What if there is more than two players?

we'll look at three different solutions:

1. The lone-divider method

2. The lone-chooser method

3. The last-diminisher method

One person is the divider, chosen randomly. The remaining players "bid" on which of the pieces cut by the divider are fair shares. The pieces are then separated into two groups: the "bid for" and the "unbid for"

Case 1: The "bid for" group has two or more pieces in it. Give each

chooser the piece they bid on and the divider the last remaining piece.

Case 2: There is only one piece in the "bid for' group. Combine this one

piece with one of the "unbid for" pieces. The remaining piece goes to the divider. The new piece is then divided using the divider- chooser method.

1. The lone-divider method --

Case 1: The "bid for" group has two or more pieces in it. Give each

chooser the piece they bid on and the divider the last

remaining piece.

Bid For:

Un-bid For:

s1

s2

s3

Hector TheoLarry

Case 2: There is only one piece in the "bid for' group. Combine this one

piece with one of the "unbid for" pieces. The remaining piece goes to the divider. The new piece is then divided using the divider-chooser method.

Bid For:

Un-bid For:

s1 s2

s3

Hector Theo Larry

Hector Theo

Larry

s3

Bid For:

s1 Un-bid For:

s2

s 1 s 2 s 3

Hector 33.3% 33.3% 33.3%

Theo 35% 10% 55%

Larry 40% 25% 35%

Example #2

Larry

Theo

Hector

s 1 s 2 s 3

Theo 33.3% 33.3% 33.3%

Hector 20% 30% 50%

Larry 10% 20% 70%

Hector

Theo

Larry

1. The lone-divider method --

Example #3

Hector

Larry

Theo

In Honor of Grandpa: Land DivisionUsing the Lone-Divider MethodLand Value=$200,000

Hector

Larry

Theo

Lisa

fair share value=

Theo

$35K$60K

$50K

$55K

Hector

$10K $15K

$105K

$70K

Larry

$60K

$35K

$65K

$40K

Lisa

$50K$50K

$50K

$50K

Hector

Larry

Theo

Lisa

Homework: p. 99 #11, 12, 15, 19

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