Shapley dalybu pavyzdziai

Triju berniuku pavyzdy
$S_1=\{\{s_1,s_2\},\{s_1,s_3\},S_2=\{s_2,s_3\}, \{s_1,s_2\},S_3=\{s_2,s_3\},\{s_1,s_3\}\}$.
Iš (197)(189) Shapley dalybos

\begin{eqnarray}\phi_i[v]=
(2-1)!(3-2)!/3! +\nonumber \\ (2-1)!(3-2)!/3!=1/3.
\end{eqnarray}


UAB pavyzdy
$S_1=\{1,2,3\}$, $S_2=\{\{1,2,3\},\{2,4\},\{1,2,4\}\}$,
$S_3=\{\{1,2,3\}, \{3,4\},\{1,3,4\}\}$,
$S_4=\{\{2,4\},\{1,2,4\},\{1,3,4\},\{2,3,4\}\}$.
I(197)(192)

\begin{eqnarray}\phi_1[v]=
(2-1)!(4-1)!/4!=1/12, \nonumber \\
\phi_2[v]= 2 (3-1...
...= 3 (3-1)!(4-3)!/4!+\nonumber \\ 2 (2-1)!(4-2)!/4!=5/12.\nonumber
\end{eqnarray}




jonas mockus 2004-03-01