Cheating Bob

A cheating Bob can always win. He just needs to pretend to have obtained $\ket{\psi_a^\perp}$ from his measurement.

#Cheating Alice By definition, if Alice cheats, she is not restricted to send one of the $\ket{\psi_a}$ states. I suspect that her optimal attack involves preparing an entangled state, sending half of it to Bob and make a measurement depending on $b$. The chosen $a$ will depend on the output value of $b$.

Let's look at a (maybe suboptimal) way for Alice to cheat.

1. Alice sends the state $\ket\phi=\frac{\ket0+\ket+}{\sqrt{2+\sqrt2}}=\frac{(1+\sqrt2)\ket0+\ket1}{\sqrt{4+2\sqrt2}}$
2. When Bob reveals $b$, Alice choses $a=b\oplus1$, to ensure $a\oplus b=1$
3. Alice sends $a$ to Bob. Let's suppose $a=0$ (the situation is obviously symmetric when $a=1$.). Bob's measurement is then $\{\ket0, \ket1\}$.

• He gets $\ket1$ with probability $\frac1{4+2\sqrt2}=14.64\%$. Bob wins in this case. - He gets $\ket0$ with probability $1-\frac1{4+2\sqrt2}=85.36\%$. Since $a\otimes b=1$, Alice always wins in this case.

#A much better trivial classical protocol As shown above, Bob's cheating probability is 100% and Alice's is at least 85%. The following fully classical protocol is better:

1. Alice randomly choses $a$ and tells it to Bob
2. Bob randomly choses $b$ ant tells it to Alice. The winner is given by $a\oplus b$. Alice cheating probability is now 50% instead of 85%, while Bob's cheating probability is no worse than in the preceding, where it was already 100% ! This protocol is therefore better than yours, even if not very useful...

#Literature on Weak Coin Flipping The protocol you describe is called weak coin flipping. Mochon has given a protocol with arbitrary small bias in arxiv:0711.4114 (Warning: hard to understand paper), involving several rounds of communication between Alice and Bob. An easier to understand protocol was proposed by Spekkens and Rudolph in arXiv:quant-ph/0202118, with a cheating probability of at most $1/\sqrt2$.

Cheating Bob

$\newcommand{\ket}[1]{\left|#1\right>} $A cheating Bob can always win. He just needs to pretend to have obtained $\ket{\psi_a^\perp}$ from his measurement.

#Cheating Alice By definition, if Alice cheats, she is not restricted to send one of the $\ket{\psi_a}$ states. I suspect that her optimal attack involves preparing an entangled state, sending half of it to Bob and make a measurement depending on $b$. The chosen $a$ will depend on the output value of $b$.

Let's look at a (maybe suboptimal) way for Alice to cheat.

1. Alice sends the state $\ket\phi=\frac{\ket0+\ket+}{\sqrt{2+\sqrt2}}=\frac{(1+\sqrt2)\ket0+\ket1}{\sqrt{4+2\sqrt2}}$
2. When Bob reveals $b$, Alice choses $a=b\oplus1$, to ensure $a\oplus b=1$
3. Alice sends $a$ to Bob. Let's suppose $a=0$ (the situation is obviously symmetric when $a=1$.). Bob's measurement is then $\{\ket0, \ket1\}$.

• He gets $\ket1$ with probability $\frac1{4+2\sqrt2}=14.64\%$. Bob wins in this case. - He gets $\ket0$ with probability $1-\frac1{4+2\sqrt2}=85.36\%$. Since $a\otimes b=1$, Alice always wins in this case.

#A much better trivial classical protocol As shown above, Bob's cheating probability is 100% and Alice's is at least 85%. The following fully classical protocol is better:

1. Alice randomly choses $a$ and tells it to Bob
2. Bob randomly choses $b$ ant tells it to Alice. The winner is given by $a\oplus b$. Alice cheating probability is now 50% instead of 85%, while Bob's cheating probability is no worse than in the preceding, where it was already 100% ! This protocol is therefore better than yours, even if not very useful...

#Literature on Weak Coin Flipping The protocol you describe is called weak coin flipping. Mochon has given a protocol with arbitrary small bias in arxiv:0711.4114 (Warning: hard to understand paper), involving several rounds of communication between Alice and Bob. An easier to understand protocol was proposed by Spekkens and Rudolph in arXiv:quant-ph/0202118, with a cheating probability of at most $1/\sqrt2$.

Cheating Bob

A cheating Bob can always win. He just needs to pretend to have obtained $\ket{\psi_a^\perp}$ from his measurement.

#Cheating Alice By definition, if Alice cheats, she is not restricted to send one of the $\ket{\psi_a}$ states. I suspect that her optimal attack involves preparing an entangled state, sending half of it to Bob and make a measurement depending on $b$. The chosen $a$ will depend on the output value of $b$..

#Literature on Weak Coin Flipping The protocol you describe is called weak coin flipping. Mochon has given a protocol with arbitrary small bias in arxiv:0711.4114 (Warning: hard to understand paper), involving several rounds of communication between Alice and Bob. An easier to understand protocol was proposed by Spekkens and Rudolph in arXiv:quant-ph/0202118, with a cheating probability of at most $1/\sqrt2$.

Cheating Bob

A cheating Bob can always win. He just needs to pretend to have obtained $\ket{\psi_a^\perp}$ from his measurement.

#Cheating Alice By definition, if Alice cheats, she is not restricted to send one of the $\ket{\psi_a}$ states. I suspect that her optimal attack involves preparing an entangled state, sending half of it to Bob and make a measurement depending on $b$. The chosen $a$ will depend on the output value of $b$.

Let's look at a (maybe suboptimal) way for Alice to cheat.

1. Alice sends the state $\ket\phi=\frac{\ket0+\ket+}{\sqrt{2+\sqrt2}}=\frac{(1+\sqrt2)\ket0+\ket1}{\sqrt{4+2\sqrt2}}$
2. When Bob reveals $b$, Alice choses $a=b\oplus1$, to ensure $a\oplus b=1$
3. Alice sends $a$ to Bob. Let's suppose $a=0$ (the situation is obviously symmetric when $a=1$.). Bob's measurement is then $\{\ket0, \ket1\}$.

• He gets $\ket1$ with probability $\frac1{4+2\sqrt2}=14.64\%$. Bob wins in this case. - He gets $\ket0$ with probability $1-\frac1{4+2\sqrt2}=85.36\%$. Since $a\otimes b=1$, Alice always wins in this case.

#A much better trivial classical protocol As shown above, Bob's cheating probability is 100% and Alice's is at least 85%. The following fully classical protocol is better:

1. Alice randomly choses $a$ and tells it to Bob
2. Bob randomly choses $b$ ant tells it to Alice. The winner is given by $a\oplus b$. Alice cheating probability is now 50% instead of 85%, while Bob's cheating probability is no worse than in the preceding, where it was already 100% ! This protocol is therefore better than yours, even if not very useful...

#Literature on Weak Coin Flipping The protocol you describe is called weak coin flipping. Mochon has given a protocol with arbitrary small bias in arxiv:0711.4114 (Warning: hard to understand paper), involving several rounds of communication between Alice and Bob. An easier to understand protocol was proposed by Spekkens and Rudolph in arXiv:quant-ph/0202118, with a cheating probability of at most $1/\sqrt2$.