Alice and Bob share a taxi:
- It costs Alice £5 to get home.
- It costs Bob £12 to get home (this will be the total fare).
- Alice lives on Bob's way home.
When Alice gets out of the cab, how much should she pay?
Cooperative Games
- N=\{1,\dots,n\} players.
- Coalitions S\subseteq N.
Examples:
-
S_1=\{2,5\}
-
S_2=\{1\}
-
S_2=N
-
S_2=\emptyset
Characteristic Function Games
A characteristic game G is given by a pair: (N,v). Where:
v:2^{N}\to\mathbb{R}
is a characteristic function which maps each coalition S\subseteq N to a real number v(S).
Taxi Problem
N=\{A,B\}
and we have:
-
v(\emptyset)=0
-
v(A)=5
-
v(B)=12
-
v(A,B)=12
Monotone Games
A characteristic function game G=(N,v) is said to be monotone if it satisfies v(S_2)\geq v(S_1) for all pairs S_1\subseteq S_2.
Superadditive Games
A characteristic function game G=(N,v) is said to be superadditive if it satisfies v(S_1\cup S_2)\geq v(S_1)+v(S_2) for all pairs S_1,S_2\subseteq N.
Solution Concepts
The 'solution' of a superadditive game corresponds to a vector: x\in\mathbb{R}^{|N|}_{\geq0} that satisfies:
\sum_{i\in N}x_i=v(N)
Fair Distribution
Efficiency: \sum_{i\in N}x_i=v(N);
Null player: if i does not contribute, then x_i=0;
Symmetry: if i and j are symmetric in then x_i=x_j;
Additivity: (look it up if you're interested).
Taxi Problem
\pi
Alice
Bob
(A,B)
5
7
(B,A)
0
12
\phi
2.5
9.5
Shapley Value
Given a characteristic function game G=(N,v) the Shapley value of a player i\in N is given by:
\phi_{i}(G)={1\over |N|!}\sum_{\pi\in\Pi_N}\Delta_{\pi}^{G}(i)
Why am I telling you this?
Applying this to group work
- Assume Amy, Bernard and Caroline work on a group project.
- Assume the marking criteria states: 70% output and 30% group work.
- Assume the total mark given to the project is 85%.
Contributions
S
v(S)
A
40
B
40
C
20
\{A,B\}
70
\{A,C\}
60
\{B,C\}
40
Shapley Value Calculation
\pi
Amy
Bernard
Caroline
(A,B,C)
40
30
30
(A,C,B)
40
40
20
(B,A,C)
30
40
30
(B,C,A)
60
40
0
(C,A,B)
40
40
20
(C,B,A)
60
20
20
\phi
45
35
20
Mark Calculation
\text{m}(i)=\text{M}\times\left(\text{OP}+\text{CW}\times{\phi(i)\over \max_j{\phi(j)}}\right)Fair Marks
Efficiency: only way to maximise group mark is equal contribution;
Null player: if i does not contribute, then m(i)=0;
Symmetry: if i and j contribute equally, they get the same mark.
\text{m}(A)=85\times(.7+.3)=85\text{m}(B)=85\times\left(.7+.3\times {35\over 45}\right)={238\over 3}\approx 79.33\text{m}(C)=85\times\left(.7+.3\times{20\over45}\right)={425\over 6}\approx 70.83The real problem...
How do we get \phi?