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I've been playing around with problems involved in introductory quantum game theory, but I am having problems figuring out strategies in this one game. For background, consider the 2x2 Pauli spin matrices $$\sigma_x = \left(\begin{array}{cc} 0&1 \\ 1&0\end{array}\right), \sigma_y = \left(\begin{array}{cc} 0&-i \\ i&0\end{array}\right), \sigma_z = \left(\begin{array}{cc} 1&0 \\ 0&-1 \end{array}\right)$$ and consider the representation of the identity 2x2 matrix to be $\mathbf{1}.$ Let the two vectors that represent a basic in the complex space to be $$u = \left(\begin{array}{c} 1\\ 0\end{array}\right), d = \left(\begin{array}{c} 0 \\ 1 \end{array}\right).$$ Note that $\sigma_x u = d, \sigma_x d = u, \sigma_z = u , \sigma_z d = -d.$

Imagine the following dynamic game of imperfect information between Alice and Bob. Consider an electron that starts in state $u$. Bob applies either the $\sigma_x$ or $\mathbf{1}$ matrix to $u$, then Alice takes a turn (without knowing Bob's move or the state of the electron) applying either $\sigma_x$ or $\mathbf{1}$ to the electron, and then Bob (not knowing Alice's move or the state of the eletron) applies either $\sigma_x$ or $\mathbf{1}$ to the electron. If the electron ends in state $u$, Bob wins $\$1$ and Alice wins $\$-1$ (i.e. loses a dollar), and if the electron ends in state $d$, Bob wins $\$-1$ and Alice wins $\$1$.

Consider a version of this game where Alice "cheats" in the sense that she has the option to initialize the electron in any superposition of the two states, where a superposition is $s = au + bd,$ such that $|a|^2 + |b|^2 = 1.$ I'm wondering, is there any superposition that Alice can set the initial game to that would get her a better outcome (Nash-Equilibrium-wise) than what she would get if she did not cheat? Any suggestions would be appreciated.

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No, there isn't.

In the game without cheating, Alice should assign equal probabilities to $\sigma_x$ and $\mathbf 1$, since anything else would allow Bob to make use of the asymmetry. Thus the value of the game is $0$ for both players.

The value of the game for Bob can't increase due to Alice's ability to cheat, so it's at most $0$. But Bob can assign equal probabilities to $\sigma_x$ and $\mathbf 1$, so the value of the game for him is at least $0$. Thus it's exactly $0$, the same as without cheating.

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