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!


1 Answer 1


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*}


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