Calculating the Shapley value in a weighted voting game.

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

• Joriki, please refer to the edit in my question. Thanks. – Mati Bot Sep 19 '12 at 14:06
• Thanks for the quick response. I've removed my comments and downvote (and in fact upvoted). I doubt that there's a nice closed form, though. The result will depend on the quota if the quota is close to $0$ or $a+2b$, but for small $b$ there will be a plateau of intermediate quota where the result doesn't change with the quota. – joriki Sep 19 '12 at 14:23