please check my working

I think i have done most the work can someone please check my working

The town of Schaerbeek has a bicameral Council. The Upper House has two mem-
bers, and the Lower House has ve members. To pas a bill, unanimity is needed in
the Upper House, and a majority is needed in the Lower House (there is no veto).
We may treat this problem as a 7-person cooperative game with N = f1; 2; : : : ; 7g,
and suppose that the rst two persons are the two Upper House members. Dene
a coalition S ⊆ N to be winning, that is, v(S) = 1, if it can pass a bill, otherwise
v(S) = 0.
You can decompose any coalition S as S = U [ L, where U is a subcoalition
containing members from the Upper House, and L is a subcoalition containing
members from the Lower House (L; U possibly empty).
(a) Describe the value of a coalition S in terms of the cardinality of U and L.
(b) Compute the power of each member of each House. To this end, compute the
Shapley value of the game using marginal vectors.


a) v(s) = { 1 if |U|=2 and |L| > 2
{0 otherwise

b) The role of the upper house members are symmetric. As are the roles of the lower house members.
Then by efficiency and symmetry
5*(shapley value for lower house members) + 2*(shapley value for upper house members) = 1
So we only need to calculate the shapley value for either a lower house member or and upper house member

I tried to calculate both to make sure they were equal but I got different results. I think the problem is with the combinations I used.

Consider the marginal contribution of a lower house member (call him P3)

v(S u {3}) - v(S) = { 1, if |U| = 2 and |L| = 2
{ 0, otherwise

There are 2! combinations such that |U| = 2
(ie the first member of the upper house is member 1 and the second member is 2 or the first member is 2 and the second member is 1)

There are (4 C 2) combinations such that |L| = 2 and 2! ways of arranging these. That is, there are 4 members (other than P3 who we are considering) and we need to chose two of them to be in our coalition. Then there are 2! ways of arranging the remaining two lower house members not in the coalition.

Therefore the shapley value of P3 = (1/7!)*(4 C 2)*2!*2!= 1/105

What you're missing is the fact that when we pick the 2 members for U and the 2 members for L it is not 2!*2!.
It should be 4! since we can pick these 4 in any order before we pick P3.
Took me a while to figure that out.
Also i'm 90% sure you are in my decision making class haha

But it says in the question that the first two members of the coalition are the upper house members. Doesn't that imply they are fixed before hand?