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For distinct players $i,j$, expand each probabilistic Banzhaf index by whether the sampled coalition contains the other player. The terms $v(T)$ and $v(T\cup \{i,j\})$ cancel on subtraction, leaving the displayed identity. In a weighted voting game with $w_i\le w_j$, every summand is nonnegative, proving that the index is monotone in the player's weight. Equal weights give equal indices, although different weights can also have equal indices.

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.

Give each permanent member weight seven, each nonpermanent member weight one, and use the strict threshold $t=38$. If a coalition omits a permanent member, its weight is at most $4\cdot 7+10=38$, so it loses. If it includes all five permanent members and $r$ others, its weight is $35+r$, which exceeds 38 exactly when $r\ge 4$. These are precisely the required winning coalitions. Thus a weighted voting game representation isThe strict-threshold convention is important: in the alternative convention requiring weight at least the quota, the quota is 39.

False. Take three players of weight one and strict threshold one, so a coalition wins exactly when it contains at least two players. Set $S=\{1,2\}$ and $T=\{2,3\}$. ThenThe convex cooperative game inequality would require $1\ge 1+1-0=2$. Thus even this elementary majority weighted voting game is not convex.

Yes. Give players $1,2,3$ weights $1,1,2$, respectively, and let the quota be three. This weighted voting game is a simple cooperative game whose minimal winning coalitions are $\{1,3\}$ and $\{2,3\}$. In orders $123,132,213,231$, player 3 supplies the winning marginal contribution. In order $312$ player 1 does, and in order $321$ player 2 does. Averaging the six marginal contribution vectors gives exactly $(1/6,1/6,4/6)$ as the vector of Shapley values.

It suffices to take four players. Both two-player coalitions $\{1,2\}$ and $\{3,4\}$ win, because two is a prime number, while the empty coalition and the four-player coalition lose.

Suppose real weights and quota $q$ represented this game. The empty coalition loses, so $0<q$. The two winning coalitions requireAdding givesso the four-player coalition must win, a contradiction. Hence prime-cardinality games cannot in general be represented as weighted voting games. The same four-player obstruction appears inside every such game with at least four players.

This proof permits negative individual weights, as the printed definition does. An argument based solely on monotonicity would require a nonnegative-weight assumption and would not suffice for that definition.

Let $W(C)={\sum }_{i\in C}{w}_i$. If the two disjoint winning coalitions exchange players, the aggregate weight is conserved:If both resulting coalitions lost, each weight would be strictly below $q$, making their sum strictly below $2q$. This is impossible. Thus every weighted voting game has swap robustness for disjoint winning coalitions, even when the weights or quota are negative.

The converse is false. The four-player prime-cardinality game from part (a) has swap robustness for disjoint winning coalitions: swapping one player each way preserves each coalition's cardinality, so in fact both winning coalitions remain winning. Part (a) proves it is not a weighted voting game.

Even imposing monotonicity would not repair the converse under the printed disjointness restriction. As a stronger counterexample, use five players and declare a coalition winning exactly when it contains player $1$ and contains either both $2,3$ or both $4,5$. This is a simple cooperative game, and all its winning coalitions contain player $1$, so no two are disjoint; the stated swap condition holds vacuously.

Nevertheless $\{1,2,3\}$ and $\{1,4,5\}$ win, while $\{1,2,4\}$ and $\{1,3,5\}$ lose. Both pairs contain every player the same total number of times. A weighted representation would therefore givea contradiction. This is a trade obstruction to weighted voting and pinpoints how weak the disjoint-only swap requirement is.

A zero-one coalitional game has this property if exchanging one player each way between two disjoint winning coalitions leaves at least one winning. Any weighted voting game satisfies it: the exchange preserves their aggregate weight, which is at least twice the quota, whereas two losing coalitions would have aggregate weight strictly below twice the quota.

The disjointness restriction is substantive. If every winning coalition contains a common player, the condition is vacuous. It is consequently weaker than restrictions allowing exchanges between overlapping winning coalitions, and does not characterize weighted voting games.

Suppose winning coalitions $W_1,\dots ,W_k$ and losing coalitions $L_1,\dots ,L_k$ contain each player the same total number of times:No real-weight, weak-quota weighted voting game represents those outcomes. Summing weights over the winning family gives at least $kq$, while summing over the losing family gives strictly less than $kq$. Equal player multiplicities make the two sums identical, a contradiction. This certificate works without assuming nonnegative weights.

For example, the winners $\{1,2,3\},\{1,4,5\}$ and losers $\{1,2,4\},\{1,3,5\}$ form a two-against-two obstruction. A monotone game whose minimal winning coalitions are the first two sets realizes this certificate, even though every winner contains player $1$.

A weighted voting game assigns nonnegative weights $w_i$ to players and a threshold $t\ge 0$, and declares a coalition winning whenNonnegative weights make the rule monotone. The empty coalition loses, and the grand coalition wins when $t<{\sum }_i{w}_i$; under that nontrivial condition this is a simple cooperative game.

If all weights are integers, the strict threshold is equivalent to the weak quota ${\sum }_{i\in S}{w}_i\ge \lfloor t\rfloor +1$. It becomes quota $t+1$ when $t$ is itself an integer. This distinction matters for noninteger thresholds.

For example, three unit weights and $t=1$ give two-of-three majority. Nonnegative weights do not imply a convex cooperative game: $v(\{1,2\})+v(\{2,3\})=2$, while $v(\{1,2,3\})+v(\{2\})=1$, violating the supermodular inequality.