Fair Shares

Cake Cutting Cake Cutting

Solution: One cuts, the other chooses

What is the goal? Envy Free Proportional Share Both players think they get at least half the cake Envy Free Both players like their slice at least as much as the other slice

Three Players (A, B, C) Envy Free Proportional Share Every players think they get at least one third of the cake Envy Free Each players like their slice at least as much as any other slice

A cuts into 3 (presumably) equal pieces Three Players (A, B, C) A cuts into 3 (presumably) equal pieces Now what?

If preferred pieces: Then assign: B C B C A What if: B/C

Record first and second preferences for B and C: B, C C, B B C

B, C C, B B C A fills in the gaps: B, C C, B A, A B,A C,A

Split each piece using “one cuts other chooses”: A fills in the gaps: B, C C, B A, A B,A C,A Split each piece using “one cuts other chooses”: B A C

Envy Free Solution A cuts: B: top two equal? Then: C chooses first, B chooses second, A gets last piece. C B A

Envy Free Solution A cuts: B: top two not equal. Then: 1. B trims largest piece to make its top two equal. 2. Set trim, T aside. 3. C chooses favorite piece. 4. B chooses, but must take trimmed piece if available. 5. A gets last piece. 6. Partition T. A C B T

Envy Free Solution A cuts: B: top two not equal. 6. Partition T: Let D be person getting trimmed piece (one of B or C). Let E be other of B and C. a. E divides T into 3 (presumably equal) parts. b. D chooses favorite part of T. c. A chooses remaining favorite part. d. E gets third part. A E D T

Cake Cutting Cake Cutting

Stable Matching Match medical residents to hospitals Match men to women Match students to schools

Unstable Pair a b C D The match of a to C and b to D is unstable if both b and C prefer each other to their current partners Definition: a stable match has no unstable pairs.

Algorithm to find a stable match Preference order A: acb a: BAC B: acb b: CBA C: cab c: ABC Capital letters (women) repeatedly propose to next best man (small letters) as needed a: b: c: a: A B b: c: C a: B b: c: C A a: B C b: c: A a: B b: C c: A

Why Stable? a: A B b: c: C a: B c: C A a: B C c: A b: C Unstable instance: Little letters (men) only improve their match over time. Capital letters (women) switch to a lower rank man only when rejected by higher ranked men. Thus b prefers C to D only if he never rejected C. But then C never approached b and so must prefer a. Therefore not unstable. a b C D

Does it run forever? No! a: A B b: c: C a: B c: C A a: B C c: A b: C Once a little letter (man) has a suitor, he will always have a suitor. A woman leaves a man only if he has another suitor. If she reaches her bottom ranked man, it follows that all other men have suitors; hence there is no competition on the last man. i.e. every woman gets matched.

How many rounds can the algorithm take?

Quality of Matches Preference order A: cba a: ACB B: acb b: BAC C: bac c: CBA Match algorithm result: a: B b: C c: A Algorithm result with reversed roles: A: a B: b C: c Best result for women Best result for men (capital letters)

Fair Division of Property among Two People

What is fair? Divide evenly what is desired by both A’s wants: No Yes Yes Yes B’s wants: Yes No Yes Yes

What is fair? Give equal value A: 0 60 40 0 B: 40 20 20 20 Sort by: A value / B value: A: 60 40 0 0 B: 20 20 20 40

What is fair? Give equal value A: 0 60 40 0 B: 40 20 20 20 Sort by: A value / B value: A: 60 40 0 0 B: 20 20 20 40 1/3 2/3

Cheating is Possible True values: A: 60 40 0 0 B: 20 20 20 40 1/3 2/3 A misleads: A: 21 21 19 39 B: 20 20 20 40

Can be too clever True values: A: 60 40 0 0 B: 20 20 20 40 1/3 2/3 A not so wise: A: 41 21 19 19 B: 40 20 20 20

“Market” Solution Give each player 100 units of money Find prices so that each player gets the same value per unit money for the items she receives A: 60 40 0 0 B: 20 20 20 40 Prices: 60 40 100/3 200/3

Comparison of Results A: 60 40 0 0 B: 20 20 20 40 Rules matter! Values with misleading A: 60 40 0 0 B: 20 20 20 40 Using values prices Rules matter!