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Fix distinct players $i,j$ and write $Ti=T\cup \{i\}$, with analogous notation for $Tj$ and $Tij$. Splitting sampled coalitions according to whether they contain the other player givesSubtracting yields the Banzhaf difference identity,In a weighted voting game, $w_j>w_i$ means that $Tj$ has greater total weight than $Ti$. Thus $v(Tj)\ge v(Ti)$ for every $T$: if the lighter coalition reaches the quota, the heavier one does too. Every summand is nonnegative, proving ${\beta }_i(v)\le {\beta }_j(v)$. Equality remains possible even when the weights differ.

The probabilistic Banzhaf index satisfies cooperative-game symmetry, the null player axiom and cooperative-game additivity, but not cooperative-game efficiency.

For cooperative-game symmetry, interchangeable players have $v(Ti)=v(Tj)$ for every $T$ omitting them. The Banzhaf difference identity therefore gives equal indices. For a null player, every marginal contribution in the defining average is zero, so its index is zero.

For cooperative-game additivity, each marginal contribution is additive:Averaging both sides over the same $2^{n-1}$ coalitions gives ${\beta }_i(v+w)={\beta }_i(v)+{\beta }_i(w)$.

For failure of cooperative-game efficiency, the indices computed in part (b) sum to $5/4$, whereas the grand-coalition value is one. This is already a counterexample. Independently, a three-player majority game gives each player index $1/2$, also exceeding the total value when summed. Pivotality averaged over independently chosen coalitions does not telescope in the way the ordering-average Shapley value does.