Now Alice, Barbara, and Candida share a (properly normalized) 3-qubit state vector of the form
$$ | \psi \rangle = a_{000} | 000 \rangle + a_{001} | 001 \rangle + \ldots + a_{110} | 110 \rangle + a_{111} | 111 \rangle. $$
The players decide what to do before measuring depending on what the referee Rosa passes them.
If Alice gets \(a = 0\), then she just measures with the bit basis. Otherwise, she applies \(H\), the Hadamard gate:
$$ H = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}. $$
Similarly for Barbara and Candida. Thus, before the players make their measurements, the state vector is given by
$$ \left( H^{a} \otimes H^{b} \otimes H^{c} \right) | \psi \rangle. $$
Here, \(H^{0} = I\) and \(H^{1} = H\).
There are four possible scenarios for the referee:
$$ 000, \qquad 011, \qquad 101, \qquad 110. $$
Thus, the probability of winning the game is given by
$$ p_{\text{win}} = \frac{1}{4} \left[ p_{\text{win}}(000) + p_{\text{win}}(011) + p_{\text{win}}(101) + p_{\text{win}}(110) \right]. $$
There are four possible ways of winning when Rosa chooses \(000\):
$$ p_{\text{win}}(000) = \left| \langle 000 | \psi \rangle \right|^{2} + \left| \langle 011 | \psi \rangle \right|^{2} + \left| \langle 101 | \psi \rangle \right|^{2} + \left| \langle 110 | \psi \rangle \right|^{2}. $$
Note that for this referee choice you have
$$ H^{a} \otimes H^{b} \otimes H^{c} = I \otimes I \otimes I. $$
There are four possible ways of winning when Rosa chooses \(011\):
$$ p_{\text{win}}(011) = \left| \langle 001 | I \otimes H \otimes H | \psi \rangle \right|^{2} + \left| \langle 010 | I \otimes H \otimes H | \psi \rangle \right|^{2} + \left| \langle 100 | I \otimes H \otimes H | \psi \rangle \right|^{2} + \left| \langle 111 | I \otimes H \otimes H | \psi \rangle \right|^{2}. $$
There are four possible ways of winning when Rosa chooses \(101\):
$$ p_{\text{win}}(101) = \left| \langle 001 | H \otimes I \otimes H | \psi \rangle \right|^{2} + \left| \langle 010 | H \otimes I \otimes H | \psi \rangle \right|^{2} + \left| \langle 100 | H \otimes I \otimes H | \psi \rangle \right|^{2} + \left| \langle 111 | H \otimes I \otimes H | \psi \rangle \right|^{2}. $$
There are four possible ways of winning when Rosa chooses \(110\):
$$ p_{\text{win}}(110) = \left| \langle 001 | H \otimes H \otimes I | \psi \rangle \right|^{2} + \left| \langle 010 | H \otimes H \otimes I | \psi \rangle \right|^{2} + \left| \langle 100 | H \otimes H \otimes I | \psi \rangle \right|^{2} + \left| \langle 111 | H \otimes H \otimes I | \psi \rangle \right|^{2}. $$
There are ways where the players can always win the game! Consider the 3-qubit state vector
$$ | \psi \rangle = \frac{1}{2} \left( |000 \rangle - |011\rangle - |101\rangle - |110\rangle \right). $$
Note that this state vector is entangled. It follows that
$$ (I \otimes H \otimes H) | \psi \rangle = \frac{1}{2} \left( |001\rangle + |010\rangle - |100\rangle + |111\rangle \right); $$
$$ (H \otimes I \otimes H) | \psi \rangle = \frac{1}{2} \left( |001\rangle - |010\rangle + |100\rangle + |111\rangle \right); $$
$$ (H \otimes H \otimes I) | \psi \rangle = -\frac{1}{2} \left( |001\rangle - |010\rangle - |100\rangle - |111\rangle \right). $$
For this particular state and measurement strategy, the probability of winning for each of the referee outcomes is 100%. It feels like cheating: the strategy guarantees that every winning outcome is equally probable. Basically, before measurement, the state vector is contained entirely in the "winning" subspace.
