Question

Find the shapely shubik power of 11, 11, 6, 2, 1 and B 12, 11, 6, 2, 1

Answer 1

We have:

[11: 11, 6, 2, 1] there are 4! = 1 x 2 x 3 x 4 = 24, then sequencial coalitions is:

`\begin{gathered} \lbrack P1,P2,P3,P4\rbrack,\lbrack P2,\bar{P1},P3,P4\rbrack,\lbrack P3,\bar{P1},P2,P4\rbrack,\lbrack P4,\bar{P1},P2,P3\rbrack^{} \\ \lbrack P1,P2,P4,P3\rbrack,\lbrack P2,\bar{P1},P4,P3\rbrack^{},\lbrack P3,\bar{P1},P4,P2\rbrack,\lbrack P4,\bar{P1},P3,P2\rbrack^{} \\ \lbrack P1,P3,P2,P4\rbrack,\lbrack P2,P3,\bar{P1},P4\rbrack^{},\lbrack P3,P2,\bar{P1},P4\rbrack,\lbrack P4,P2,\bar{P1,}P3\rbrack^{} \\ \lbrack P1,P3,P4,P2\rbrack,\lbrack P2,P3,P4,P1\rbrack^{},\lbrack P3,P2,P4,P1\rbrack,\lbrack P4,P2,P3,P1\rbrack^{} \\ \lbrack P1,P4,P2,P3\rbrack,\lbrack P2,P4,\bar{P1},P3\rbrack^{},\lbrack P3,P4,\bar{P1,}P2\rbrack,\lbrack P4,P3,\bar{P1},P2\rbrack^{} \\ \lbrack P1,P4,P3,P2\rbrack,\lbrack P2,P4,P3,P1\rbrack,\lbrack P3,P4,P2,P1\rbrack,\lbrack P4,P3,P2,P1\rbrack \end{gathered}`

the shapley shubik power distribution:

`P1=(12)/(24)=(1)/(2)`