504,000 – ($159,000 + $160,000 + $161,000) = $21,000 left in the fund.Solution, cont’d: After the distributions, there is $504,000 – ($159,000 + $160,000 + $161,000) = $21,000 left in the fund. The leftover money is distributed equally to the three sisters in the amount of $7000 each.The leftover money is distributed equally to the three sisters in the amount of $7000 each.

71. Example 1, cont’d Solution, cont’d: The final shares are:Solution, cont’d: The final shares are: Maura receives $166,000 and no property.Maura receives $166,000 and no property. Nessa receives the cottage, for which she paid a net amount of $33,000.Nessa receives the cottage, for which she paid a net amount of $33,000. Odelia receives the family home, for which she paid a net amount of $133,000.Odelia receives the family home, for which she paid a net amount of $133,000.

72. Example 1, cont’d Solution, cont’d: Note that the division is proportional because each sister receives what she considers to be a fair share.Solution, cont’d: Note that the division is proportional because each sister receives what she considers to be a fair share.

73. Method of Points Three players will share three items.Three players will share three items. 1)Each player assigns points to each item, so that the points for each player total to 100. 2)List all 6 possible arrangements of players and items, along with the point values.

74. Method of Points, cont’d 3)Note the smallest number of points assigned to an item in each arrangement. Choose the arrangement with the largest value of the smallest number.Choose the arrangement with the largest value of the smallest number. If there is not only one such arrangement, keep all the arrangements with that smallest point value and go to step 4.If there is not only one such arrangement, keep all the arrangements with that smallest point value and go to step 4.

75. Method of Points, cont’d 4)For each arrangement kept in step 3, note the middle point value. Choose the arrangement with the largest value of the middle number.Choose the arrangement with the largest value of the middle number. If there is not only one such arrangement, keep all the arrangements with that middle point value and go to step 5.If there is not only one such arrangement, keep all the arrangements with that middle point value and go to step 5.

76. Method of Points, cont’d 5)For each arrangement kept in step 4, note the largest point value. Choose any arrangement with the largest value of the largest number.Choose any arrangement with the largest value of the largest number.

77. Example 2 Three couples, A, B, and C, need to decide who gets which room in a hotel.Three couples, A, B, and C, need to decide who gets which room in a hotel. The couples assign points to each room as shown below.The couples assign points to each room as shown below.

78. Example 2, cont’d Solution:Solution:

79. Example 2, cont’d Solution, cont’d: The largest minimum point value is 40, in row 4.Solution, cont’d: The largest minimum point value is 40, in row 4.

80. Example 2, cont’d Solution, cont’d: The arrangement selected is:Solution, cont’d: The arrangement selected is: Couple A, Room 2Couple A, Room 2 Couple B, Room 3Couple B, Room 3 Couple C, Room 1Couple C, Room 1 Two couples got their first choice and one got their second choice.Two couples got their first choice and one got their second choice.

81. Example 3 Three couples, A, B, and C, need to decide who gets which room in a cabin.Three couples, A, B, and C, need to decide who gets which room in a cabin. The couples assign points to each room as shown below.The couples assign points to each room as shown below.

82. Example 3, cont’d

83. Solution, cont’d: The largest minimum point value is 12, which occurs in 4 arrangements.Solution, cont’d: The largest minimum point value is 12, which occurs in 4 arrangements. Table from previous slideTable from previous slide

84. Example 3, cont’d Solution, cont’d: Keep those 4 arrangements and look at the middle point values.Solution, cont’d: Keep those 4 arrangements and look at the middle point values.

85. Example 3, cont’d Solution, cont’d: The largest middle point value is 20, which occurs in 3 arrangements.Solution, cont’d: The largest middle point value is 20, which occurs in 3 arrangements.

86. Example 3, cont’d Solution, cont’d: Keep those 3 arrangements and look at the largest point values.Solution, cont’d: Keep those 3 arrangements and look at the largest point values. The maximum largest number is 71.The maximum largest number is 71.

87. Example 3, cont’d Solution, cont’d: The arrangement selected is:Solution, cont’d: The arrangement selected is: Couple A, Room 2Couple A, Room 2 Couple B, Room 1Couple B, Room 1 Couple C, Room 3Couple C, Room 3

88. Question: The assignments were Couple A, Room 2; Couple B, Room 1; and Couple C, Room 3. Considering the point values each couple assigned to the three rooms, is this division proportional? a. yes b. no

89. Example 3, cont’d Solution, cont’d: Sometimes the method of points produces a division that is not proportional. However it is still the best division that can be made.Solution, cont’d: Sometimes the method of points produces a division that is not proportional. However it is still the best division that can be made.

90. Mixed Fair Division Recall that mixed fair division problems involve sharing a mixture of discrete and continuous objects.Recall that mixed fair division problems involve sharing a mixture of discrete and continuous objects. A method for solving mixed fair-division problems is the adjusted winner procedure.A method for solving mixed fair-division problems is the adjusted winner procedure.

91. Adjusted Winner Procedure Two players are to fairly divide any number of items.Two players are to fairly divide any number of items. The items may be discrete and/or continuous.The items may be discrete and/or continuous. Ownership of some items may be shared.Ownership of some items may be shared. 1)Each player assigns points to each item, for a total of 100 points.

92. Adjusted Winner Procedure, cont’d 2)Each player tentatively receives items to which he or she assigned the most points. The points are added to the player’s total.The points are added to the player’s total. If 2 players tie for an item, the item goes to the player with the lowest point total so far.If 2 players tie for an item, the item goes to the player with the lowest point total so far.

93. Adjusted Winner Procedure, cont’d 3)Look at the players’ points. If the point totals are equal, the process is done.If the point totals are equal, the process is done. If Player X has more points than Player Y, then give Player Y the item for which the ratio of the number of points assigned by X to the number of points assigned by Y is the smallest.If Player X has more points than Player Y, then give Player Y the item for which the ratio of the number of points assigned by X to the number of points assigned by Y is the smallest.

94. Adjusted Winner Procedure, cont’d 4)Re-examine the players’ points. a)If the point totals are equal, the process is done. b)If Player X still has more points than Player Y, repeat step 3. c)If Player Y now has more points than Player X, move a fraction of the last item moved back to X.

95. Adjusted Winner Procedure, cont’d 4)Cont’d: The formula for case C is as follows, where q = fraction of item to be moved, T X = Player X’s point total without this item, T Y = Player Y’s point total without this item, P X = number of points Player X assigned to this item, P Y = number of points Player Y assigned to this item.

96. Question: Three players will divide a car, a boat, and an RV. Their point assignments are shown in the table below. Complete step 2 in the adjusted winner procedure. a. Player 1: boat; Player 2: RV; Player 3: car b. Player 1: boat, RV; Player 2: nothing; Player 3: car c. Player 1: boat; Player 2: car, RV; Player 3: nothing Player 1Player 2Player 3 car153035 boat3520 RV50 45

97. Example 4 Two players, A and B, need to divide a house, a boat, a cabin, and a condominium.Two players, A and B, need to divide a house, a boat, a cabin, and a condominium. Both have assigned points as shown below.Both have assigned points as shown below.

98. Example 4, cont’d Solution: The tentative assignment of items, along with the current point totals, is shown in the table below.Solution: The tentative assignment of items, along with the current point totals, is shown in the table below.

99. Example 4, cont’d Solution, cont’d: Player A has more points than Player B.Solution, cont’d: Player A has more points than Player B. Determine which item to move from A to B by considering the ratios of both items currently assigned to Player A.Determine which item to move from A to B by considering the ratios of both items currently assigned to Player A.

100. Example 4, cont’d Solution, cont’d: Move the house, which has the smaller ratio, from Player A to Player B.Solution, cont’d: Move the house, which has the smaller ratio, from Player A to Player B.

101. Example 4, cont’d Solution, cont’d: Now Player B has more points than Player A.Solution, cont’d: Now Player B has more points than Player A. A fraction of the house must be given back to Player A.A fraction of the house must be given back to Player A. The values for the formula are:The values for the formula are: T X = T A = 30T X = T A = 30 T Y = T B = 45T Y = T B = 45 P X = P A = 45P X = P A = 45 P Y = P B = 35P Y = P B = 35

102. Example 4, cont’d Solution, cont’d: The calculation is:Solution, cont’d: The calculation is: Player B keeps 3/8 of the house and 5/8 of the house goes back to Player A.Player B keeps 3/8 of the house and 5/8 of the house goes back to Player A.

103. Example 4, cont’d Solution, cont’d: Re-examine the points totals.Solution, cont’d: Re-examine the points totals. Player A has 30 + 45(5/8) = 58.125 points.Player A has 30 + 45(5/8) = 58.125 points. Player B has 25 + 20 + 35(3/8) = 58.125 points.Player B has 25 + 20 + 35(3/8) = 58.125 points. The division is now proportional.The division is now proportional.

104. 4.2 Initial Problem Solution Zack and Zeke need to share a pickup truck, a horse, and a cow.Zack and Zeke need to share a pickup truck, a horse, and a cow. Solution: They could use the adjusted winner procedure to share the items. Suppose they assign points as shown.Solution: They could use the adjusted winner procedure to share the items. Suppose they assign points as shown.

105. Initial Problem Solution, cont’d Tentatively, the assignments are:Tentatively, the assignments are: Zack: truck and cow, 73 pointsZack: truck and cow, 73 points Zeke: horse, 35 points.Zeke: horse, 35 points. Zack has more points, so something must be given to Zeke. Zack has more points, so something must be given to Zeke. Check the ratios for the truck and the cow.Check the ratios for the truck and the cow.

106. Initial Problem Solution, cont’d The ratios are shown in the table.The ratios are shown in the table. The ratio for the truck is smaller.The ratio for the truck is smaller. Move the truck from Zack to Zeke.Move the truck from Zack to Zeke.

107. Initial Problem Solution, cont’d Now:Now: Zack has 40 points.Zack has 40 points. Zeke has 65 points.Zeke has 65 points. Since the points are not equal, a fraction of the truck must be given back to Zack.Since the points are not equal, a fraction of the truck must be given back to Zack. The values for the formula are:The values for the formula are: T X = T Zack = 40; P X = P Zack = 33T X = T Zack = 40; P X = P Zack = 33 T Y = T Zeke = 35; P Y = P Zeke = 30T Y = T Zeke = 35; P Y = P Zeke = 30

108. Initial Problem Solution, cont’d Solution, cont’d: The calculation is:Solution, cont’d: The calculation is: Zeke keeps 60.3% of the truck and 39.7% of the truck goes back to Zack.Zeke keeps 60.3% of the truck and 39.7% of the truck goes back to Zack.

109. Initial Problem Solution, cont’d Solution, cont’d: Re-examine the points totals.Solution, cont’d: Re-examine the points totals. Zack has 40 + 33(0.397) = 53.095 points.Zack has 40 + 33(0.397) = 53.095 points. Zeke has 35 + 30(0.603) = 53.095 points.Zeke has 35 + 30(0.603) = 53.095 points. For a proportional division to be created, the truck must be shared.For a proportional division to be created, the truck must be shared.

110. Section 4.3 Envy-Free Division GoalsGoals Study envy-free divisionStudy envy-free division Continuous envy-free division for 3 playersContinuous envy-free division for 3 players

111. 4.3 Initial Problem Dylan, Emery, and Fordel will share a cake that is half chocolate and half yellow.Dylan, Emery, and Fordel will share a cake that is half chocolate and half yellow. Dylan likes chocolate cake twice as much as yellow.Dylan likes chocolate cake twice as much as yellow. Emery likes chocolate and yellow cake equally well.Emery likes chocolate and yellow cake equally well. Fordel likes yellow cake twice as much as chocolate.Fordel likes yellow cake twice as much as chocolate. How should they divide the cake?How should they divide the cake? The solution will be given at the end of the section.The solution will be given at the end of the section.

112. Envy-Free Division A division, among n players, is considered envy-free if each player feels that:A division, among n players, is considered envy-free if each player feels that: He or she has received at least 1/n of the total valueHe or she has received at least 1/n of the total value No other player’s share is more valuable than his or her own.No other player’s share is more valuable than his or her own. Note: A proportional division is not always envy-free.Note: A proportional division is not always envy-free.

113. Question: Ice cream was divided among three players. The players’ values of the shares are shown in the table below. The division was Emma, Portion 1; Fay, Portion 2; Grace, Portion 3. Is the division envy free? a. yes b. no

114. Continuous Envy-Free Division This section covers envy-free divisions for continuous fair-division problems involving three players.This section covers envy-free divisions for continuous fair-division problems involving three players. The procedure is split into two parts:The procedure is split into two parts: Part 1 distributes the majority of the item.Part 1 distributes the majority of the item. Part 2 distributes the excess.Part 2 distributes the excess.

115. Envy-Free Division Part 1 Players A, B, and C are to share a cake (or some other item).Players A, B, and C are to share a cake (or some other item). 1)Player A (the divider) divides the cake into 3 pieces he or she considers to be of equal value. 2)Player B chooses the one most valuable piece of these 3 pieces.

116. Envy-Free Division Part 1, cont’d 3)Player B (the trimmer) trims the most valuable piece so it is equal to the second most valuable piece. The excess is set aside.The excess is set aside. 4)Player C (the chooser) chooses the piece he or she considers to have the greatest value.

117. Envy-Free Division Part 1, cont’d 5)Player B chooses next. Player B gets the trimmed piece if it is available.Player B gets the trimmed piece if it is available. If the trimmed piece is gone, B chooses the most valuable piece from the 2 remaining pieces.If the trimmed piece is gone, B chooses the most valuable piece from the 2 remaining pieces. 6)Player A gets the 1 remaining piece.

118. Example 1 Gabi, Holly, and Izzy will share a cake that is ¼ chocolate, ¼ white, ¼ yellow, and ¼ spice cake.Gabi, Holly, and Izzy will share a cake that is ¼ chocolate, ¼ white, ¼ yellow, and ¼ spice cake.

119. Example 1, cont’d The girls’ preference ratios are given below:The girls’ preference ratios are given below:

120. Example 1, cont’d Solution: Let Gabi be the divider, Holly the trimmer, and Izzy the chooser.Solution: Let Gabi be the divider, Holly the trimmer, and Izzy the chooser. Gabi assigns point values to slices of cake.Gabi assigns point values to slices of cake.

121. Example 1, cont’d Solution, cont’d: Gabi could divide the cake as shown below into 3 pieces of equal value to her.Solution, cont’d: Gabi could divide the cake as shown below into 3 pieces of equal value to her.

122. Example 1, cont’d Solution, cont’d: Holly is the trimmer.Solution, cont’d: Holly is the trimmer.

123. Example 1, cont’d Solution, cont’d: Holly trims piece 3 to make it equal in value to pieces 1 and 2.Solution, cont’d: Holly trims piece 3 to make it equal in value to pieces 1 and 2.

124. Example 1, cont’d

125. Solution, cont’d:Solution, cont’d: Izzy chooses piece 2.Izzy chooses piece 2. Holly gets the trimmed piece, piece 3.Holly gets the trimmed piece, piece 3. Gabi gets the last piece, piece 1.Gabi gets the last piece, piece 1.

126. Example 1, cont’d

127. Envy-Free Division Part 2 Players A, B, and C are to share a cake (or some other item) and have completed Part 1.Players A, B, and C are to share a cake (or some other item) and have completed Part 1. 1)Of Players B and C, the player who received the trimmed piece becomes the second chooser. The other player becomes the second divider.The other player becomes the second divider. 2)The second divider divides the excess into 3 pieces of equal value.

128. Envy-Free Division Part 2 3)The second chooser takes the piece he or she considers to be of the greatest value. 4)Player A chooses the remaining piece that he or she considers to be of the greatest value. 5)The second divider gets the last remaining piece.

129. Question: In Part 1 of an envy-free division, the results were as follows: Player A (the divider) got share 2; Player B (the trimmer) got share 1; and Player C (the chooser) got share 3, which had been trimmed. Who is the second divider for Part 2? a. Player Ab. Player Bc. Player C

130. Example 2 Gabi, Holly, and Izzy will complete the division of the cake.Gabi, Holly, and Izzy will complete the division of the cake. The excess portion is shown below.The excess portion is shown below.

131. Example 2, cont’d Solution: Recall that Gabi was the first divider and Holly received the trimmed piece.Solution: Recall that Gabi was the first divider and Holly received the trimmed piece. Holly will be the second chooser and Izzy will be the second divider.Holly will be the second chooser and Izzy will be the second divider.

132. Example 2, cont’d Solution, cont’d: Izzy divides the excess into 3 pieces.Solution, cont’d: Izzy divides the excess into 3 pieces.

133. Example 2, cont’d

134. Solution, cont’d:Solution, cont’d: Since the pieces are all of equal value to Holly, she arbitrarily chooses a piece, say piece 3.Since the pieces are all of equal value to Holly, she arbitrarily chooses a piece, say piece 3. Gabi, the first divider, chooses either equally valuable piece, say piece 1.Gabi, the first divider, chooses either equally valuable piece, say piece 1. Izzy receives the last piece, piece 2.Izzy receives the last piece, piece 2.

135. Example 2, cont’d

138. Solution, cont’d: The division is envy-free because each player feels she received a share worth as much or more as every other player’s share.Solution, cont’d: The division is envy-free because each player feels she received a share worth as much or more as every other player’s share.

139. Example 3 Jenny, Kara, and Lindsey need to divide 10 yards each of beige linen, red silk, and yellow gingham.Jenny, Kara, and Lindsey need to divide 10 yards each of beige linen, red silk, and yellow gingham. Each assigns points per yard as shown below.Each assigns points per yard as shown below.

140. Example 3, cont’d Solution: Jenny divides the fabric.Solution: Jenny divides the fabric.

141. Example 3, cont’d

144. Solution, cont’d: Lindsey took the trimmed share, so Kara cannot have it.Solution, cont’d: Lindsey took the trimmed share, so Kara cannot have it. Kara chooses the most valuable remaining share, share 2.Kara chooses the most valuable remaining share, share 2. The remaining piece, share 1, goes to Jenny.The remaining piece, share 1, goes to Jenny. Note: This completes Part 1.Note: This completes Part 1.

145. Example 3, cont’d Solution, cont’d: The excess is divided into 3 equal pieces.Solution, cont’d: The excess is divided into 3 equal pieces. Since the excess pieces are all the same, there is no real choosing to do in Part 2.Since the excess pieces are all the same, there is no real choosing to do in Part 2. Each sister gets 2/3 yard of silk as her share of the excess.Each sister gets 2/3 yard of silk as her share of the excess.

146. Example 3, cont’d

147. 4.3 Initial Problem Solution Dylan, Emery, and Fordel will share a cake that is half chocolate and half yellow.Dylan, Emery, and Fordel will share a cake that is half chocolate and half yellow. Dylan likes chocolate cake twice as much as yellow.Dylan likes chocolate cake twice as much as yellow. Emery likes chocolate and yellow cake equally well.Emery likes chocolate and yellow cake equally well. Fordel likes yellow cake twice as much as chocolate.Fordel likes yellow cake twice as much as chocolate. How should they divide the cake?How should they divide the cake?

148. Initial Problem Solution, cont’d Let Dylan be the divider, Emery the trimmer, and Fordel the chooser.Let Dylan be the divider, Emery the trimmer, and Fordel the chooser.

149. Initial Problem Solution, cont’d Part 1: Dylan evaluates the cake:Part 1: Dylan evaluates the cake:

150. Initial Problem Solution, cont’d Dylan creates 3 equal shares:Dylan creates 3 equal shares:

151. Initial Problem Solution, cont’d

152. Emery trims piece 3 so that it is equal in value to pieces 1 and 2:Emery trims piece 3 so that it is equal in value to pieces 1 and 2:

153. Initial Problem Solution, cont’d

154. Fordel chooses piece 3, the trimmed piece.Fordel chooses piece 3, the trimmed piece. The remaining pieces are identical, so Emery and Dylan each take one.The remaining pieces are identical, so Emery and Dylan each take one. Part 2: The excess is all yellow cake, so it can merely be divided into 3 equal-sized pieces and shared among the players.Part 2: The excess is all yellow cake, so it can merely be divided into 3 equal-sized pieces and shared among the players. The final division is shown on the next slides.The final division is shown on the next slides.

155. Initial Problem Solution, cont’d