= Solution
Where its denominator is nonzero, normalization enforces \b[<cooperative-game efficiency>] and preserves \b[<cooperative-game symmetry> and the <null player> axiom]. The violated axiom is \b[<cooperative-game additivity>].
For an exact counterexample on three players, let $v$ be the majority game, and let $w$ be the dictatorship in which a <coalition> wins precisely when it contains player one. Their <probabilistic Banzhaf indices> are
$$
\beta(v)=(1/2,1/2,1/2),\qquad\beta(w)=(1,0,0),
$$
with $v(N)=w(N)=1$. Hence
$$
\beta'(v)=(1/3,1/3,1/3),\qquad\beta'(w)=(1,0,0).
$$
Unnormalized indices add, so $\beta(v+w)=(3/2,1/2,1/2)$. Their <sum> is $5/2$, while $(v+w)(N)=2$. Therefore
$$
\boxed{\beta'(v+w)=(6/5,2/5,2/5)\ne(4/3,1/3,1/3)=\beta'(v)+\beta'(w).}
$$
The sum game is a valid <transferable utility game>, although it is not binary; that is the natural domain of the additivity axiom. The failure comes from a game-dependent scaling factor. A zero denominator, as for the zero game, makes the displayed normalization undefined unless a separate convention is supplied; none of the counterexample's denominators vanish.
Back to article page
