Given a special case of WVG (Weighted Voting Game) of $a$ 1s and $b$ 2s and a quota q, $ [q:1,1,1,1..1,2,2,..2] $. I need help with calculating the Shapley value of a player with a weight of $2$ and a player with a weight of $1$ as a function of $a$,$b$ and $q$.

I know how to calculate the Shapely value in general, but I would like to get a simple closed form for the result. I tried to think about what happens if there are only players with weights of $1$ (simple scenario $ a=2k-1,b=0,q=k $ I think you will agree that $ \phi_{1}(v)=\frac{1}{a}=\frac{1}{2k-1} $. the simple calculation is $ \binom{2k-2}{k-1}(k-1)!(k-1)!=\frac{1}{2k-1}=\frac{1}{a} $ which is $ \frac{1}{2k-1} $. The calculation for a player of weight 1 (when there are 1s and 2s) would be a sum of choosing i players of weight 1 and $ \frac{q-i}{2} $ players of weight 2. similar calculation for a player of weight 2.

I tagged binomial coefficients because I thought they could be useful for counting the occurrences.

Thanks in advance, Mati

also posted here: http://math.stackexchange.com/posts/199080

Thanks in advance, Mati