# Shapley value in a weighted majority game.

I was given the following question:

We have a weighted majority game which consists of:

$$[\frac{1}{2};\alpha,\frac{1-\alpha}{n-1},...,\frac{1-\alpha}{n-1}]$$ (we have a total of $$n$$ players)

Where $$0<\alpha\leq\frac{1}{2}$$

I was asked to find the shapley value for the $$\alpha$$ player but I couldn't work my way into a solution.

I would love to see the solution for this, thanks!

This seems to me like a nice riddle. I guess the person who asked you knows the solution? I will give it a go.

Some definitions and observations up front:

• The set of all players $$i=1,2,\dots,n$$ is denoted by $$N$$.
• The vector of weights is defined by $$w^T=(\alpha, \frac{1-\alpha}{n-1},\dots, \frac{1-\alpha}{n-1})$$.
• The sum of all weights of a coalition $$S\subseteq N$$ is $$w(S)=\sum\limits_{i\in S}{w_i}$$. Note, that \begin{align*} w(N) &= \alpha + (n-1)\frac{1-\alpha}{n-1} = 1. \end{align*}
• $$W=\{S\subseteq N: w(S)\geq \frac{1}{2}\}$$ is the set of winning coalitions and $$\overline{W}=2^N\setminus W$$ is the set of losing coalitions.
• There are cases for $$\alpha$$ where $$w(S)=w(N\setminus S)=\frac{1}{2}$$.

The most fun part to me thinking about this question was the following observation: Let's say we have a weighted game $$[q;w_1,1,\dots,1]$$ where $$q, w_1$$ are whole numbers and $$w_1\leq q\leq n-1$$. How many coalitions $$S\subseteq \{2,3,.\dots,n\}$$ exist with weights $$q-w_1, q-w_1+1,\dots, q-1$$? These are the coaltions where player 1 is decisive. The answer is $$\binom{n-1}{q-w_1}, \binom{n-1}{q-w_1+1}\dots,\binom{n-1}{q-1}$$. In such kind of games the computation of the Shapley-Shubik power index of player 1 simplifies to: \begin{align*} \phi_1 &= \sum_{\substack{S\subseteq N,\\ S\in \overline{W}, \\ S\cup\{i\}\in W}} \frac{1}{\binom{n}{|S|}(n-|S|)} \\ &= \frac{\binom{n-1}{q-w_1}}{\binom{n}{q-w_1}(n-q+w_1)} +\frac{\binom{n-1}{q-w_1+1}}{\binom{n}{q-w_1+1}(n-q+w_1-1)} +\dots +\frac{\binom{n-1}{q-1}}{\binom{n}{q-1}(n-q+1)} \\ &= \frac{n-q+w_1}{n}\frac{1}{n-q+w_1} +\frac{n-q+w_1-1}{n}\frac{1}{n-q+w_1-1} +\dots +\frac{n-q+1}{n}\frac{1}{n-q+1} \\ &= \frac{w_1}{n}. \end{align*}

We need to distinguish between even and odd $$n$$. For each case, we can determine intervals for $$\alpha$$ where the game does not change and find a suitable game representation for each interval, where we can use the formula above to calculate the Shapley-Shubik power index. I will not post everything here (it is mostly technical and not exciting) and keep it short:

For $$n$$ even, if $$0 < \alpha < \frac{1}{n}$$ then $$\phi_1 = 0$$. Otherwise, for $$x\in \{1,2,\dots, \frac{n}{2}-1\}$$ \begin{align*} \phi_1 &= \begin{cases} \frac{2x}{n} & \frac{1+2(x-1)}{n+2(x-1)}< \alpha < \frac{1+2x}{n+2x} \\ \frac{2x+1}{n} & \alpha = \frac{1+2x}{n+2x}. \end{cases} \end{align*}

For $$n$$ odd, if $$0< \alpha < \frac{2}{n+1}$$ then $$\phi_1 = \frac{1}{n}$$. Otherwise, for $$x\in \{1,2,\dots, \frac{n-1}{2}\}$$ \begin{align*} \phi_1 &= \begin{cases} \frac{2x+1}{n} & \frac{2x}{n-1 + 2x} < \alpha < \frac{2(x+1)}{n-1 + 2(x+1)} \\ \frac{2x}{n} & \alpha = \frac{2x}{n-1 + 2x}. \end{cases} \end{align*}