# 5.4: Lone Divider

- Page ID
- 34200

The lone divider method works for any number of parties – we will use *N* for the number of parties. One participant is randomly designated the divider, and the rest of the participants are designated as choosers.

The Lone Divider method proceeds as follows:

1) The divider divides the item into \(N\) pieces, which we’ll label \(S_{1}, S_{2}, \ldots, S_{N}\).

2) Each of the choosers will separately list which pieces they consider to be a fair share. This is called their **declaration**, or **bid**.

3) The lists are examined. There are two possibilities:

a. If it is possible to give each party a piece they declared then do so, and the divider gets the remaining piece.

b. If two or more parties both want the same pieces and no others, then give a non-contested piece to the divider. The rest of the pieces are combined and repeat the entire procedure with the remaining parties. If there are only two parties left, they can use divider-chooser.

Consider the example from earlier, in which the pieces were valued as:

\(\begin{array}{|l|l|l|l|}

\hline & \textbf { Piece 1 } & \textbf { Piece 2 } & \textbf { Piece 3 } \\

\hline \textbf { Chooser 1 } & 40 \% & 30 \% & 30 \% \\

\hline \textbf { Chooser 2 } & 45 \% & 30 \% & 25 \% \\

\hline \textbf { Divider } & 33.3 \% & 33.3 \% & 33.3 \% \\

\hline

\end{array}\)

**Solution**

Each chooser makes a declaration of which pieces they value as a fair share. In this case,

Chooser 1 would make the declaration: Piece 1

Chooser 2 would make the declaration: Piece 1

Since both choosers want the same piece, we cannot immediately allocate the pieces. The lone divider method specifies that we give a non-contested piece to the divider. Both pieces 2 and 3 are uncontested, so we flip a coin and give Piece 2 to the divider. Piece 1 and 3 are then recombined to make a piece worth 70% to Chooser 1, and 70% to Chooser 2. Since there are only two players left, they can divide the recombined pieces using divider-chooser. Each is guaranteed a piece they value as at least 35%, which is a fair share.

Use the Lone Divider method to complete the fair division given the values below.

\(\begin{array}{|l|l|l|l|}

\hline & \textbf { Piece 1 } & \textbf { Piece 2 } & \textbf { Piece 3 } \\

\hline \textbf { Chooser 1 } & 40 \% & 30 \% & 30 \% \\

\hline \textbf { Chooser 2 } & 40 \% & 35 \% & 25 \% \\

\hline \textbf { Divider } & 33.3 \% & 33.3 \% & 33.3 \% \\

\hline

\end{array}\)

**Answer**-
Chooser 1 would make the declaration: Piece 1

Chooser 2 would make the declaration: Piece 1, Piece 2

We can immediately allocate the pieces, giving Piece 2 to Chooser 2, Piece 1 to Chooser 1, and Piece 3 to the Divider. All players receive a piece they value as a fair share.

Suppose that Abby, Brian, Chris, and Dorian are dividing a plot of land. Dorian was selected to be the divider through a coin toss. Each person’s valuation of each piece is shown below.

\(\begin{array}{|l|l|l|l|l|}

\hline & \textbf { Piece 1 } & \textbf { Piece 2 } & \textbf { Piece 3 } & \textbf { Piece 4 } \\

\hline \textbf { Abby } & 15 \% & 30 \% & 20 \% & 35 \% \\

\hline \textbf { Brian } & 30 \% & 35 \% & 10 \% & 25 \% \\

\hline \textbf { Chris } & 20 \% & 45 \% & 20 \% & 15 \% \\

\hline \textbf { Dorian } & 25 \% & 25 \% & 25 \% & 25 \% \\

\hline

\end{array}\)

**Solution**

Based on this, their declarations should be:

Abby: Piece 2, Piece 4

Brian: Piece 1, Piece 2, Piece 4

Chris: Piece 2

This case can be settled simply – by awarding Piece 2 to Chris, Piece 4 to Abby, Piece 1 to Brian, and Piece 3 to Dorian. Each person receives a piece that they value as at least a fair share (25% value).

Suppose the valuations in the previous problem were:

\(\begin{array}{|l|l|l|l|l|}

\hline & \textbf { Piece 1 } & \textbf { Piece 2 } & \textbf { Piece 3 } & \textbf { Piece 4 } \\

\hline \textbf { Abby } & 15 \% & 30 \% & 20 \% & 35 \% \\

\hline \textbf { Brian } & 20 \% & 35 \% & 10 \% & 35 \% \\

\hline \textbf { Chris } & 20 \% & 45 \% & 20 \% & 15 \% \\

\hline \textbf { Dorian } & 25 \% & 25 \% & 25 \% & 25 \% \\

\hline

\end{array}\)

**Solution**

The declarations would be:

Abby: Piece 2, Piece 4

Brian: Piece 2, Piece 4

Chris: Piece 2

Notice in this case that there is no simple settlement. So, the piece no one else declared, Piece 3, is awarded to the original divider Dorian, and the procedure is repeated with the remaining three players.

Suppose that on the second round of this method Brian is selected to be the divider, three new pieces are cut, and the valuations are as follows:

\(\begin{array}{|l|l|l|l|}

\hline & \textbf { Piece 1 } & \textbf { Piece 2 } & \textbf { Piece 3 } \\

\hline \textbf { Abby } & 40 \% & 30 \% & 30 \% \\

\hline \textbf { Brian } & 33.3 \% & 33.3 \% & 33.3 \% \\

\hline \textbf { Chris } & 50 \% & 20 \% & 30 \% \\

\hline

\end{array}\)

The declarations here would be:

Abby: Piece 1

Chris: Piece 1

Once again we have a standoff. Brian can be awarded either of Piece 2 or Piece 3, and the remaining pieces can be recombined. Since there are only two players left, they can divide the remaining land using the basic divider-chooser method.

Four investors are dividing a piece of land valued at $320,000. One was chosen as the divider, and their values of the division (in thousands) are shown below. Who was the divider? Describe the outcome of the division.

\(\begin{array}{|l|l|l|l|l|}

\hline & \textbf { Piece 1 } & \textbf { Piece 2 } & \textbf { Piece 3 } & \textbf { Piece 4 } \\

\hline \textbf { Sonya } & \$ 90 & \$ 70 & \$ 80 & \$ 80 \\

\hline \textbf { Cesar } & \$ 80 & \$ 80 & \$ 80 & \$ 80 \\

\hline \textbf { Adrianna } & \$ 60 & \$ 70 & \$ 100 & \$ 90 \\

\hline \textbf { Raquel } & \$ 70 & \$ 50 & \$ 90 & \$ 110 \\

\hline

\end{array}\)

**Answer**-
A fair share would be $80,000. Cesar was the divider.

Their declarations would be:

Sonya: Pierce 1, Piece 3, Piece 4.

Adrianna: Piece 3, Piece 4.

Raquel: Piece 3, Piece 4.