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For deterministic classical responses, denote the two possible local answers by $x_0,x_1$, $y_0,y_1$ and $z_0,z_1$. Certainty in the GHZ game would requireTaking the exclusive or of all four equations gives zero on the left, since every answer occurs twice, but one on the right. This contradiction rules out deterministic certainty. Pre-shared randomness cannot cure it: conditioning on the random seed fixes a deterministic response table, and none succeeds on every allowed input. Thus noncommunicating classical players cannot win with certainty. No uniform distribution on the referee's questions is needed for this worst-case claim.

For the quantum strategy, share the GHZ state and measure the Pauli X gate observable when the local question is 0, or the Pauli Y gate observable when it is 1. Return bit 0 for measurement eigenvalue $+1$ and bit 1 for eigenvalue $-1$. The four relevant operator identities areFor the first identity, the three bit flips exchange $|000⟩$ and $|111⟩$. For each remaining identity, the two Y actions contribute $i^2=-1$ on $|000⟩$ and $(-i{)}^2=-1$ on $|111⟩$, so they exchange the terms with a minus sign. Local measurements on separate systems commute, and the product of their observed eigenvalues is therefore fixed by the displayed joint eigenvalue. That product equals $(-1{)}^{x+y+z}$, giving exactly the required output parity for each promised input. Hence the quantum strategy wins with certainty without communication. This is the GHZ theorem in nonlocal-game form; it does not let a player control another player's individual outcome.

Let $p=|\alpha {|}^2$ and $1-p=|\beta {|}^2$. The first basic measurement has outcome probabilities $p$ and $1-p$, obtained by conjugating the amplitudes in the bras. Conditional on its first outcome, the next computational-basis measurement returns 0 with probability $p$; conditional on its second outcome, that probability is $1-p$. The law of total probability therefore givesConsequently the minimum is $1/2$, attained exactly when $|\alpha |=|\beta |=1/\sqrt{2}$. The relative complex phases do not affect this result. The alternatives after the first measurement form a classical mixture; their amplitudes must not be added coherently as if no measurement occurred.

A basic measurement is specified by an orthonormal basis $\{|b_j⟩{\}}_{j=0}^{d-1}$, with outcome $j$ associated with the rank-one orthogonal projection $P_j=|b_j⟩⟨b_j|$. For a normalized pure state $|\psi ⟩$, the Born rule givesThe probabilities sum to one because ${\sum }_j{P}_j=I$. Conditional on an outcome with $p(j)>0$, the post-measurement state iswhich represents $|b_j⟩$ up to a global phase. An outcome of zero probability has no conditional state to normalize. For a density matrix $\rho$, the corresponding expressions are $p(j)=\mathrm{Tr}(P_j\rho )$ and $P_j\rho P_j/p(j)$. If the outcome is ignored, the nonselective projective measurement instead produces ${\sum }_j{P}_j\rho P_j$, generally a mixed state.^[ref]

Where its denominator is nonzero, normalization enforces cooperative-game efficiency and preserves cooperative-game symmetry and the null player axiom. The violated axiom is 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 arewith $v(N)=w(N)=1$. HenceUnnormalized indices add, so $\beta (v+w)=(3/2,1/2,1/2)$. Their sum is $5/2$, while $(v+w)(N)=2$. ThereforeThe 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.

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.

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.

Use the inclusive quota: a coalition wins at weight at least $25$. For each player, count losing predecessor coalitions that become winning when that player joins. Group these swing counts by predecessor cardinality $t=0,1,2,3$:

For example, player one is pivotal for predecessor coalitions $\{2\}$, $\{2,3\}$, $\{2,4\}$, $\{3,4\}$ and $\{2,3,4\}$. Player three is pivotal only for $\{1,4\}$; player four has the symmetric counterpart.

Each swing contributes $1/8$ to the probabilistic Banzhaf index. For the Shapley value, a swing with $t$ predecessors contributes $t!(3-t)!/4!$, which is $1/4,1/12,1/12,1/4$ for the four possible sizes. Applying these weights to the table givesThe Shapley values sum to one. The unnormalized Banzhaf indices sum to $5/4$; they must not be replaced by normalized shares in this part.