A Mathematical View of Our World

A Mathematical View A Mathematical View of Our Worldof Our World

11stst ed. ed.

Parks, Musser, Trimpe, Parks, Musser, Trimpe, Maurer, and MaurerMaurer, and Maurer

Chapter 4Chapter 4

Fair DivisionFair Division

Section 4.1Section 4.1Divide and Choose MethodsDivide 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

4.1 Initial Problem4.1 Initial Problem• The brothers Drewvan, Oswald, and The brothers Drewvan, Oswald, and

Granger are to share their family’s 3600-Granger are to share their family’s 3600-acre estate.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.

4.1 Initial Problem, cont’d4.1 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 How can the brothers fairly divide the estate?estate?• The solution will be given at the end of the The solution will be given at the end of the

section.section.

Fair-Division ProblemsFair-Division Problems• Fair-division problemsFair-division problems involve fairly dividing involve fairly dividing

something between two or more people, something between two or more people, without the aid of an outside arbitrator.without the aid of an outside arbitrator.

• The people who will share the object are The people who will share the object are called called playersplayers..

• The solution to a problem is called a The solution to a problem is called a fair-fair-division proceduredivision procedure or a or a fair-division schemefair-division scheme. .

Types of Fair-Division ProblemsTypes of Fair-Division Problems

• ContinuousContinuous fair-division problems: fair-division problems: • The object(s) can be divided into pieces of The object(s) can be divided into pieces of

any size with no loss of value.any size with no loss of value.• An example is dividing a cake or an amount of An example is dividing a cake or an amount of

money among two or more people.money among two or more people.

Types of Fair-Division, cont’dTypes of Fair-Division, cont’d

• DiscreteDiscrete fair-division problems: 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 We assume the players do not want to sell

everything and divide the proceeds.everything and divide the proceeds.• However, sometimes money must be used However, sometimes money must be used

when no other fair division is possiblewhen no other fair division is possible• An example is dividing a car, a house, and a An example is dividing a car, a house, and a

boat among two or more people.boat among two or more people.


• MixedMixed fair-division problems: fair-division problems: • Some objects to be shared can be divided Some objects to be shared can be divided

and some cannot.and some cannot.• This type is a combination of continuous This type is a combination of continuous

and discrete fair division.and discrete fair division.• An example is dividing an estate consisting of An example is dividing an estate consisting of

money, a house, and a car among two or money, a house, and a car among two or more people.more people.

Question:Question:

Three cousins will share an Three cousins will share an inheritance. The estate includes a inheritance. The estate includes a house, a car, and cash. What type of house, a car, and cash. What type of fair-division problem is this?fair-division problem is this?

a. continuous a. continuous b. discreteb. discretec. mixedc. mixed


• This section will consider only This section will consider only continuous fair-division problems.continuous fair-division problems.

• We make the assumption that the value We make the assumption that the value of a player’s share is determined by his of a player’s share is determined by his or her values.or her values.• Different players may value the same Different players may value the same

share differently. share differently.

Value of a ShareValue of a Share• In a fair-division problem with In a fair-division problem with nn players, players,

a player has received a a player has received a fair sharefair share if that if that player considers his or her share to be player considers his or her share to be worth at least 1/worth at least 1/nn of the total value of the total value being shared. being shared.

• A division that results in every player A division that results in every player receiving a fair share is called receiving a fair share is called proportionalproportional..

Value of a Share, cont’dValue of a Share, cont’d

• We assume that a player’s values in a We assume that a player’s values in a fair-division problem cannot change fair-division problem cannot change based on the results of the division. based on the results of the division.

• We also assume that no player has any We also assume that no player has any knowledge of any other player’s values.knowledge of any other player’s values.

Fair Division for Two PlayersFair Division for Two Players

• The standard procedure for a The standard procedure for a continuous fair-division problem with continuous fair-division problem with two players is called the two players is called the divide-and-divide-and-choose methodchoose method..

• This method is described as dividing a This method is described as dividing a cake, but it can be used to fairly divide cake, but it can be used to fairly divide any continuous object. any continuous object.

Divide-And-Choose MethodDivide-And-Choose Method• Two players, X and Y, are to divide a cake.Two players, X and Y, are to divide a cake.

1)1) Player X divides the cake into 2 pieces that he Player X divides the cake into 2 pieces that he or she considers to be of equal value.or she considers to be of equal value.

• Player X is called the Player X is called the dividerdivider..

2)2) Player Y picks the piece he or she considers Player Y picks the piece he or she considers to be of greater value.to be of greater value.

• Player Y is called the Player Y is called the chooserchooser..

3)3) Player X gets the piece that player Y did not Player X gets the piece that player Y did not choose. choose.

Divide-And-Choose Method, cont’dDivide-And-Choose Method, cont’d

• This method produces a proportional This method produces a proportional division.division.

• The divider thinks both pieces are equal, The divider thinks both pieces are equal, so the divider gets a fair share.so the divider gets a fair share.

• The chooser will find at least one of the The chooser will find at least one of the pieces to be a fair share or more than a pieces to be a fair share or more than a fair share. The chooser selects that fair share. The chooser selects that piece, and gets a fair share. piece, and gets a fair share.

Example 1Example 1

• Margo and Steven will share a $4 Margo and Steven will share a $4 pizza that is half pepperoni and half pizza that is half pepperoni and half Hawaiian.Hawaiian.

• Margo likes both kinds of pizza equally.Margo likes both kinds of pizza equally.• Steven likes pepperoni 4 times as much Steven likes pepperoni 4 times as much

as Hawaiian.as Hawaiian.

Example 1, cont’dExample 1, cont’d• Margo cuts the pizza Margo cuts the pizza

into 6 pieces and into 6 pieces and arranges them as arranges them as shown.shown.

a)a) What monetary What monetary value would Margo value would Margo and Steven each and Steven each place on the original place on the original two halves of the two halves of the pizza?pizza?

Example 1, cont’dExample 1, cont’d• Solution: The whole pizza is worth $4.Solution: The whole pizza is worth $4.

• Margo values both kinds of pizza Margo values both kinds of pizza equally. To her each half is worth half of equally. To her each half is worth half of the total value, or $2.the total value, or $2.

• Steven values pepperoni 4 times as Steven values pepperoni 4 times as much as Hawaiian. To him the much as Hawaiian. To him the pepperoni half is worth 4/5 of the total pepperoni half is worth 4/5 of the total value, or $3.20. The Hawaiian half is value, or $3.20. The Hawaiian half is worth 1/5 of the total, or $0.80.worth 1/5 of the total, or $0.80.

Example 1, cont’dExample 1, cont’db)b) What value What value

would each would each person place on person place on each of the two each of the two plates of pizza?plates of pizza?

c)c) What plate will What plate will Steven choose?Steven choose?

Example 1, cont’dExample 1, cont’d

b)b) Solution: The whole pizza is worth $4.Solution: The whole pizza is worth $4.• Margo values both kinds of pizza Margo values both kinds of pizza

equally. To her each plate of pizza is equally. To her each plate of pizza is worth half of the total value, or $2.worth half of the total value, or $2.

Example 1, cont’dExample 1, cont’db)b) Solution, cont’d:Solution, cont’d:

• Steven values pepperoni 4 times as Steven values pepperoni 4 times as much as Hawaiian. much as Hawaiian.

• To him each pepperoni slice is worth To him each pepperoni slice is worth $3.20/3 = $1.067 and each Hawaiian slice is $3.20/3 = $1.067 and each Hawaiian slice is worth $0.80/3 = $0.267.worth $0.80/3 = $0.267.

• The first plate is worth 2($0.267) + The first plate is worth 2($0.267) + 1($1.067) = $1.60.1($1.067) = $1.60.

• The second plate is worth 1($0.267) + The second plate is worth 1($0.267) + 2($1.067) = $2.402($1.067) = $2.40

Example 1, cont’dExample 1, cont’dc)c) Solution:Solution:

• Steven will choose the second plate, Steven will choose the second plate, with one slice of Hawaiian and two slices with one slice of Hawaiian and two slices of pepperoni.of pepperoni.

• Margo gets a plate of pizza that she Margo gets a plate of pizza that she feels is worth half the value. feels is worth half the value.

• Steven gets a plate of pizza that he feels Steven gets a plate of pizza that he feels is worth more than half the value.is worth more than half the value.

Example 2Example 2• Caleb and Diego will drive 6 hours during Caleb and Diego will drive 6 hours during

the day and 4 hours at night.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 How should they divide the driving into 2

shifts if Caleb is the divider and Diego is shifts if Caleb is the divider and Diego is the chooser?the chooser?

Example 2, cont’dExample 2, cont’d• Solution:Solution:

• Caleb can assign 2 points to each hour Caleb can assign 2 points to each hour of night driving and 1 point to each hour of night driving and 1 point to each hour of day driving.of day driving.

• Caleb values the entire drive at 1(6) + 2(4) Caleb values the entire drive at 1(6) + 2(4) = 14 points.= 14 points.

• To Caleb a fair share will be worth half the To Caleb a fair share will be worth half the total value, or 7 points.total value, or 7 points.


• Solution, cont’d:Solution, cont’d:• A possible fair division for Caleb is to A possible fair division for Caleb is to

create shifts of:create shifts of:• 6 hours of daytime driving and 0.5 hours of 6 hours of daytime driving and 0.5 hours of

nighttime driving.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.

Example 2, cont’dExample 2, cont’d• Solution, cont’d:Solution, cont’d:

• Diego can assign 1 point to each hour of Diego can assign 1 point to each hour of night driving and 1 point to each hour of night driving and 1 point to each hour of day driving.day driving.

• Diego values the entire drive at 1(6) + 1(4) Diego values the entire drive at 1(6) + 1(4) = 10 points.= 10 points.

• To Diego a fair share will be worth half the To Diego a fair share will be worth half the total value, or 5 points.total value, or 5 points.


• Diego values the first shift at 1(6) + Diego values the first shift at 1(6) + 1(0.5) = 6.5 points.1(0.5) = 6.5 points.

• Diego values the second shift at 1(3.5) = Diego values the second shift at 1(3.5) = 3.5 points.3.5 points.

• Diego will choose the first shift, because Diego will choose the first shift, because it is worth more to him. it is worth more to him.

Two Players, cont’dTwo Players, cont’d• Notice that in both of the previous Notice that in both of the previous

examples:examples:• The divider got a share he or she felt was The divider got a share he or she felt was

equal to exactly half of the total value.equal to exactly half of the total value.• The chooser got a share he or she felt was The chooser got a share he or she felt was

equal to more than half of the total value. equal to more than half of the total value. • It is often advantageous to be the chooser, It is often advantageous to be the chooser,

so the roles should be randomly chosen.so the roles should be randomly chosen.

Fair Division for Three PlayersFair Division for Three Players

• In a continuous fair-division problem In a continuous fair-division problem with 3 players, it is still possible to with 3 players, it is still possible to have one player divide the object and have one player divide the object and the other players choose.the other players choose.

• This method is also called the This method is also called the lone-lone-divider methoddivider method. .

Divide-And-Choose MethodDivide-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)1) Player X (the divider) divides the cake into 3 Player X (the divider) divides the cake into 3 pieces that he/she considers to be of equal pieces that he/she considers to be of equal value.value.

2)2) Players Y and Z (the choosers) each decide Players Y and Z (the choosers) each decide which pieces are worth at least 1/3 of the total which pieces are worth at least 1/3 of the total valuevalue..

• These pieces are said to be acceptable.These pieces are said to be acceptable.3)3) The choosers announce their acceptable The choosers announce their acceptable

piecespieces..


3)3) There are 2 possibilities:There are 2 possibilities:a)a) If at least 1 piece is unacceptable to both Y If at least 1 piece is unacceptable to both Y

and Z, Player X gets that piece. and Z, Player X gets that piece. • If Y and Z can each choose acceptable If Y and Z can each choose acceptable

pieces, they do so.pieces, they do so.• If Y and Z cannot each choose acceptable If Y and Z cannot each choose acceptable

pieces, they put the remaining pieces pieces, they put the remaining pieces back together and use the two player back together and use the two player method to re-divide.method to re-divide.


3)3) Cont’d:Cont’d:b)b) If every piece is acceptable to both Y If every piece is acceptable to both Y

and Z, they each take an acceptable and Z, they each take an acceptable piece. Player X gets the leftover piece. piece. Player X gets the leftover piece.

• Note: The divide-and-choose method can Note: The divide-and-choose method can be extended to more than 3 players. The be extended to more than 3 players. The more players, the more complicated the more players, the more complicated the process becomes.process becomes.

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

a. Player A gets share 3, Player B gets share 1, and a. Player A gets share 3, Player B gets share 1, and Player C gets share 2.Player C gets share 2.b. Player A gets share 1, Player B gets share 3, and b. Player A gets share 1, Player B gets share 3, and Player C gets share 2.Player C gets share 2.c. Player A gets share 2, Player B gets share 3, and c. Player A gets share 2, Player B gets share 3, and Player C gets share 1.Player C gets share 1.d. Player A gets share 1, Player B gets share 2, and d. Player A gets share 1, Player B gets share 2, and Player C gets share 3.Player C gets share 3.

Example 3Example 3• Emma, Fay, and Grace will divide 24 ounces of ice Emma, Fay, and Grace will divide 24 ounces of ice

cream, which is made up of equal amounts of cream, which is made up of equal amounts of vanilla, chocolate, and strawberry.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 Fay prefers chocolate 2 to 1 over either other

flavor and prefers vanilla and strawberry equally flavor and prefers vanilla and strawberry equally well.well.

• Grace prefers vanilla to chocolate to strawberry Grace prefers vanilla to chocolate to strawberry in the ratio 1 to 2 to 3. in the ratio 1 to 2 to 3.

• If Emma is the divider, what are the results of the If Emma is the divider, what are the results of the divide-and-choose method for 3 players?divide-and-choose method for 3 players?

Example 3, cont’dExample 3, cont’d• Solution: Suppose Emma divides the ice cream Solution: Suppose Emma divides the ice cream

into 3 equal parts, each consisting of one of the into 3 equal parts, each consisting of one of the flavors. flavors.

Example 3, cont’dExample 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.

Example 3, cont’dExample 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

Example 3, cont’dExample 3, cont’d• Solution, cont’d: All of the players’ values are Solution, cont’d: All of the players’ values are

summarized in the table below.summarized in the table below.

Example 3, cont’dExample 3, cont’d• Solution, cont’d: Solution, cont’d:

• Portion 1 is unacceptable to both Fay and Portion 1 is unacceptable to both Fay and Grace. As the divider, Emma will receive Grace. As the divider, Emma will receive portion 1.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: The division is Emma: portion 1; Fay: portion 2; Grace: portion 3.portion 2; Grace: portion 3.

Last-Diminisher MethodLast-Diminisher Method

• A method for continuous fair-division A method for continuous fair-division problems with 3 or more players is called problems with 3 or more players is called the the last-diminisher methodlast-diminisher method..

• Suppose any number of players X, Y, … Suppose any number of players X, Y, … are dividing a cake. are dividing a cake.

1)1) Player X cuts a piece of cake that he or she Player X cuts a piece of cake that he or she considers to be a fair share.considers to be a fair share.

Last-Diminisher Method, cont’dLast-Diminisher Method, cont’d

2)2) Each player, in turn, judges the fairness Each player, in turn, judges the fairness of the piece.of the piece.

a)a) If a player considers the piece fair or less If a player considers the piece fair or less than fair, it passes to the next player.than fair, it passes to the next player.

b)b) If a player considers the piece more than If a player considers the piece more than fair, the player trims the piece to make it fair, the player trims the piece to make it fair, returning the trimming to the undivided fair, returning the trimming to the undivided portion and passing the trimmed piece to portion and passing the trimmed piece to the next player.the next player.

Last-Diminisher Method, cont’dLast-Diminisher Method, cont’d

3)3) The last player to trim the piece, gets the The last player to trim the piece, gets the piece as his or her share.piece as his or her share.

• If no player trimmed the piece, player X gets If no player trimmed the piece, player X gets the piece.the piece.

4)4) After one player gets a piece of cake, After one player gets a piece of cake, the process begins again without that the process begins again without that player and that piece.player and that piece.

• When only 2 players remain, they use the When only 2 players remain, they use the divide-and-choose method.divide-and-choose method.

Example 4Example 4• Hector, Isaac, and James will divide 24 Hector, Isaac, and James will divide 24

ounces of ice cream, which is equal parts ounces of ice cream, which is equal parts vanilla, chocolate, and strawberry.vanilla, chocolate, and strawberry.

• Hector values vanilla to chocolate to Hector values vanilla to chocolate to strawberry 1 to 2 to 3.strawberry 1 to 2 to 3.

• Isaac likes the 3 flavors equally.Isaac likes the 3 flavors equally.• James values vanilla to chocolate to James values vanilla to chocolate to

strawberry 1 to 2 to 1.strawberry 1 to 2 to 1.


• Using the last-diminisher method with Using the last-diminisher method with Hector as the first divider and Isaac as the Hector as the first divider and Isaac as the first judge, find the results of the division.first judge, find the results of the division.

• Solution: Solution: • Hector assigns 1 point to each ounce of Hector assigns 1 point to each ounce of

vanilla, 2 points to each ounce of chocolate, vanilla, 2 points to each ounce of chocolate, and 3 points to each ounce of strawberry. and 3 points to each ounce of strawberry.

Example 4, cont’dExample 4, cont’d• Solution, cont’d: A fair share of ice cream, Solution, cont’d: A fair share of ice cream,

to Hector, is worth 48/3 = 16 points.to Hector, is worth 48/3 = 16 points.


• One possible fair share for Hector would One possible fair share for Hector would be all 8 ounces of vanilla plus 4 ounces be all 8 ounces of vanilla plus 4 ounces of chocolate.of chocolate.

• This share is worth 1(8) + 2(4) = 16 This share is worth 1(8) + 2(4) = 16 points to Hector, so he would be happy points to Hector, so he would be happy with this share. with this share.

• Next, Isaac must decide whether the Next, Isaac must decide whether the share is fair, according to his values.share is fair, according to his values.


• Isaac assigns 1 point to each ounce Isaac assigns 1 point to each ounce of vanilla, 1 point to each ounce of of vanilla, 1 point to each ounce of chocolate, and 1 point to each chocolate, and 1 point to each ounce of strawberry. ounce of strawberry.

• Isaac values all of the ice cream at Isaac values all of the ice cream at 1(8) + 1(8) + 1(8) = 24 points. 1(8) + 1(8) + 1(8) = 24 points.

• A fair share to Isaac is 8 points.A fair share to Isaac is 8 points.


• Isaac’s value for Hector’s serving is 1(8) Isaac’s value for Hector’s serving is 1(8) + 1(4) = 12 points.+ 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 Isaac trims off 4 points worth of ice cream.cream.

• Suppose he trims off the 4 ounces of Suppose he trims off the 4 ounces of chocolate.chocolate.


• Next, James must judge the share.Next, James must judge the share.• James assigns 1 point to each ounce of James assigns 1 point to each ounce of

vanilla, 2 points to each ounce of vanilla, 2 points to each ounce of chocolate, and 1 point to each ounce of chocolate, and 1 point to each ounce of strawberry. strawberry.

• James values all of the ice cream at 1(8) James values all of the ice cream at 1(8) + 2(8) + 1(8) = 32 points.+ 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.


• The existing share is now just 8 The existing share is now just 8 ounce of vanilla.ounce of vanilla.

• To James, the share is worth 1(8) = To James, the share is worth 1(8) = 8 points.8 points.

• James thinks this is less than a fair James thinks this is less than a fair share.share.

• James will not trim the share.James will not trim the share.


• Isaac was the last-diminisher, and gets Isaac was the last-diminisher, and gets the share of ice cream.the share of ice cream.

• Hector and James will divide the Hector and James will divide the remaining ice cream using the divide-remaining ice cream using the divide-and-choose method. and-choose method.

• Note: This is only one of many different Note: This is only one of many different possible solutions.possible solutions.

4.1 Initial Problem Solution4.1 Initial Problem Solution• The brothers Drewvan, Oswald, and Granger are The brothers Drewvan, Oswald, and Granger are

to share their family’s estate, which is 1200 acres to share their family’s estate, which is 1200 acres each of vineyards, woodlands, and fields.each of vineyards, woodlands, and fields.

• Drewvan prefers vineyards to woodlands to fields Drewvan prefers vineyards to woodlands to fields 3 to 2 to 1.3 to 2 to 1.

• Oswald prefers vineyards to woodlands to fields Oswald prefers vineyards to woodlands to fields 2 to 3 to 1.2 to 3 to 1.

• Granger prefers vineyards to woodlands to fields Granger prefers vineyards to woodlands to fields 2 to 1 to 3.2 to 1 to 3.

Initial Problem Solution, cont’dInitial Problem Solution, cont’d

• Use the divide-and-choose method for 3 Use the divide-and-choose method for 3 players.players.• Let Drewvan be the divider.Let Drewvan be the divider.

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

Initial Problem Solution, cont’dInitial Problem Solution, cont’d• To Drewvan, a fair share is worth 7200/3 = To Drewvan, a fair share is worth 7200/3 =

2400 points.2400 points.• One possible fair division is shown below.One possible fair division is shown below.

Initial Problem Solution, cont’dInitial Problem Solution, cont’d• Next, the two choosers will consider this Next, the two choosers will consider this

division.division.• Granger and Oswald both value the Granger and Oswald both value the

entire estate at 7200 points also.entire estate at 7200 points also.• To Oswald, a fair share is worth 2400 To Oswald, a fair share is worth 2400

points.points.• To Granger, a fair share is worth 2400 To Granger, a fair share is worth 2400

points.points.


• Oswald considers piece 1 to be unacceptable. Oswald considers piece 1 to be unacceptable.


• Granger considers pieces 1 and 2 to be Granger considers pieces 1 and 2 to be unacceptable. unacceptable.


• Both choosers think piece 1 is Both choosers think piece 1 is unacceptable, so Drewvan gets piece 1.unacceptable, so Drewvan gets piece 1.

• Granger thinks only piece 3 is Granger thinks only piece 3 is acceptable, so he gets that piece.acceptable, so he gets that piece.

• Oswald thinks pieces 2 and 3 are Oswald thinks pieces 2 and 3 are acceptable, so Oswald gets piece 2.acceptable, so Oswald gets piece 2.

Section 4.2Section 4.2Discrete and Mixed Discrete and Mixed Division ProblemsDivision 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

4.2 Initial Problem4.2 Initial Problem• When twins Zack and Zeke turned 16 they When twins Zack and Zeke turned 16 they

received: received: • A pickup truckA pickup truck• A horseA horse• A cow A 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 The solution will be given at the end of the

section.section.

Discrete Fair DivisionDiscrete Fair Division

• Recall that discrete fair division Recall that discrete fair division problems involve sharing objects that problems involve sharing objects that cannot be divided without losing value.cannot be divided without losing value.

• Two methods for solving discrete fair-Two methods for solving discrete fair-division problems are:division problems are:• The method of sealed bids.The method of sealed bids.• The method of points.The method of points.

Method of Sealed BidsMethod of Sealed Bids• Any number of players, Any number of players, NN, are to share , are to share

any number of items.any number of items.• If necessary, money will be used to insure If necessary, money will be used to insure

fairness.fairness.

1)1) All players submit sealed bids, stating All players submit sealed bids, stating monetary values for each item.monetary values for each item.

2)2) Each item goes to the highest bidder.Each item goes to the highest bidder.• The highest bidder places the dollar amount of The highest bidder places the dollar amount of

his or her bid into a compensation fund.his or her bid into a compensation fund.

Method of Sealed Bids, cont’dMethod of Sealed Bids, cont’d

3)3) From the compensation fund, each From the compensation fund, each player receives 1/player receives 1/NN of his or her bid of his or her bid on each item.on each item.

4)4) Any money leftover in the fund is Any money leftover in the fund is distributed equally to all players. distributed equally to all players.

• Note: This method is also called the Note: This method is also called the Knaster Inheritance Procedure.Knaster Inheritance Procedure.

Example 1Example 1• Three sisters Maura, Nessa, and Odelia Three sisters Maura, Nessa, and Odelia

will share a house and a cottage. will share a house and a cottage. • Apply the method of sealed bids to divide Apply the method of sealed bids to divide

the property, using the bids shown below.the property, using the bids shown below.


• Solution: Each piece of property goes to Solution: Each piece of property goes to the highest bidder.the highest bidder.

• Odelia gets the family home and places Odelia gets the family home and places $301,000 into the compensation fund.$301,000 into the compensation fund.

• Nessa gets the cottage and places $203,000 Nessa gets the cottage and places $203,000 into the compensation fund.into the compensation fund.


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

Example 1, cont’dExample 1, cont’d• Solution, cont’d: Each sister receives 1/3 Solution, cont’d: Each sister receives 1/3

of her total bids from the compensation of her total bids from the compensation fund.fund.

• Maura receivesMaura receives

• Nessa receivesNessa receives

• Odelia receivesOdelia receives

286000 203000 $163,0003

289000 188000 $159,0003

301000 182000 $161,0003

Example 1, cont’dExample 1, cont’d• Solution, cont’d: After the Solution, cont’d: After the

distributions, there is $504,000 – distributions, there is $504,000 – ($159,000 + $160,000 + $161,000) ($159,000 + $160,000 + $161,000) = $21,000 left in the fund.= $21,000 left in the fund.

• The leftover money is distributed The leftover money is distributed equally to the three sisters in the equally to the three sisters in the amount of $7000 each.amount of $7000 each.

Example 1, cont’dExample 1, cont’d• Solution, cont’d: The final shares Solution, cont’d: The final shares

are:are:• Maura receives $166,000 and no Maura receives $166,000 and no

property.property.• Nessa receives the cottage, for which Nessa receives the cottage, for which

she paid a net amount of $33,000.she paid a net amount of $33,000.• Odelia receives the family home, for Odelia receives the family home, for

which she paid a net amount of which she paid a net amount of $133,000.$133,000.

Example 1, cont’dExample 1, cont’d• Solution, cont’d: Note that the division is Solution, cont’d: Note that the division is

proportional because each sister receives what proportional because each sister receives what she considers to be a fair share. she considers to be a fair share.

Method of PointsMethod of Points• Three players will share three items.Three players will share three items.1)1) Each player assigns points to each Each player assigns points to each

item, so that the points for each item, so that the points for each player total to 100.player total to 100.

2)2) List all 6 possible arrangements of List all 6 possible arrangements of players and items, along with the players and items, along with the point values.point values.

Method of Points, cont’dMethod of Points, cont’d3)3) Note the smallest number of points Note the smallest number of points

assigned to an item in each assigned to an item in each arrangement.arrangement.

• Choose the arrangement with the Choose the arrangement with the largest value of the smallest number.largest value of the smallest number.

• If there is not only one such If there is not only one such arrangement, keep all the arrangement, keep all the arrangements with that smallest point arrangements with that smallest point value and go to step 4.value and go to step 4.

Method of Points, cont’dMethod of Points, cont’d4)4) For each arrangement kept in step For each arrangement kept in step

3, note the middle point value.3, note the middle point value.• Choose the arrangement with the Choose the arrangement with the

largest value of the middle number.largest value of the middle number.• If there is not only one such If there is not only one such

arrangement, keep all the arrangement, keep all the arrangements with that middle point arrangements with that middle point value and go to step 5.value and go to step 5.

Method of Points, cont’dMethod of Points, cont’d

5)5) For each arrangement kept in step For each arrangement kept in step 4, note the largest point value.4, note the largest point value.

• Choose any arrangement with the Choose any arrangement with the largest value of the largest number.largest value of the largest number.

Example 2Example 2• Three couples, A, B, and C, need to decide Three couples, A, B, and C, need to decide

who gets which room in a hotel.who gets which room in a hotel.• The couples assign points to each room as The couples assign points to each room as

shown below. shown below.

Example 2, cont’dExample 2, cont’d• Solution:Solution:


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

Example 2, cont’dExample 2, cont’d• Solution, cont’d: The arrangement Solution, cont’d: The arrangement

selected is: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 Two couples got their first choice and one got their second choice.and one got their second choice.

Example 3Example 3• Three couples, A, B, and C, need to decide Three couples, A, B, and C, need to decide

who gets which room in a cabin.who gets which room in a cabin.• The couples assign points to each room as The couples assign points to each room as

shown below. shown below.



• Solution, cont’d: The largest minimum point value Solution, cont’d: The largest minimum point value is 12, which occurs in 4 arrangements.is 12, which occurs in 4 arrangements.

• Table from Table from previous slideprevious slide

Example 3, cont’dExample 3, cont’d• Solution, cont’d: Keep those 4 arrangements Solution, cont’d: Keep those 4 arrangements

and look at the middle point values.and look at the middle point values.


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

Example 3, cont’dExample 3, cont’d• Solution, cont’d: Keep those 3 arrangements Solution, cont’d: Keep those 3 arrangements

and look at the largest point values.and look at the largest point values.• The maximum largest number is 71.The maximum largest number is 71.


• Solution, cont’d: The arrangement Solution, cont’d: The arrangement selected is:selected is:

• Couple A, Room 2Couple A, Room 2• Couple B, Room 1Couple B, Room 1• Couple C, Room 3Couple C, Room 3

Question:Question:The assignments were Couple A, Room 2; The assignments were Couple A, Room 2; Couple B, Room 1; and Couple C, Room 3. Couple B, Room 1; and Couple C, Room 3. Considering the point values each couple Considering the point values each couple assigned to the three rooms, is this assigned to the three rooms, is this division proportional? division proportional?

a. yesa. yesb. nob. no


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

Mixed Fair DivisionMixed Fair Division

• Recall that mixed fair division problems Recall that mixed fair division problems involve sharing a mixture of discrete involve sharing a mixture of discrete and continuous objects.and continuous objects.

• A method for solving mixed fair-division A method for solving mixed fair-division problems is the adjusted winner problems is the adjusted winner procedure. procedure.

Adjusted Winner ProcedureAdjusted Winner Procedure• Two players are to fairly divide any Two players are to fairly divide any

number of items.number of items.• The items may be discrete and/or The items may be discrete and/or

continuous.continuous.• Ownership of some items may be Ownership of some items may be

shared.shared.

1)1) Each player assigns points to each Each player assigns points to each item, for a total of 100 points.item, for a total of 100 points.

Adjusted Winner Procedure, cont’dAdjusted Winner Procedure, cont’d

2)2) Each player tentatively receives Each player tentatively receives items to which he or she assigned items to which he or she assigned the most points.the most points.

• The points are added to the player’s The points are added to the player’s total.total.

• If 2 players tie for an item, the item If 2 players tie for an item, the item goes to the player with the lowest point goes to the player with the lowest point total so far.total so far.


3)3) Look at the players’ points.Look at the players’ points.• If the point totals are equal, the If the point totals are equal, the

process is done.process is done.• If Player X has more points than Player If Player X has more points than Player

Y, then give Player Y the item for Y, then give Player Y the item for which the ratio of the number of points which the ratio of the number of points assigned by X to the number of points assigned by X to the number of points assigned by Y is the smallest. assigned by Y is the smallest.


4)4) Re-examine the players’ points.Re-examine the players’ points.a)a) If the point totals are equal, the If the point totals are equal, the

process is done.process is done.

b)b) If Player X still has more points than If Player X still has more points than Player Y, repeat step 3.Player Y, repeat step 3.

c)c) If Player Y now has more points than If Player Y now has more points than Player X, move a fraction of the last Player X, move a fraction of the last item moved back to X.item moved back to X.

Adjusted Winner Procedure, cont’dAdjusted Winner Procedure, cont’d4)4) Cont’d: The formula for case C is as Cont’d: The formula for case C is as

follows, where follows, where qq = fraction of item to be = fraction of item to be moved, moved, TTXX = Player X’s point total without = Player X’s point total without this item, this item, TTY Y = Player Y’s point total without = Player Y’s point total without this item, this item, PPXX = number of points Player X = number of points Player X assigned to this item, assigned to this item, PPYY = number of points = number of points Player Y assigned to this item.Player Y assigned to this item.

Y X Y

X Y

T T PqP P

Question:Question:Three players will divide a car, a boat, and an RV. Three players will divide a car, a boat, and an RV. Their point assignments are shown in the table below. Their point assignments are shown in the table below. Complete step 2 in the adjusted winner procedure. Complete step 2 in the adjusted winner procedure.

a. Player 1: boat; Player 2: RV; Player 3: cara. Player 1: boat; Player 2: RV; Player 3: carb. Player 1: boat, RV; Player 2: nothing; Player 3: carb. Player 1: boat, RV; Player 2: nothing; Player 3: carc. Player 1: boat; Player 2: car, RV; Player 3: nothing c. Player 1: boat; Player 2: car, RV; Player 3: nothing

Player 1 Player 2 Player 3car 15 30 35boat 35 20 20RV 50 50 45

Example 4Example 4• Two players, A and B, need to divide a house, Two players, A and B, need to divide a house,

a boat, a cabin, and a condominium.a boat, a cabin, and a condominium.• Both have assigned points as shown below. Both have assigned points as shown below.

Example 4, cont’dExample 4, cont’d• Solution: The tentative assignment Solution: The tentative assignment

of items, along with the current point of items, along with the current point totals, is shown in the table below. totals, is shown in the table below.


Player A has more Player A has more points than Player B.points than Player B.

• Determine which item Determine which item to move from A to B to move from A to B by considering the by considering the ratios of both items ratios of both items currently assigned to currently assigned to Player A. Player A.

Example 4, cont’dExample 4, cont’d• Solution, cont’d: Move the house, Solution, cont’d: Move the house,

which has the smaller ratio, from which has the smaller ratio, from Player A to Player B.Player A to Player B.

Example 4, cont’dExample 4, cont’d• Solution, cont’d: Now Player B has Solution, cont’d: Now Player B has

more points than Player A.more points than Player A.• A fraction of the house must be given A fraction of the house must be given

back to Player A.back to Player A.• The values for the formula are:The values for the formula are:

• TTXX = T = TAA = 30= 30

• TTYY = T = TBB = 45= 45

• PPX X = P= PAA = 45= 45

• PPY Y = P= PBB = 35= 35


• Solution, cont’d: The calculation is:Solution, cont’d: The calculation is:

• Player B keeps 3/8 of the house and Player B keeps 3/8 of the house and 5/8 of the house goes back to Player 5/8 of the house goes back to Player A.A.

45 30 35 50 545 35 80 8

B A B

A B

T T PqP P


• Solution, cont’d: Re-examine the Solution, cont’d: Re-examine the points totals.points totals.

• Player A has 30 + 45(5/8) = 58.125 Player A has 30 + 45(5/8) = 58.125 points.points.

• Player B has 25 + 20 + 35(3/8) = Player B has 25 + 20 + 35(3/8) = 58.125 points.58.125 points.

• The division is now proportional.The division is now proportional.

4.2 Initial Problem Solution4.2 Initial Problem Solution• Zack and Zeke need to share a pickup truck, a Zack and Zeke need to share a pickup truck, a

horse, and a cow.horse, and a cow.• Solution: They could use the adjusted winner Solution: They could use the adjusted winner

procedure to share the items. Suppose they procedure to share the items. Suppose they assign points as shown.assign points as shown.


• 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 Zack has more points, so something must be given to Zeke.must be given to Zeke.• Check the ratios for the truck and the cow.Check the ratios for the truck and the cow.

Initial Problem Solution, cont’dInitial Problem Solution, cont’d• The ratios are The ratios are

shown in the shown in the table.table.• The ratio for the The ratio for the

truck is smaller.truck is smaller.• Move the truck Move the truck

from Zack to from Zack to Zeke.Zeke.

Initial Problem Solution, cont’dInitial 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 Since the points are not equal, a fraction of the truck must be given back to Zack.the truck must be given back to Zack.• The values for the formula are:The values for the formula are:

• TTXX = = TTZackZack = 40; = 40; PPX X = P= PZackZack = 33= 33

• TTYY = T = TZekeZeke = 35; = 35; PPY Y = P= PZekeZeke = 30= 30


• Solution, cont’d: The calculation is:Solution, cont’d: The calculation is:

• Zeke keeps 60.3% of the truck and Zeke keeps 60.3% of the truck and 39.7% of the truck goes back to 39.7% of the truck goes back to Zack.Zack.

35 40 30 25 39.7%33 30 63

Zeke Zack Zeke

Zack Zeke

T T PqP P


• Solution, cont’d: Re-examine the Solution, cont’d: Re-examine the points totals.points totals.

• Zack has 40 + 33(0.397) = 53.095 Zack has 40 + 33(0.397) = 53.095 points.points.

• Zeke has 35 + 30(0.603) = 53.095 Zeke has 35 + 30(0.603) = 53.095 points.points.

• For a proportional division to be For a proportional division to be created, the truck must be shared.created, the truck must be shared.

Section 4.3Section 4.3Envy-Free DivisionEnvy-Free Division

• GoalsGoals• Study envy-free divisionStudy envy-free division

• Continuous envy-free division for 3 playersContinuous envy-free division for 3 players

4.3 Initial Problem4.3 Initial Problem• Dylan, Emery, and Fordel will share a Dylan, Emery, and Fordel will share a

cake that is half chocolate and half yellow.cake that is half chocolate and half yellow.• Dylan likes chocolate cake twice as much as Dylan likes chocolate cake twice as much as

yellow.yellow.• Emery likes chocolate and yellow cake Emery likes chocolate and yellow cake

equally well.equally well.• Fordel likes yellow cake twice as much as Fordel likes yellow cake twice as much as

chocolate.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.

Envy-Free DivisionEnvy-Free Division• A division, among A division, among nn players, is considered players, is considered

envy-freeenvy-free if each player feels that: if each player feels that:• He or she has received at least 1/He or she has received at least 1/nn of the total of the total

valuevalue• No other player’s share is more valuable than his No other player’s share is more valuable than his

or her own.or her own.• Note: A proportional division is not always Note: A proportional division is not always

envy-free.envy-free.

Question:Question:Ice cream was divided among three players. The Ice cream was divided among three players. The players’ values of the shares are shown in the table players’ values of the shares are shown in the table below. below.

The division was Emma, Portion 1; Fay, Portion 2; The division was Emma, Portion 1; Fay, Portion 2; Grace, Portion 3.Grace, Portion 3.Is the division envy free? Is the division envy free? a. yes b. noa. yes b. no

Continuous Envy-Free DivisionContinuous Envy-Free Division

• This section covers envy-free divisions This section covers envy-free divisions for continuous fair-division problems for continuous fair-division problems involving three players.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.

Envy-Free Division Part 1Envy-Free Division Part 1

• Players A, B, and C are to share a Players A, B, and C are to share a cake (or some other item).cake (or some other item).

1)1) Player A (the divider) divides the cake Player A (the divider) divides the cake into 3 pieces he or she considers to be into 3 pieces he or she considers to be of equal value.of equal value.

2)2) Player B chooses the one most Player B chooses the one most valuable piece of these 3 pieces.valuable piece of these 3 pieces.

Envy-Free Division Part 1, cont’dEnvy-Free Division Part 1, cont’d

3)3) Player B (the trimmer) trims the most Player B (the trimmer) trims the most valuable piece so it is equal to the valuable piece so it is equal to the second most valuable piece.second most valuable piece.

• The excess is set aside.The excess is set aside.

4)4) Player C (the chooser) chooses the Player C (the chooser) chooses the piece he or she considers to have the piece he or she considers to have the greatest value.greatest value.

Envy-Free Division Part 1, cont’dEnvy-Free Division Part 1, cont’d

5)5) Player B chooses next.Player B chooses next.• Player B gets the trimmed piece if it is Player B gets the trimmed piece if it is

available.available.• If the trimmed piece is gone, B chooses If the trimmed piece is gone, B chooses

the most valuable piece from the 2 the most valuable piece from the 2 remaining pieces.remaining pieces.

6)6) Player A gets the 1 remaining piece. Player A gets the 1 remaining piece.

Example 1Example 1• Gabi, Holly, and Izzy will share a cake Gabi, Holly, and Izzy will share a cake

that is ¼ chocolate, ¼ white, ¼ yellow, that is ¼ chocolate, ¼ white, ¼ yellow, and ¼ spice cake.and ¼ spice cake.

Example 1, cont’dExample 1, cont’d• The girls’ preference ratios are given The girls’ preference ratios are given

below:below:

Example 1, cont’dExample 1, cont’d• Solution: Let Gabi be the divider, Holly the Solution: Let Gabi be the divider, Holly the

trimmer, and Izzy the chooser.trimmer, and Izzy the chooser.• Gabi assigns point values to slices of cake.Gabi assigns point values to slices of cake.

Example 1, cont’dExample 1, cont’d• Solution, cont’d: Gabi could divide the Solution, cont’d: Gabi could divide the

cake as shown below into 3 pieces of cake as shown below into 3 pieces of equal value to her.equal value to her.

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

Example 1, cont’dExample 1, cont’d• Solution, cont’d: Holly trims piece 3 to Solution, cont’d: Holly trims piece 3 to

make it equal in value to pieces 1 and 2.make it equal in value to pieces 1 and 2.



• 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.


Envy-Free Division Part 2Envy-Free Division Part 2• Players A, B, and C are to share a cake (or Players A, B, and C are to share a cake (or

some other item) and have completed Part some other item) and have completed Part 1.1.

1)1) Of Players B and C, the player who Of Players B and C, the player who received the trimmed piece becomes the received the trimmed piece becomes the second choosersecond chooser..

• The other player becomes the The other player becomes the second dividersecond divider..

2)2) The second divider divides the excess into The second divider divides the excess into 3 pieces of equal value. 3 pieces of equal value.

Envy-Free Division Part 2Envy-Free Division Part 2

3)3) The second chooser takes the piece The second chooser takes the piece he or she considers to be of the he or she considers to be of the greatest value.greatest value.

4)4) Player A chooses the remaining piece Player A chooses the remaining piece that he or she considers to be of the that he or she considers to be of the greatest value.greatest value.

5)5) The second divider gets the last The second divider gets the last remaining piece. remaining piece.

Question:Question:

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

a. Player Aa. Player A b. Player Bb. Player B c. Player Cc. Player C

Example 2Example 2• Gabi, Holly, and Izzy will complete the Gabi, Holly, and Izzy will complete the

division of the cake.division of the cake.• The excess portion is shown below.The excess portion is shown below.

Example 2, cont’dExample 2, cont’d• Solution: Recall that Gabi was the first Solution: Recall that Gabi was the first

divider and Holly received the trimmed piece.divider and Holly received the trimmed piece.• Holly will be the second chooser and Izzy will Holly will be the second chooser and Izzy will

be the second divider.be the second divider.

Example 2, cont’dExample 2, cont’d• Solution, cont’d: Izzy divides the excess Solution, cont’d: Izzy divides the excess

into 3 pieces. into 3 pieces.



• Solution, cont’d: Solution, cont’d: • Since the pieces are all of equal value to Since the pieces are all of equal value to

Holly, she arbitrarily chooses a piece, say Holly, she arbitrarily chooses a piece, say piece 3.piece 3.

• Gabi, the first divider, chooses either Gabi, the first divider, chooses either equally valuable piece, say piece 1.equally valuable piece, say piece 1.

• Izzy receives the last piece, piece 2. Izzy receives the last piece, piece 2.




Example 2, cont’dExample 2, cont’d• Solution, cont’d: The division is envy-free because Solution, cont’d: The division is envy-free because

each player feels she received a share worth as each player feels she received a share worth as much or more as every other player’s share.much or more as every other player’s share.

Example 3Example 3• Jenny, Kara, and Lindsey need to divide 10 Jenny, Kara, and Lindsey need to divide 10

yards each of beige linen, red silk, and yards each of beige linen, red silk, and yellow gingham.yellow gingham.

• Each assigns points per yard as shown below.Each assigns points per yard as shown below.

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




Example 3, cont’dExample 3, cont’d• Solution, cont’d: Lindsey took the Solution, cont’d: Lindsey took the

trimmed share, so Kara cannot have it.trimmed share, so Kara cannot have it.• Kara chooses the most valuable remaining Kara chooses the most valuable remaining

share, share 2.share, share 2.• The remaining piece, share 1, goes to The remaining piece, share 1, goes to

Jenny.Jenny.• Note: This completes Part 1.Note: This completes Part 1.


• Solution, cont’d: The excess is divided Solution, cont’d: The excess is divided into 3 equal pieces.into 3 equal pieces.• Since the excess pieces are all the same, Since the excess pieces are all the same,

there is no real choosing to do in Part 2.there is no real choosing to do in Part 2.• Each sister gets 2/3 yard of silk as her Each sister gets 2/3 yard of silk as her

share of the excess.share of the excess.


4.3 Initial Problem Solution4.3 Initial Problem Solution• Dylan, Emery, and Fordel will share a Dylan, Emery, and Fordel will share a

cake that is half chocolate and half yellow.cake that is half chocolate and half yellow.• Dylan likes chocolate cake twice as much as Dylan likes chocolate cake twice as much as

yellow.yellow.• Emery likes chocolate and yellow cake Emery likes chocolate and yellow cake

equally well.equally well.• Fordel likes yellow cake twice as much as Fordel likes yellow cake twice as much as

chocolate.chocolate. • How should they divide the cake?How should they divide the cake?


• Let Dylan be the divider, Emery the Let Dylan be the divider, Emery the trimmer, and Fordel the choosertrimmer, and Fordel the chooser. .


• Part 1: Dylan evaluates the cake:Part 1: Dylan evaluates the cake:

Initial Problem Solution, cont’dInitial Problem Solution, cont’d• Dylan creates 3 equal shares:Dylan creates 3 equal shares:


Initial Problem Solution, cont’dInitial Problem Solution, cont’d• Emery trims piece 3 so that it is equal in Emery trims piece 3 so that it is equal in

value to pieces 1 and 2:value to pieces 1 and 2:


Initial Problem Solution, cont’dInitial Problem Solution, cont’d• Fordel chooses piece 3, the trimmed piece.Fordel chooses piece 3, the trimmed piece.• The remaining pieces are identical, so The remaining pieces are identical, so

Emery and Dylan each take one.Emery and Dylan each take one.• Part 2: The excess is all yellow cake, so it Part 2: The excess is all yellow cake, so it

can merely be divided into 3 equal-sized can merely be divided into 3 equal-sized pieces and shared among the players. pieces and shared among the players.

• The final division is shown on the next The final division is shown on the next slides.slides.




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A Mathematical View of Our World