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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!

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1 Answer 1

up vote 2 down vote accepted

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 $i\in T$.

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just to be sure; in a game with n players, the shapley value would be n-1!/n!=1/n!? –  Bob Dec 10 '12 at 17:02
    
@Bob That depends on the game, obviously. –  Michael Greinecker Dec 10 '12 at 17:10
    
same restrictions as here.. only now we have n players. –  Bob Dec 10 '12 at 18:08
    
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 $n>2$. –  Michael Greinecker Dec 10 '12 at 18:44

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