$\newcommand{\ket}[1]{\left|#1\right>}$ I have the next protocol:

1. $A$ tosses a fair coin $a\in \{0,1\}$, if $a=0$, $A$ sends to $B$ $\ket{\psi_0}=\ket0$, if $a=1$ $A$ sends to $B$, $\ket{\psi_1}=\ket{+}$.

2. $B$ now picks randomly $b\in \{0,1\}$.

3. $A$ sends to $B$, the value of $a$.

After step 1., $B$ measures the state he gets in the basis $\{\psi_a, \psi^{\perp}_a\}$, if he doesn't get $\psi_a$ he wins, otherwise the score is $a\oplus b$, i.e if it's $0$ then $B$ wins, if it's $1$ then $A$ wins.

The question is to find the best strategy for $B$ to win if he's dishonest and $A$ is honest, and it's probability, and when B is honest and $A$ is dishonest to show that there exist a constant positive prob for which $B$ can win the game.

My attempt at solution is as follows:

For the first question: a. Well, as far as I can see $B$ can only cheat on choosing between b=0 or 1, but still he's left with 50/50 chance of winning, I don't see better strategy.

b. Here when $A$ cheats and $B$ is honest, if $A$ picks $a = 0 \ or \ 1$, and $\psi$ he sends is different than $\phi_a$, then by the criterion of the game $B$ wins. In case $A$ doesn't cheat $B$ has 0.5 chance of winning the game. So the probability should be (prob. A is dishonest)(prob B to win)+(prob A is honest)(prob B to win)= 0.51+0.50.5=0.75.

But then again, I might be wrong here. :-(

$\newcommand{\ket}[1]{\left|#1\right>} $I have the next protocol:

1. $A$ tosses a fair coin $a\in \{0,1\}$, if $a=0$, $A$ sends to $B$ $\ket{\psi_0}=\ket0$, if $a=1$ $A$ sends to $B$, $\ket{\psi_1}=\ket{+}$.

2. $B$ now picks randomly $b\in \{0,1\}$.

3. $A$ sends to $B$, the value of $a$.

After step 1., $B$ measures the state he gets in the basis $\{\psi_a, \psi^{\perp}_a\}$, if he doesn't get $\psi_a$ he wins, otherwise the score is $a\oplus b$, i.e if it's $0$ then $B$ wins, if it's $1$ then $A$ wins.

The question is to find the best strategy for $B$ to win if he's dishonest and $A$ is honest, and it's probability, and when B is honest and $A$ is dishonest to show that there exist a constant positive prob for which $B$ can win the game.

My attempt at solution is as follows:

For the first question: a. Well, as far as I can see $B$ can only cheat on choosing between b=0 or 1, but still he's left with 50/50 chance of winning, I don't see better strategy.

b. Here when $A$ cheats and $B$ is honest, if $A$ picks $a = 0 \ or \ 1$, and $\psi$ he sends is different than $\phi_a$, then by the criterion of the game $B$ wins. In case $A$ doesn't cheat $B$ has 0.5 chance of winning the game. So the probability should be (prob. A is dishonest)(prob B to win)+(prob A is honest)(prob B to win)= 0.51+0.50.5=0.75.

But then again, I might be wrong here. :-(

$\newcommand{\ket}[1]{\left|#1\right>}$ I have the next protocol:

1. $A$ tosses a fair coin $a\in \{0,1\}$, if $a=0$, $A$ sends to $B$ $\ket{\psi_0}=\ket0$, if $a=1$ $A$ sends to $B$, $\ket{\psi_1}=\ket{+}$.

2. $B$ now picks randomly $b\in \{0,1\}$.

3. $A$ sends to $B$, the value of $a$.

After step 1., $B$ measures the state he gets in the basis $\{\psi_a, \psi^{\perp}_a\}$, if he doesn't get $\psi_a$ he wins, otherwise the score is $a\oplus b$, i.e if it's $0$ then $B$ wins, if it's $1$ then $A$ wins.

The question is to find the best strategy for $B$ to win if he's honest and $A$ isn't, and it's probability, and when B is dishonest and $A$ is honest to show that there exist a constant positive prob for which $B$ can win the game.

My attempt at solution is as follows:

For the first question: a. Well, as far as I can see $B$ can only cheat on choosing between b=0 or 1, but still he's left with 50/50 chance of winning, I don't see better strategy.

b. Here when $A$ cheats and $B$ is honest, if $A$ picks $a = 0 \ or \ 1$, and $\psi$ he sends is different than $\phi_a$, then by the criterion of the game $B$ wins. In case $A$ doesn't cheat $B$ has 0.5 chance of winning the game. So the probability should be (prob. A is dishonest)(prob B to win)+(prob A is honest)(prob B to win)= 0.51+0.50.5=0.75.

But then again, I might be wrong here. :-(

$\newcommand{\ket}[1]{\left|#1\right>}$ I have the next protocol:

1. $A$ tosses a fair coin $a\in \{0,1\}$, if $a=0$, $A$ sends to $B$ $\ket{\psi_0}=\ket0$, if $a=1$ $A$ sends to $B$, $\ket{\psi_1}=\ket{+}$.

2. $B$ now picks randomly $b\in \{0,1\}$.

3. $A$ sends to $B$, the value of $a$.

After step 1., $B$ measures the state he gets in the basis $\{\psi_a, \psi^{\perp}_a\}$, if he doesn't get $\psi_a$ he wins, otherwise the score is $a\oplus b$, i.e if it's $0$ then $B$ wins, if it's $1$ then $A$ wins.

The question is to find the best strategy for $B$ to win if he's dishonest and $A$ is honest, and it's probability, and when B is honest and $A$ is dishonest to show that there exist a constant positive prob for which $B$ can win the game.

My attempt at solution is as follows:

For the first question: a. Well, as far as I can see $B$ can only cheat on choosing between b=0 or 1, but still he's left with 50/50 chance of winning, I don't see better strategy.

b. Here when $A$ cheats and $B$ is honest, if $A$ picks $a = 0 \ or \ 1$, and $\psi$ he sends is different than $\phi_a$, then by the criterion of the game $B$ wins. In case $A$ doesn't cheat $B$ has 0.5 chance of winning the game. So the probability should be (prob. A is dishonest)(prob B to win)+(prob A is honest)(prob B to win)= 0.51+0.50.5=0.75.

But then again, I might be wrong here. :-(