= Solution
Let $v:2^N\to\mathbb R$, $v(\varnothing)=0$, be a <coalitional game> with $n=|N|$. The <Shapley value> is
$$
\phi_i(N,v)=\sum_{S\subseteq N\setminus\{i\}}
\frac{|S|!(n-|S|-1)!}{n!}\bigl(v(S\cup\{i\})-v(S)\bigr).
$$
Equivalently, choose a uniformly random player ordering and average each player's <marginal contribution> when that player joins its predecessors. The coefficient counts precisely the orderings with predecessor set $S$.
The standard axioms are efficiency, $\sum_i\phi_i=v(N)$; symmetry, giving equal values to players with identical <marginal contributions>; the dummy-player property, giving $\phi_i=v(\{i\})$ if every <marginal contribution> of $i$ equals $v(\{i\})$; and additivity, $\phi_i(v+w)=\phi_i(v)+\phi_i(w)$. The null-player form of the dummy axiom gives zero to a player contributing nothing. These properties characterize the <Shapley value>. They also follow directly from its formula: efficiency telescopes along each ordering, symmetry relabels orderings, constant dummy contributions average to themselves, and the sum is linear in $v$.
For <balanced contributions of Shapley values>, remove player $j$ from a uniformly random ordering. The resulting ordering of $N\setminus\{j\}$ is uniform, so it computes $i$'s <Shapley value> in the restricted game. If $j$ comes after $i$, removing it does not change $i$'s <marginal contribution>. If $j$ precedes $i$, and $S$ consists of $i$'s other predecessors, the difference is
$$
\Delta_{ij}v(S)=v(S\cup\{i,j\})-v(S\cup\{j\})-v(S\cup\{i\})+v(S).
$$
There are $(|S|+1)!(n-|S|-2)!$ full orderings with exactly $S\cup\{j\}$ before $i$. Consequently
$$
\phi_i(N,v)-\phi_i(N\setminus\{j\},v)
=\sum_{S\subseteq N\setminus\{i,j\}}
\frac{(|S|+1)!(n-|S|-2)!}{n!}\Delta_{ij}v(S).
$$
Both the summand and coefficient are unchanged when $i,j$ are interchanged. Thus
$$
\boxed{\phi_i(N,v)-\phi_i(N\setminus\{j\},v)
=\phi_j(N,v)-\phi_j(N\setminus\{i\},v)}.
$$
The restriction keeps all <coalition> values for the remaining players, exactly matching the payoff comparison after a departure. The identity does not require the differences to be positive.
Now consider the <coverage coalitional game> formed by the finite book sets. For disjoint <coalitions> $S,T$, put $B(S)=\bigcup_{i\in S}B_i$. Then
$$
v(S\cup T)=v(S)+v(T)-|B(S)\cap B(T)|\leq v(S)+v(T).
$$
Thus coverage is generally subadditive. It is a <superadditive coalitional game> precisely when the book sets are pairwise disjoint: disjointness makes the displayed formula additive, while a common book of players $i,j$ gives $v(\{i,j\})<v(\{i\})+v(\{j\})$. For example, $B_1=B_2=\{b\}$ gives values $1,1,1$ and disproves unconditional superadditivity. The printed clause must therefore have its disjointness condition apply to superadditivity as well as to <core> nonemptiness.
For the <core of a coverage game>, an allocation must satisfy efficiency $\sum_i x_i=v(N)$ and $\sum_{i\in S}x_i\geq v(S)$ for every <coalition>. In particular $x_i\geq|B_i|$. Summing gives
$$
\left|\bigcup_iB_i\right|=\sum_ix_i\geq\sum_i|B_i|.
$$
The reverse inequality always holds, with equality exactly when no book appears in two sets. Therefore overlap makes the <core> empty. If the sets are disjoint, $x_i=|B_i|$ gives every <coalition> exactly its value and is a <core> allocation; the singleton lower bounds and efficiency make it the only one. Hence \b[superadditivity and a nonempty core each occur exactly in the pairwise-disjoint case].
Finally write the game as a sum over individual books. For a book $b$, let $K_b=\{k:b\in B_k\}$. Its contribution to <coalition> value is one if the <coalition> meets $K_b$, and zero otherwise. In a random ordering, player $i$ contributes this book precisely when $i\in K_b$ and is earliest among its $|K_b|$ knowers. Each knower is equally likely to be earliest. Additivity of the <Shapley value> therefore gives the <Shapley value of a coverage game>
$$
\boxed{\phi_i(v)=\sum_{b\in B_i}\frac1{|K_b|}}.
$$
This formula remains valid when books overlap and the <core> is empty.
Back to article page
