Another (quick) question;

Let $T \subset N$ be a coalition. The unanimity game on $T$ is the game $(N, u_T)$ where $u_T(S)=1$ if $T \subset S$ and $u_T(S)=0$ if $T\S$. In other words, a coalition $S$ has worth $1$ (is winning) if it contains all players of $T$, and worth $0$ (is loosing) if this is not the case.
Calculate the core and the Shapley value for $(N, u_T)$

Then the core consists of $x_n-m \geq 0$ with $m \in [0,n-1]$

And then we know $x_n - 0 + x_n - 1 + \dots + xn - (n-1) = 1$ (efficiency)

So we could denote the core as $x_n - m + x_n - m' \geq 0$ + the efficiency

Am I right thinking the Shapley value should be

\begin{array}{|1|} \hline \frac{1}{n-1!} \cdot (\frac{1}{n},\frac{1}{n},\dots,\frac{1}{n})=(\frac{1}{n!},\frac{1}{n!},\dots,\frac{1}{n!}) \\\tag{1} \hline \end{array}

Is this ok? Thanks!


A payoff-vector is in the core, as long as it is not blocked by $T$, the only coalition that can block. We have $v(N)=v(T)=1$ and everything in which $\sum_{i\in T}x_1< 1$ can be blocked by $T$. On the other hand, if $\sum_{i\in T}x_i=1$, the only way to make a member of $T$ better off is by making another member of $T$ worse off. So everything that gives the whole pie to $T$ is in the core.

Everyone outside $T$ is a null-player and should therefore get a value of zero. By symmetry, everyone in $T$ gets the same value. By efficiency, these values should add up to $v(N)$. Hence, $v(i)=v(N)/|T|$ for all $i\in T$.

  • $\begingroup$ just to be sure; in a game with n players, the shapley value would be n-1!/n!=1/n!? $\endgroup$ – Bob Dec 10 '12 at 17:02
  • $\begingroup$ @Bob That depends on the game, obviously. $\endgroup$ – Michael Greinecker Dec 10 '12 at 17:10
  • $\begingroup$ same restrictions as here.. only now we have n players. $\endgroup$ – Bob Dec 10 '12 at 18:08
  • $\begingroup$ As I said, every player outside $T$ is a dummy player and will get a value of zero. So the answer is no. Also, by the symmetry axiom, the sum of all values has to add up to $v(N)$, so there is no game in which every player gets a value of $1/n!$ if $n>2$. $\endgroup$ – Michael Greinecker Dec 10 '12 at 18:44

Shapley value: $s_{i}=\begin{cases} \begin{array}{c} \frac{1}{k}\: \: i\in T\\ 0\: \: i\notin T \end{array}\end{cases} $ , $k=|T|>0$

Core: $C=\{x\in\mathbb{R}_{+0}^{n}:\sum_{i}x_{i}=1\:\&\:\forall_{i}(i\notin T\Rightarrow x_{i}=0)\}$, i.e set of all probability distributions on $T$

Shapley value: from equivalent "probabilistic" definition, Shapley value of a game $v$ is the average of the marginal vectors of the game. For a random order of the set of players, player $i$ is a last member of $T$ with probability $\frac{1}{k}$ if $i\in T$ and with probability $0$ if $i\notin T$.

Core: For a set $S$ and probability distribution $x$ over $T$, $$ \sum_{S}x_{i}=\sum_{S\cap T}x_{i}\begin{cases} \begin{array}{cc} =1 & \: T\subseteq S\\ \geqq0 & \: T\nsubseteq S \end{array}\end{cases} $$


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