up to $N=8000$, see Fig. \[Results Figure\]. We see that the [*fairness condition*]{} $\mbox{sup}_m \bar{P}_{ch}(m) << 1$ is satisfied (for $N=600$ we got $\mbox{sup}_m \bar{P}_{ch}(m) = \bar{P}_{ch}(92)= 0.0811$, while for $N_A=N_R$ we have $\mbox{sup}_m \bar{P}_{ch}(m) = \bar{P}_{ch}(1455)= 0.0247$ for $N=8000$), which is explicitly shown on Fig. \[Results Figure\]. Moreover, the numerical fit gives $\mbox{sup}_m \bar{P}_{ch}(m) \propto N^{-1/2}$ behavior, giving us the scaling of the complexity of the protocol, with respect to the number $N$ of exchanged messages between Alice and Bob.

![(color online) The expected probability to cheat $\bar{P}_{ch}(m)$ (upper row) and the maximal expected probability to cheat $\mbox{sup}_m \bar{P}_{ch}(m)$ (lower row) for the uniform $p(\alpha)$ on $I_{\alpha}=[0.9,0.99]$. The plots from the left column represents results for our protocol, while the right ones are for the restricted “typical” case of $N_A=N_R$. Note the scaling behavior $\mbox{sup}_m \bar{P}_{ch}(m) \propto N^{-1/2}$. \
[]{data-label="Results Figure"}](Fig34.pdf){width="8.5cm" height="6.0cm"}

V. Fairness of the protocol: General measurements and noise {#sec:fairness-general_measurements}
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First, we consider only one-qubit orthogonal measurements. Since Alice is an honest client, $X^{\cal{A}} = \hat{A}^{\otimes m}$, we have that $P^{\cal{A}}_A (m, \alpha, X^{\cal{A}}) = 1$ and $P^{\cal{A}}_R (m, \alpha, X^{\cal{A}}) = P_R(m, \alpha)$.

Bob is a dishonest client and his strategy $X^{\cal{B}}$ consists of measuring $(m-k)$ times the Accept observable $\hat{A}$ and $k$ times observable $\hat{K}= 0\cdot |m\rangle\langle m| + 1\cdot |m^\bot\rangle\langle m^\bot|$, where $$\!\! |m\rangle \! = \! \cos\frac{\theta}{2}|0\rangle + e^{\varphi}\!\sin\frac{\theta}{2}|1\rangle \! = \! \cos\frac{\theta^\prime}{2}|-\rangle  +  e^{\varphi^\prime}\!\!\sin\frac{\theta^\prime}{2}|+\rangle.$$ Let $m = m_a + m_r$, where $m_a$ is the number of the Accept and $m_r$ th