Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

8.3: Continuous Methods 2 - Lone Chooser and Last Diminisher Methods

  • Page ID
    22354
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    The Lone Chooser method, like the Lone Divider method, is an extension of the Divider/Chooser method. The Lone Chooser method for three players involves two dividers and one chooser. It can be extended to N players with N-1 dividers and one chooser. We will focus on the three player Lone Chooser method in this book.

    The Last Diminisher is a very different method from the Divider/Chooser methods we discuss in this book. In a sense, everyone is a divider and everyone is a chooser. The Last Diminisher method works well when many players must divide a continuous object like a cake or a piece of land.

    Lone Chooser Method

    In the Lone Chooser method for three players, there are two dividers and one chooser. The basic idea is that the two dividers use the Divider/Chooser method to divide the object into two pieces. At this point each of the dividers believes that he/she has at least half the value of the object. Next each divider divides his/her piece into three smaller pieces for a total of six pieces. The chooser then picks one piece from each of the dividers’ pieces leaving all three players with two pieces each.

    Summary of the Lone Chooser method
    1. Randomly choose one player to be the chooser, C. The other two players are dividers, D1 and D2.
    2. The dividers D1 and D2 use the Divider/Chooser method to divide the object into two pieces.
    3. Each of the dividers D1 and D2 subdivide his/her piece into three pieces of equal value. Use the same ideas as we used in the Divider/Chooser method for determining the value of each piece.
    4. The chooser assigns a value to each of the six pieces according to his/her value system. The chooser then picks the piece with the greatest value from each of the dividers. Each of the dividers keep his/her other two pieces.
    Example \(\PageIndex{1}\): Lone Chooser Method for Three Players

    Fred, Gloria, and Harvey wish to split a three-flavored cake worth $36 that is one-third chocolate, one-third strawberry, and one-third vanilla. Fred does not like chocolate, but likes strawberry and vanilla equally well. Gloria likes chocolate twice as much as vanilla and likes strawberry three times as much as vanilla. Harvey likes chocolate and strawberry equally but likes vanilla twice as much as chocolate or strawberry. Use the Lone-Chooser method to find a fair division for the cake.

    KPx-27fYwf7W_KHKJZFNEDRCkfKso5_NombG96OL0lSFjI8UkC8ertCh_y3G5Gz8OSC6qAZErd75Sqe3eAdKZm_qThT5-TFF6GrBLQn9z3KvijcCvf3Cq8zY9fCikas70EV633I
    Figure \(\PageIndex{1}\): Three-Flavored Cake

    First let’s figure out how each player values each of the pieces of the cake.

    Fred sees the chocolate part of the cake as having a value of $0 since he does not like chocolate. All of the value of the cake is in the strawberry and vanilla parts. Since he likes them equally well, the strawberry part and the vanilla part are both worth $36/2=$18.

    For Gloria, let x = the value of the vanilla part of the cake. Then the chocolate part is worth 2x and the strawberry part is worth 3x.

    To Gloria the vanilla part is worth $6, the chocolate part is worth $12 and the chocolate part is worth $18.

    For Harvey, let y = the value of the chocolate part of the cake. Then the value of the strawberry part is also y and the value of the vanilla part is 2y.

    To Harvey the chocolate and strawberry parts are each worth $9 and the vanilla part is worth $18.

    Assume the players draw cards to pick the chooser and Henry wins. Then Fred and Gloria must first do the Divider/Chooser method on the cake. They draw cards again and Fred is chosen as the divider. He needs to divide the cake into two pieces each worth $18. There are many possible ways for Fred to cut the cake. Let’s assume he cuts it as shown in figure 8.3.2.

    6hMFl_E6PLI7MKZZerpPL36VILdoZ6eAhfxYot9X2M3xiXWo3nQ6rZVhJzEQhlmhHNh1pYeh6Wg5xXPG8JQdHBIEdk1tx667i6fMLnDSZ0ybyszirobLSz3QUJg9HucG5tABDGw
    Figure \(\PageIndex{2}\): How Fred Cuts the Cake

    Before Gloria can choose a piece, she must find the value of each of Fred’s pieces in her value system.

    H8XYTLZl0yszQ9TbGLWFr_M9TcgCT3YqdqtkZA05GiLZI1Ifzw1nNOBRpZo1-R1KnmFrFBXa-DGXMBsMqOHOyLdpU2XdCyAY-I1cwk9GuHC2NzWZd6UcKdCDEfCYFme6YlVwudM
    Figure \(\PageIndex{3}\): How Gloria Sees the Cake

    Gloria should choose the strawberry and chocolate piece since it has the highest value to her.

    Now Fred and Gloria independently divide their pieces into three pieces of equal value in their respective value systems. Remember that each flavor of cake takes up 120◦. Since Fred sees the chocolate as having no value, he needs to divide the 120◦ of vanilla into three equal pieces of 40◦, each worth $6. One of the pieces will also include the 60◦ of chocolate. Gloria’s piece is worth a total of $24 so she needs to divide it into three pieces each worth $8. Start with the Strawberry part. tHrXyOONAqcYSiNNIz98ehJfgqT5pl-CTGqVb0DHvw2e-fVMiQJHlsBNHLAjdfpK4xWukwaPs-p32JZh7FBWg0ouGbi6ECIYtNPcpiZsZ5KwDAOZDYhawTMmVeP0FvBasVAHCyI She cuts two pieces of strawberry that have an angle of 53.3◦. The remaining amount of strawberry is TWgjIDoMUv67hNOPxXAgBiOn2oh4wQ0j_p_kT9g8sGkSGgOzOBZRauiSuBu7YrrNl5BjlSROXWz1yMif5hlukZ-aI4UUGGKv8yBjJ2crLOm7oboeg1DwBSEf9UTJc5pOeRhLt0g Combining the small strawberry piece with the chocolate piece gives a larger piece with an angle of 73.4◦ and a value of vP15Tu89PqeejbnNiqHKFxbBYeIf5mhNG26beEIEfVToWCPSQmrhG-ODHLtrn8_DahGm-O1Cdv-xDMpF9KfsEUp9Hts1XOLj4AmUgEDYVhP8LL1lCY6lDbsAf58n7mQG8NEDc4A The pieces are numbered for convenience.

    kYIK9f8b55zNAAlIrW-r3JvqKD_sePr17hUNO0J4Q-jjllrUFpCzGn8-lGGCrhhloZgw3q666bO4lqoPRocXVQC0dvpobsDJqW3tGT2oF8NJO-cD0yxBE_ZYI8t6BK7-42QK2t0
    Figure \(\PageIndex{4}\): The Subdivisions of Each Piece

    At this point, Harvey joins the game. He assigns a value to each piece according to his value system.

    -MU7H34gtS9nK0Bzud6DybgZhla0OXeFHB7dltuGfdEalGWH1Uy-atmYIKkDzNe66xQtCX49DHeJB4Umo6BO7_ZaYf-19nNBMjh2qLaVaDHtNQFUXns1nV7Qtqv80-TtBHMmEv4
    Figure \(\PageIndex{5}\): How Harvey Sees the Pieces

    Harvey should pick the piece from each divider that he sees as having the greatest value. Fred and Gloria each keep the two pieces from their parts that Harvey does not pick. That gives each of the players two of the six pieces.

    Now let’s look at the final division. Harvey gets pieces 3 and 4 for a total value of $16.00, more than a fair share to him. Gloria gets pieces 5 and 6 for a total value of $16, more than a fair share to her. The original divider Fred gets pieces 1 and 2 for a total value of $12, exactly a fair share to him.

    Last Diminisher Method

    The Last Diminisher method can be used to divide a continuous object among many players. The concept is fairly simple. A player cuts a piece of the object. Each of the other players gets to decide if the piece is a fair share or too big. Any player that thinks the piece is too big can cut it smaller (diminish it). The key to the method is that the last person to cut/diminish a piece has to keep it.

    Summary of the Last Diminisher method

    1. The players use a random method to choose an order. The players continue in the same order throughout the game. Call the n players, in order, P1, P2, P3, …, Pn.
    2. Player 1 cuts a piece he/she considers a fair share. In order, each of the remaining players either passes (says the piece is a fair share and P1 can have it) or diminishes the piece. The last player to diminish the piece keeps the piece and leaves the game. If no one diminishes the piece, P1 keeps the piece and leaves the game.
    3. The lowest numbered player still in the game cuts a piece of the object. Each of the remaining players can either pass or diminish the piece. The last player to cut/diminish the piece keeps it and leaves the game.
    4. Repeat step three until only two players remain. These players use the Divider/Chooser method to finish the game.
    Example \(\PageIndex{2}\): Last Diminisher Method, #1

    Suppose six players want to divide a piece of land using the Last Diminisher method. They draw cards to choose an order. Assume the players in order are denoted P1, P2, P3, P4, P5, and P6.

    In round one, P1 cuts a piece by drawing lines on a map of the land. Assume players P2 through P6 pass on the piece. Since P1 is the last player to cut or diminish the piece, P1 keeps the piece and leaves the game.

    In round two, players P2 through P6 remain in the game. P2 is the lowest numbered player so P2 cuts a piece of the land. Assume that P3 and P4 pass on the piece. P5 thinks that the piece is more than a fair share so P5 diminishes the piece by redrawing the lines on the map to make the piece smaller. Assume P6 passes. Since P5 is the last player to diminish the piece, P5 keeps the piece and leaves the game.

    In round three, players P2, P3, P4, and P6 remain in the game. P2 is still the lowest numbered player so P2 cuts a piece of the land. This time assume that P3 diminishes, P4 passes and P6 diminishes the piece. P6 keeps the piece and leaves the game.

    In round four, players P2, P3, and P4 remain in the game. Once again P2 cuts a piece. Assume both P3 and P4 pass on the piece. P2 keeps the piece and leaves the game.

    For round five, since only P3 and P4 are left, they do Divider/Chooser on the remaining land. P3 cuts and P4 chooses.

    Example \(\PageIndex{3}\): Last Diminisher Method, #2

    Eight players want to divide a piece of land using the Last Diminisher method. They draw straws to determine an order. Assume the players in order are denoted by P1, P2, P3, P4, P5, P6, P7, and P8. P3 and P5 are the only diminishers in round one. No one diminishes in rounds two, three and six. P8 is the only diminisher in round four. Both P4 and P6 diminish in round five. Describe the fair division round by round.

    Round one: P1 cuts a piece, P2 passes, P3 diminishes, P4 passes, P5 diminishes, P6, P7, and P8 pass. P5 is the last diminisher so P5 keeps the piece and leaves the game.

    Round two: P1 cuts a piece and everyone else passes so P1 keeps the piece and leaves the game.

    Round three: P2 is now the lowest numbered player so P2 cuts a piece. Everyone else passes so P2 keeps the piece and leaves the game.

    Round four: P3 is now the lowest numbered player so P3 cuts a piece. P4, P6, and P7 pass. P8 diminishes the piece making P8 the last diminisher so P8 keeps the piece and leaves the game.

    Round five: P3 is still the lowest numbered player so P3 cuts a piece. P4 and P6 both diminish the piece but P7 passes. P6 is the last diminisher so P6 keeps the piece and leaves the game.

    Round six: P3 cuts a piece again and everyone else passes, so P3 keeps the piece and leaves the game.

    Round seven: P4 and P7 are the only players left so they use the Divider/Chooser method to divide the remaining land. P4 divides and P7 chooses.