Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 42 / 5 / b / Solution

Put $w(S)=|S|!(n-|S|-1)!/n!$ for $S\subseteq N\setminus \{i\}$. Under the bijection $S\mapsto R=(N\setminus \{i\})\setminus S$, the weights satisfy $w(S)=w(R)$ and $N\setminus S=R\cup \{i\}$. ConsequentlySubtracting ${\sum }_{S}w(S)v(S)$ proves the alternative Shapley value expressionFor the dual coalitional game $v^{\prime }(S)=v(N)-v(N\setminus S)$, its marginal contribution at predecessor set $S$ isThe same weight-preserving complement bijection therefore gives ${\varphi }_i(v^{\prime })={\varphi }_i(v)$, and both equal the displayed expression. The Shapley value is invariant under coalitional duality. This Shapley self-duality calculation complements within $N\setminus \{i\}$, not within $N$ without accounting for the distinguished player.