unanimity game, calculate core/shapley value 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 $i\in T$.