Tripartite Dynamic Zero-Sum Quantum Games

Abstract

The Nash equilibrium plays a crucial role in game theory. Most of results are based on classical resources. Our goal in this paper is to explore multipartite zero-sum game with quantum settings. We find that in two different settings there is no strategy for a tripartite classical game being fair. Interestingly, this is resolved by providing dynamic zero-sum quantum games using single quantum state. Moreover, the gains of some players may be changed dynamically in terms of the committed state. Both quantum games are robust against the preparation noise and measurement errors.

1. Introduction

The quantum state as an important resource has been widely used for accomplishing difficult or impossible tasks with classical resources [1]. Quantum game theory as one of important applications is to investigate strategic behavior of agents using quantum resources. It is closely related to quantum computing and Bell theory [2,3,4]. In most cases, distributive tasks can be regarded as equivalent quantum games [5,6]. So far, it has been widely used in Bell tests [3,7], quantum network verification [8], distributed computation [8,9,10,11,12,13], parallel testing [14,15,16,17], device-independent quantum key distribution [18,19,20,21].

Different from those applications, Marinatto and Weber present a two quantum game using Nash strategy which gives more reward than classical [22]. Eisert et al. resolve the prisoner’s dilemma with quantum settings by providing higher gains than its with classical settings [23]. Sekiguchi et al. have prove that the uniqueness of Nash equilibria in quantum Cournot duopoly game [24]. Brassard et al. recast colorredMermin’s multi-player game in terms of quantum pseudo-telepathy [25]. Meyer introduces the quantum strategy for coin flipping game [26]. In the absence of a fair third party, Zhang et al. prove that a space separated two party game can achieve fairness by combining Nash equilibrium theory with quantum game theory [27]. All of these quantum games show different supremacy to classical games. Nevertheless, there are specific games without quantum advantage. One typical example is guessing your neighbor’s input game (GYNI) [28] or its generalization [8]. Hence, it is interesting to find games with different features.

Every game contains three elements: player, strategy and gain function [6,22]. There are various classification according to different benchmarks. According to the participants’ understanding of other participants, one game may be a complete information game [3] or a incomplete information game [28]. From the time sequence of behavior, it may be divided into static game [22] and dynamic game [27,29]. Another case is cooperating game [3] or competing game (non-cooperating game) [22]. Non cooperative game can be further divided into complete information static game, complete information dynamic game, incomplete information static game and incomplete information dynamic game [22,29]. As the basis of non-cooperative game [29], Nash theory is composed of the optimal strategies of all participants such that no one is willing to break the equilibrium. Moreover, it is a zero-sum game if the total gains is zero for any combination of strategies [27]. Otherwise, it is a non zero-sum game. So far, most of games are focused on cooperative games such as Bell game [3]. Our goal in this paper is to find dynamic zero-sum games with Nash equilibrium. Dynamic game refers to that the actions of different players have a sequence, and the post actor can observe the actions chosen by the actor in front of him [29]. Different from bipartite zero-sum game [27], we provide tripartite quantum fair zero-sum games which cannot be realized in classical scenarios even if it is difficult to evaluating Nash equilibrium [30]. The interesting feature is that the present quantum game uses of only clean qubit without entanglement.

The rest of paper is organized as follows. In Section 2, we introduce a tripartite zero-sum game with two different settings inspired by bipartite game [27]. We show that there is no strategy for achieving a fair game using classical resources. Both models can be regarded as complete information dynamic zero-sum game. In Section 3, we present quantum zero-sum games with the same settings using quantum single-photon states. Both games are asymptotically fair in terms of some free parameters. Although any quantum pure state is unitarily equivalent to a classical state, our results show that this kind of resources are also useful for special quantum tasks going beyond classical scenarios. In Section 4, we show that the robustness of two quantum games. The last section concludes the paper.

2. Classical Tripartite Dynamic Zero-Sum Games

2.1. Game Model

We firstly present some definitions as follows.

Definition 1.

Dynamic zero-sum game means that the actions of different players have a sequence, the later participant can observe the formers actions, and the sum of payments for all players is zero for any combination of strategies.

Definition 2.

Fair game means that the game does not favor any player. Games in this paper are all zero-sum games, thus the fairness in this paper means everyone’s average gain is zero.

Definition 3.

Asymptotically fair game means that under the given initial conditions, the game may not be fair, but with the increase of variable parameter, the degree of deviation from the fair game becomes smaller and smaller and the game is fair when the variable parameter approaches infinity.

Inspired by bipartite game [27], we present a tripartite game $\mathsf{G}$, as shown in Figure 1.

In the present games, we assume that the actions of different participants have a sequence, where the latter participant can observe the former’s action. There are four stages in the present model as follows.

S1.

Alice randomly puts a ball into one of three black boxes, $\diamond A$, $\diamond B$ and $\diamond C$. Alice sends the box $\diamond C$ to Rice, and sends $\diamond B$ to Bob.

S2.

Bob gives his own choice, i.e., he chooses to open $\diamond B$ or asks Alice to open $\diamond A$, but he does not take any action.

S3.

Rice chooses her own strategy and takes action, i.e., she opens $\diamond C$ or lets Alice open $\diamond A$. If Rice opens $\diamond C$ and there is no ball in the box, the game enters the fourth stage.

S4.

It is Bob’s turn to take action according the strategy he has chosen in S2, i.e., he opens box $\diamond B$ or asks Alice to open box $\diamond A$.

Moreover, for any strategy combination, we assume that the total payment of all participants is zero. In the following, we will explore this kind of games with two settings with different gains.

The classical game tree is shown in Figure 2. In the stages of $\mathsf{G}$, the wining rules of the game are given by

W1.

Bob and Rice win if Rice chooses to open $\diamond A$ and finds the ball in S3.

W2.

Alice wins if Rice does not find the ball in box $\diamond A$ in S3.

W3.

Rice wins if Rice opens $\diamond C$ and finds the ball in S3.

W4.

Bob and Rice win if Bob opens $\diamond A$ and finds the ball in S4.

W5.

Alice wins if Bob opens $\diamond A$ and does not find the ball in $\diamond A$ in S4.

W6.

Bob wins if Bob opens box $\diamond B$ and finds the ball in S4.

W7.

Alice wins if Bob opens box $\diamond B$ and does not find the ball in S4.

Now, for convenience, we define the following probabilities.

P1.

Denote $P_1$ as the probability that Alice puts the ball into $\diamond A$.

P2.

Denote $P_2$ as the probability that Alice puts the ball into $\diamond B$ or $\diamond C$.

P3.

Denote $P_{\diamond C}$ as the probability that Rice chooses to open $\diamond C$.

P4.

Denote $P_{\diamond B}$ as the probability that Bob chooses to open $\diamond B$.

Here, we assume that Alice has equal probability to put the ball into $\diamond B$ or $\diamond C$. It follows that $P_1+2P_2=1$, with $P_{\diamond B},P_{\diamond C}\in [0,1]$.

From winning rules W1–W7 of $\mathsf{G}$, it is easy to get the winning probability $P_{Ric{e}_1}$ for Rice from W3 (i.e., Rice finds the ball after opening $\diamond C$) is given by

$$\begin{array}{c}\hfill P_{Ric{e}_1}=P_2{P}_{\diamond C}\end{array}$$

(1)

Moreover, the winning probability for Bob from W6 (i.e., Rice chooses to open $\diamond C$ and does not find the ball, but Bob finds the ball after opening $\diamond B$) is given by

$$\begin{array}{c}\hfill P_{Bo{b}_1}=P_2{P}_{\diamond C}{P}_{\diamond B}(1-P_2)\end{array}$$

(2)

For Alice, there are three subcases W2, W5 and W7 for winning. It follows that

$$\begin{array}{ccc}\hfill P_{Alice}& =& 1-P_1+P_{\diamond B}{P}_{\diamond C}{P}_2^2-P_{\diamond B}{P}_{\diamond C}{P}_2-P_{\diamond C}{P}_2\hfill \\ & & +P_{\diamond B}{P}_{\diamond C}{P}_1+P_{\diamond C}{P}_1{P}_2-P_{\diamond B}{P}_{\diamond C}{P}_1{P}_2\hfill \end{array}$$

(3)

The winning probability for Bob from W1 and W4 is given by

$$\begin{array}{ccc}\hfill P_{Bo{b}_2}& =& P_1-P_{\diamond B}{P}_{\diamond C}{P}_1-P_{\diamond C}{P}_1{P}_2\hfill \\ & & +P_{\diamond B}{P}_{\diamond C}{P}_1{P}_2\hfill \end{array}$$

(4)

The same result holds for Rice from W1 and W4, i.e., $P_{Ric{e}_2}=P_{Bo{b}_2}$.

Here, we analyze players’ strategies for the present game $\mathsf{G}$ to show the main idea. For Alice, she does not know which box Rice or Bob would choose to open before she prepares. Alice may lose the game if she puts the ball in one of the three boxes with a higher probability. Thus, there is a tradeoff for Alice to choose her strategy (the probability $P_1$). For Bob, he does not know which box Rice would to open when he gives the probability $P_{\diamond B}$. So, he should consider a known parameter $P_1$ given by Alice and an unknown parameter $P_{\diamond C}$ given by Rice. Similar analysis can be applied to Rice’s strategy, Rice needs to choose her own parameter based on what she knows before she takes action.

2.2. The First Tripartite Classical Game

It is well known that every player will maximize its own interests in a non-cooperative game. In this section, we present the first game ${\mathsf{G}}_1$ with the gain setting given in

Table 1. The gain settings of players in the first classical game ${\mathsf{G}}_1$. Here, $g_{c(b,a)}$ denotes the gain of Rice (or Bob, or Alice). ${\mathsf{R}}_{succ}(\diamond C)$ means that Rice finds the ball after opening box $\diamond C$. ${\mathsf{B}}_{succ}(\diamond B)$ means that Bob finds the ball after opening box $\diamond B$. ${\mathsf{R}/\mathsf{B}}_{fail}(·)$ means that neither Rice or Bob finds the ball. ${\mathsf{R}/\mathsf{B}}_{succ}(\diamond A)$ means that either Rice or Bob successes by opening box $\diamond A$. $\eta \ge 1$ in the first classical game.

Cases∖Gains	${\mathit{g}}_{\mathit{c}}$	${\mathit{g}}_{\mathit{b}}$	${\mathit{g}}_{\mathit{a}}$
${\mathsf{R}}_{succ}(\diamond C)$	2	−1	−1
${\mathsf{B}}_{succ}(\diamond B)$	−1	2	−1
$\mathsf{R}/{\mathsf{B}}_{fail}(·)$	−1	−1	2
$\mathsf{R}/{\mathsf{B}}_{succ}(\diamond A)$	$\eta$	1	−$\eta$−1

. Our goal is to show the no-fairness of this game with classical resources.

2.2.1. The Average Gain of Rice

From Table 1, the average gain of Rice is given by

$$\begin{array}{ccc}\hfill G_{Rice}& =& 2P_{Ric{e}_1}-P_{Bo{b}_1}-P_{Alice}+\eta P_{Ric{e}_2}\hfill \\ & =& 3P_2{P}_{\diamond C}+(\eta +1)P_1-(\eta +1)P_1{P}_{\diamond B}{P}_{\diamond C}\hfill \\ & & -(\eta +1)P_1{P}_2{P}_{\diamond C}+(\eta +1)P_1{P}_2{P}_{\diamond B}{P}_{\diamond C}-1\hfill \end{array}$$

(5)

From Equation (6), we get that the partial derivative of $G_{Rice}$ with respect to $P_{\diamond C}$ is given by

$$\begin{array}{ccc}\hfill \frac{\partial G_{Rice}}{\partial P_{\diamond C}}& =& 3P_2-(\eta +1)P_1{P}_{\diamond B}\hfill \\ & & +(\eta +1)P_{\diamond B}{P}_1{P}_2-(\eta +1)P_1{P}_2\hfill \end{array}$$

(6)

From Equation (6), if $\frac{\partial G_{Rice}}{\partial P_{\diamond C}}<0$, $G_{Rice}$ is a deceasing function. In this case, Rice has to set $P_{\diamond C}=0$ to maximize her gains. If $\frac{\partial G_{Rice}}{\partial P_{\diamond C}}>0$, i.e., $G_{Rice}$ is increasing function, Rice will choose $P_{\diamond C}=1$ to maximize her gains. Moreover, when $\frac{\partial G_{Rice}}{\partial P_{\diamond C}}=0$, i.e., $G_{Rice}$ is a constant, $P_{\diamond C}$ can be any probability.

2.2.2. The Average Gain of Bob

Similar to Equation (6), we get that the average gain of Bob is given by

$$\begin{array}{ccc}\hfill G_{Bob}& =& -P_{Ric{e}_1}+2P_{Bo{b}_1}-P_{Alice}+P_{Bo{b}_2}\hfill \\ & =& 3P_{\diamond B}{P}_{\diamond C}{P}_2-3P_{\diamond B}{P}_{\diamond C}{P}_2^2+2P_1\hfill \\ & & -2P_{\diamond B}{P}_{\diamond C}{P}_1-2P_{\diamond C}{P}_1{P}_2\hfill \\ & & +2P_{\diamond B}{P}_{\diamond C}{P}_1{P}_2-1\hfill \end{array}$$

(7)

There are several cases to maximize $G_{Bob}$. We only present one case in the following for explaining the main idea. The other cases are included in Appendix A.

For $\frac{\partial G_{Rice}}{\partial P_{\diamond C}}\le 0$, i.e., $\frac{3P_2-(\eta +1)P_1{P}_2}{(\eta +1)P_1(1-P_2)}\le P_{\diamond B}\le 1$, we get that $P_{\diamond C}=0$. From Equation (8), we obtain

$$\begin{array}{c}\hfill G_{Bo{b}_1}=2P_1-1\end{array}$$

(8)

C1.

If $0\le \frac{3P_2-(\eta +1)P_1{P}_2}{(\eta +1)P_1(1-P_2)}\le 1$, we get that $\frac{3}{2\eta +5}\le P_1\le \frac{3}{\eta +1}$. In this case, Bob chooses $G_{Bo{b}_1}$ such that $P_{\diamond B}=X_0$, where $X_0>\frac{3P_2-(\eta +1)P_1{P}_2}{(\eta +1)P_1(1-P_2)}$.

C2.

If $\frac{3P_2-(\eta +1)P_1{P}_2}{(\eta +1)P_1(1-P_2)}<0$, i.e., $\frac{3}{\eta +1}<P_1\le 1$. Owing to $P_{\diamond B}\ge 0$, we get that $P_{\diamond B}$ can be any probability.

All the results are given in

Table 2. The values of $P_{\diamond B}$ and $P_{\diamond C}$. $P_{\diamond B\left(\diamond C\right)}$ depends on the different cases in the game, where $X_1=\frac{\eta +8-\sqrt{{\eta }^2+4\eta +52}}{2\eta +2}$. Here, ${\Delta }_i$ denotes the intervals given by ${\Delta }_1=[0,\frac{3}{2\eta +5})$, ${\Delta }_2=[\frac{3}{2\eta +5},X_1)$, ${\Delta }_3=[X_1,\frac{3}{\eta +1}]$ and ${\Delta }_4=(\frac{3}{\eta +1},1]$.

$\eta$		$1\le \eta \le 2$	$\eta >2$
$P_1$	$P_1\in {\Delta }_1$	$P_1\in {\Delta }_2$	$P_1\in {\Delta }_2$	$P_1\in {\Delta }_3$	$P_1\in {\Delta }_4$
$P_{\diamond B}$	1	$X_0$	$\frac{3P_2-(\eta +1)P_1{P}_2}{P_1(\eta +1)(1-P_2)}$	$X_0$	$[0,1]$
$P_{\diamond C}$	1	0	1	0	0

.

2.2.3. The Average Gain of Alice

In this subsection, we calculate the average gain of Alice, which is denoted by $G_{Alice}$. It is easy to write the expression for $G_{Alice}$ according to Table 1 as follows.

$$\begin{array}{ccc}\hfill G_{Alice}& =& -P_{Ric{e}_1}-P_{Bo{b}_1}+2P_{Alice}-(\eta +1)P_{Ric{e}_2}\hfill \\ & =& -3P_{\diamond C}{P}_2-3P_{\diamond B}{P}_{\diamond C}{P}_2+3P_{\diamond B}{P}_{\diamond C}{P}_2^2\hfill \\ & & +(\eta +3)P_{\diamond C}{P}_1{P}_2+(\eta +3)P_{\diamond B}{P}_{\diamond C}{P}_1\hfill \\ & & -(\eta +3)P_1-(\eta +3)P_{\diamond B}{P}_{\diamond C}{P}_1{P}_2+2\hfill \end{array}$$

(9)

For the case of $0\le P_1<\frac{3}{2\eta +5}$, it follows from Table 2 that $P_{\diamond C}=1$ and $P_{\diamond B}=1$. Equation (9) is then rewritten into

$$\begin{array}{c}\hfill G_{Alice}=3P_2^2-6P_2+2\end{array}$$

(10)

where $P_2=\frac{1-P_1}{2}$. It is easy to prove that $G_{Alice}$ achieves the maximum when $P_1=\frac{3}{2\eta +5}$. Denote $P_1=\frac{3}{2\eta +5}-ϵ$, where $ϵ$ is a small constant satisfying $ϵ>0$. It follows from Equation (10) that

$$\begin{array}{c}\hfill G_{Alice}=\frac{-{\eta }^2+4\eta +23}{{(2\eta +5)}^2}+O\left(ϵ\right)\end{array}$$

(11)

By using the same method for the rest of cases (see Appendix B for details), we can get

Table 3. $G_{Alice}^m$ denotes the maximum of $G_{Alice}$, and $P_{max}$ denotes the corresponding point of $P_1$.

$P_1$	$0\le P_1<\frac{3}{2\eta +5}$	$\frac{3}{2\eta +5}\le P_1\le 1$
$P_{max}$	$\frac{3}{2\eta +5}-ϵ$	$\frac{3}{2\eta +5}$
$G_{Alice}^m$	$\frac{-{\eta }^2+4\eta +23}{{(2\eta +5)}^2}+O\left(ϵ\right)$	$\frac{\eta +1}{2\eta +5}$

. From Table 3, we get that $\frac{-{\eta }^2+4\eta +23}{{(2\eta +5)}^2}+O\left(ϵ\right)>\frac{\eta +1}{2\eta +5}$ for $1\le \eta <2$, and $\frac{-{\eta }^2+4\eta +23}{{(2\eta +5)}^2}+O\left(ϵ\right)<\frac{\eta +1}{2\eta +5}$ for $\eta \ge 2$. Thus, Alice will choose $P_1=\frac{3}{2\eta +5}-ϵ$ to maximize her gain when $1\le \eta <2$, while $P_1=\frac{3}{2\eta +5}$ when $\eta \ge 2$.

2.2.4. Fair Zero-Sum Game

From Section 2.2.3, we get that $P_1=\frac{3}{2\eta +5}-ϵ$ for $1\le \eta <2$. By using induction we know that $P_{\diamond B}=1$ and $P_{\diamond C}=1$ from Table 2. From Equations (6), (8) and (9), the expressions of $G_{Alice}$, $G_{Bob}$ and $G_{Rice}$ with respect to $\eta$ are shown in

Table 4. The average gains of players in the first classical game ${\mathsf{G}}_1$.

Gain∖$\mathit{\eta }$	$\mathbf{1}\le \mathit{\eta }<\mathbf{2}$	$\mathit{\eta }\ge \mathbf{2}$
$G_{Alice}$	$\frac{-{\eta }^2+4\eta +23}{{(2\eta +5)}^2}+O\left(ϵ\right)$	$\frac{1+\eta }{2\eta +5}$
$G_{Bob}$	$\frac{-{\eta }^2-5\eta -13}{{(2\eta +5)}^2}+O\left(ϵ\right)$	$\frac{1-2\eta }{2\eta +5}$
$G_{Rice}$	$\frac{\eta -2}{2\eta +5}+O\left(ϵ\right)$	$\frac{\eta -2}{2\eta +5}$

.

Result 1 The tripartite game ${\mathsf{G}}_1$ is unfair for any $\eta \ge 1$.

Proof. 

The numeric evaluations of $G_{Alice}$, $G_{Bob}$ and $G_{Rice}$ are shown in Figure 3 for $\eta \le 100$ and $ϵ={10}^{-5}$. It shows that the tripartite classical game ${\mathsf{G}}_1$ is unfair. Formally, since the tripartite game ${\mathsf{G}}_1$ is a zero-sum game, i.e., the summation of the average gains of all players is zero. It is sufficient to prove that the gain of one player is strictly greater than that of the other. The proof is completed by two cases.

C1.

For $1\le \eta <2$, from Table 4, we get that

$$\begin{array}{c}\hfill G_{Alice}-G_{Bob}=\frac{9\eta +36}{{(2\eta +5)}^2}+O\left(ϵ\right)>0\end{array}$$

(12)

C2.

For $\eta \ge 2$, from Table 4, we obtain that

$$\begin{array}{c}\hfill G_{Alice}-G_{Bob}=\frac{3\eta }{2\eta +5}>0\end{array}$$

(13)

if $ϵ$ is very small. This completes the proof. □

2.3. The Second Tripartite Classical Game

In this section, we present the second game ${\mathsf{G}}_2$ with different settings shown in

Table 5. The gain settings of the second game ${\mathsf{G}}_2$. Here, $g_{c(b,a)}$ denotes gain of Rice (or Bob, or Alice). We assume that $\mu \ge 2$ in this second game.

Cases∖Gains	${\mathit{g}}_{\mathit{c}}$	${\mathit{g}}_{\mathit{b}}$	${\mathit{g}}_{\mathit{a}}$
${\mathsf{R}}_{succ}(\diamond C)$	$\mu$	−1	1 − $\mu$
${\mathsf{B}}_{succ}(\diamond B)$	−1	2	−1
$\mathsf{R}/{\mathsf{B}}_{fail}(·)$	−1	−1	2
$\mathsf{R}/{\mathsf{B}}_{succ}(\diamond A)$	1	1	−2

. The first game and the second game are the same except for the gain table, i.e., both games adopt the game model in Section 2.1. Similar to the first game ${\mathsf{G}}_1$, our goal is to prove its unfair.

Similar to Section 2.2, we can evaluate the gains of all parties, as shown in

Table 6. The average gains of players in the second classical tripartite game ${\mathsf{G}}_2$.

Gain∖$\mu$	$2\le \mu \le \frac{41+\sqrt{1345}}{14}$	$\mu >\frac{41+\sqrt{1345}}{14}$
$G_{Alice}$	$\frac{54-14\mu }{49}$	$\frac{6-2\mu }{\mu +5}$
$G_{Bob}$	$-\frac{19}{49}$	$\frac{\mu -3}{\mu +5}$
$G_{Rice}$	$\frac{2\mu -5}{7}$	$\frac{\mu -3}{\mu +5}$

. The details are shown in Appendix C, Appendix D and Appendix E. From Table 6, we can prove the following theorem.

Result 2 The tripartite classical game ${\mathsf{G}}_2$ is unfair for any $\mu \ge 2$.

Proof. 

The numeric evaluations of $G_{Alice}$, $G_{Bob}$ and $G_{Rice}$ are shown in Figure 4 for $\mu \le 100$. It shows that the tripartite classical game ${\mathsf{G}}_2$ is unfair. This can be proved formally as follows. From the assumption, the second classical game ${\mathsf{G}}_2$ is zero-sum. It is sufficient to prove that there is no $\mu$ such that $G_{Alice}$, $G_{Bob}$ and $G_{Rice}$ equal to zero. The proof is completed by two cases.

C1.

For $2\le \mu \le \frac{41+\sqrt{1345}}{14}$, from Table 6, we get that

$$\begin{array}{c}\hfill G_{Bob}=-\frac{19}{49}\ne 0\end{array}$$

(14)

C2.

For $\mu >\frac{41+\sqrt{1345}}{14}$, from Table 6, we obtain that

$$\begin{array}{c}\hfill G_{Bob}=\frac{\mu -3}{\mu +5}\ne 0\end{array}$$

(15)

This completes the proof. □

3. Zero-Sum Quantum Games

In this section, by quantizing the classical game shown in Figure 1, we get that there are also four stages in quantum game, as shown in Figure 5. The correspondence between the classical and quantum game are: the classical game is to put a ball into three ordinary black boxes, while the quantum game is to put a particle into three quantum boxes. In the classical games, Bob and Rice can selectively let Alice open the box $\diamond A$ to prevent Alice from putting the ball into the box $\diamond A$ so that Bob and Rice cannot find the ball. In quantum game, they can prevent this same problem by setting the committed state before the game, i.e., Alice, Bob and Rice agree on which state Alice should set the photon to. Alice has three quantum boxes, $\diamond A$, $\diamond B$ and $\diamond C$ used to store a photon. The state of the photon in the boxes are denoted $|a\rangle$, $|b\rangle$ and $|c\rangle$. The quantum game is given by the following four stages S1–S4.

S1.

Alice randomly puts the single photon into one of the three quantum boxes, $\diamond A$, $\diamond B$ and $\diamond C$. Alice sends the box $\diamond C$ to Rice, and sends $\diamond B$ to Bob.

S2.

Bob gives his own strategy, and he opens box $\diamond B$.

S3.

Rice chooses her own strategy and takes action, i.e., she opens $\diamond C$. If neither Rice nor Bob finds the photon, the game enters the fourth stage.

S4.

Bob(Rice) asks Alice to send him(her) box $\diamond A$ to verify whether the state of photon prepared by Alice is the same as the committed state.

Moreover, for any strategy combination, we assume that the total payment of all participants is zero and the latter participant can observe the former’s action. In the following, we will explore this kind of games with two settings with different gains.

The quantum game tree is shown in Figure 6. In the stages of $\mathsf{G}$, the wining rules of the game is given by

W1.

Rice wins if Rice finds the photon after opening $\diamond C$.

W2.

Bob wins if Bob finds the photon after opening $\diamond B$.

W3.

Bob and Rice win if neither Rice nor Bob finds the photon but Alice is not honest.

W4.

Alice wins if neither Rice nor Bob finds the photon and Alice is honest.

We consider the dynamic zero-sum quantum game with the same setting parameters given in Table 1 and Table 5, but the difference is that each symbol has a slightly different meaning, i.e., in quantum games, ${\mathsf{R}}_{succ}(\diamond C)$ means that Rice finds the photon after opening the box $\diamond C$. ${\mathsf{B}}_{succ}(\diamond B)$ means that Bob finds the photon after opening the box $\diamond B$. ${\mathsf{R}/\mathsf{B}}_{fail}(·)$ means that neither Rice nor Bob finds the photon and Alice is honest. ${\mathsf{R}/\mathsf{B}}_{succ}(\diamond A)$ means that neither Rice nor Bob finds the photon but Alice is not honest.

3.1. The Winning Rules of The Quantum Game

The winning rules of quantum game is similar to W1–W4. For convenience we define the following probabilities.

P1.

Denote ${\alpha }_1$ as the probability that Alice puts the photon into the box $\diamond A$.

P2.

Denote ${\alpha }_2$ as the probability that Alice puts the photon into box $\diamond B$ or box $\diamond C$.

P3.

Denote $1-\gamma$ as the probability that Rice chooses to open the box $\diamond C$ when she received $\diamond C$.

P4.

Denote $1-\beta$ as the probability that Bob chooses to open the box $\diamond B$ when he received $\diamond B$.

Similar to classical game shown in Figure 1, we have ${\alpha }_1+2{\alpha }_2=1$ and $\gamma ,\beta \in [0,1]$.

In quantum scenarios, the box of $\diamond A$, $\diamond B$, or $\diamond C$ is realized by a quantum state $|a\rangle ,|b\rangle$ or $|c\rangle$. The statement of one party finding the photon by opening one box ($\diamond A$ for example) means that one party find the photon in the state $|a\rangle$ after projection measurement under the basis $\left\{\right|a\rangle ,|b\rangle ,|c\rangle \}$. With these assumption, we get an experimental quantum game as follows.

Alice’s preparation. Alice prepares the single photon in the following supposition state

$$\begin{array}{c}\hfill |\psi \rangle =\sqrt{{\alpha }_1}|a\rangle +\sqrt{{\alpha }_2}|b\rangle +\sqrt{{\alpha }_2}|c\rangle \end{array}$$

(16)

where $|a\rangle ,|b\rangle$ and $|c\rangle$ can be realized by using different paths, ${\alpha }_1+2{\alpha }_2=1$ and ${\alpha }_1\in [0,1]$ is a parameter controlled by Alice.

Bob’s operation. Bob splits the box $\diamond B$ into box $\diamond B$ and box $\diamond B^{\prime }$ according the following transformation

$$\begin{array}{c}\hfill |b\rangle \mapsto \sqrt{1-\beta }|b\rangle +\sqrt{\beta }|b^{\prime }\rangle \end{array}$$

(17)

where $\beta \in [0,1]$ is a parameter controlled by Bob.

Rice’s operation. Rice splits the box $\diamond C$ into two parts $\diamond C$ and $\diamond C^{\prime }$ according to the following transformation

$$\begin{array}{c}\hfill |c\rangle \mapsto \sqrt{1-\gamma }|c\rangle +\sqrt{\gamma }|c^{\prime }\rangle \end{array}$$

(18)

where $\gamma \in [0,1]$ is a parameter controlled by Rice.

Similar to classical games, Alice may choose a large ${\alpha }_1$ to increase the probability of the photon appearing in box $\diamond A$, which will then reduce the probability of Bob and Rice finding the photon. However, this strategy may result in losing the game with a high probability for Alice in the verification stage. Similar intuitive analysis holds for others. Hence, it should be important to find reasonable parameters for them. We give the detailed process in the following.

Suppose that Alice prepares the photon in the following state

$$\begin{array}{c}\hfill |\psi \rangle =\sqrt{{\alpha }_1}|a\rangle +\sqrt{{\alpha }_2}|b\rangle +\sqrt{{\alpha }_2}|c\rangle \end{array}$$

(19)

using path encoding. When Bob (Rice) receives its box $\diamond B$($\diamond C$) and splits it into two parts, the final state of the photon is given by

$$\begin{array}{ccc}\hfill |{\psi }_0\rangle & =& \sqrt{{\alpha }_1}|a\rangle +\sqrt{{\alpha }_2(1-\beta )}|b\rangle +\sqrt{{\alpha }_2\beta }|b^{\prime }\rangle \hfill \\ & & +\sqrt{{\alpha }_2(1-\gamma )}|c\rangle +\sqrt{{\alpha }_2\gamma }|c^{\prime }\rangle \hfill \end{array}$$

(20)

From Equation (20), the probability that Rice finds the photon in $\diamond C$ (using single photon detector) is

$$\begin{array}{c}\hfill P_{Ric{e}_1}={\left|\langle c|{\psi }_0\rangle \right|}^2={\alpha }_2(1-\gamma )\end{array}$$

(21)

Moreover, the probability that Bob finds the photon in $\diamond B$ is given by

$$\begin{array}{c}\hfill P_{Bo{b}_1}={\left|\langle b|{\psi }_0\rangle \right|}^2={\alpha }_2(1-\beta )\end{array}$$

(22)

If neither Rice nor Bob finds the photon, the state in Equation (20) will collapse into

$$\begin{array}{ccc}\hfill |{\psi }_1\rangle & =& \sqrt{\frac{{\alpha }_1}{{\alpha }_1+{\alpha }_2\beta +{\alpha }_2\gamma }}|a\rangle +\sqrt{\frac{{\alpha }_2\beta }{{\alpha }_1+{\alpha }_2\beta +{\alpha }_2\gamma }}|b^{\prime }\rangle \hfill \\ & & +\sqrt{\frac{{\alpha }_2\gamma }{{\alpha }_1+{\alpha }_2\beta +{\alpha }_2\gamma }}|c^{\prime }\rangle \hfill \end{array}$$

(23)

If Alice did prepare the photon in the committed state $|\varphi \rangle =\sqrt{{\omega }_1}|a\rangle +\sqrt{{\omega }_2}|b\rangle +\sqrt{{\omega }_2}|c\rangle$ initially, it is easy to prove that the state at this stage should be

$$\begin{array}{ccc}\hfill |{\psi }_C\rangle & =& \sqrt{\frac{{\omega }_1}{{\omega }_1+{\omega }_2\beta +{\omega }_2\gamma }}|a\rangle +\sqrt{\frac{{\omega }_2\beta }{{\omega }_1+{\omega }_2\beta +{\omega }_2\gamma }}|b^{\prime }\rangle \hfill \\ & & +\sqrt{\frac{{\omega }_2\gamma }{{\omega }_1+{\omega }_2\beta +{\omega }_2\gamma }}|c^{\prime }\rangle \hfill \end{array}$$

(24)

where ${\omega }_1+2{\omega }_2=1$. By performing a projection measurement on $|{\psi }_1\rangle$, Rice or Bob gets an ideal state $|{\psi }_C\rangle$ with the probability

$$\begin{array}{c}\hfill |\langle {\psi }_1|{\psi }_C\rangle {|}^2=\frac{{\left(\sqrt{{\alpha }_1{\omega }_1}+x\sqrt{{\alpha }_2{\omega }_2}\right)}^2}{({\alpha }_1+{\alpha }_{2}x)({\omega }_1+{\omega }_{2}x)}\end{array}$$

(25)

where $x=\beta +\gamma$.

The probability that neither Rice nor Bob finds the photon when Alice did prepare the photon in the committed state $|\varphi \rangle$ is give by

$$\begin{array}{ccc}\hfill P_{Alice}& =& (1-P_{Ric{e}_1}-P_{Bo{b}_1}){\left|\langle {\psi }_1|{\psi }_C\rangle \right|}^2\hfill \\ & =& \frac{{(\sqrt{{\alpha }_1{\omega }_1}+x\sqrt{{\alpha }_2{\omega }_2})}^2}{{\omega }_1+{\omega }_{2}x}\hfill \end{array}$$

(26)

Moreover, the probability that neither Rice nor Bob finds the photon but Rice or Bob detects forge state prepared by Alice is denote by $P_{Ric{e}_2}$ or $P_{Bo{b}_2}$, which is given by

$$\begin{array}{ccc}\hfill P_{Ric{e}_2}& =& (1-P_{Ric{e}_1}-P_{Bo{b}_1})(1-|\langle {\psi }_1|{\psi }_C\rangle {|}^2)\hfill \\ & =& -\frac{{\left(\sqrt{{\alpha }_1{\omega }_1}+x\sqrt{{\alpha }_2{\omega }_2}\right)}^2}{{\omega }_1+{\omega }_{2}x}+{\alpha }_1+{\alpha }_{2}x\hfill \end{array}$$

(27)

with $P_{Ric{e}_2}=P_{Bo{b}_2}$ from winning rule W3.

3.2. The First Tripartite Quantum Game

In this subsection, we introduce the quantum implementation of the first game ${\mathsf{G}}_1$ with the gain setting shown in Table 1. Here, each symbol in Table 1 has a slightly different meaning, i.e., in quantum game, ${\mathsf{R}}_{succ}(\diamond C)$ means that Rice finds the photon after opening the box $\diamond C$. ${\mathsf{B}}_{succ}(\diamond B)$ means that Bob finds the photon after opening the box $\diamond B$. ${\mathsf{R}/\mathsf{B}}_{fail}(·)$ means that neither Rice or Bob finds the photon and Alice is honest. ${\mathsf{R}/\mathsf{B}}_{succ}(\diamond A)$ means that neither Rice or Bob finds the photon but Alice is not honest.

3.2.1. The Average Gain of Rice

Denote $G_{Rice}$ as Rice’s average gain. We can easily get $G_{Rice}$ according to Table 1 as follows

$$\begin{array}{ccc}\hfill G_{Rice}& =& 2P_{Ric{e}_1}+\eta P_{Ric{e}_2}-P_{Bo{b}_1}-P_{Alice}\hfill \\ & =& -\frac{(\eta +1){(\sqrt{{\alpha }_1{\omega }_1}+x\sqrt{{\alpha }_2{\omega }_2})}^2}{{\omega }_1+{\omega }_{2}x}\hfill \\ & & +(\eta -2){\alpha }_2\gamma +(\eta +1){\alpha }_2\beta +\eta {\alpha }_1+{\alpha }_2\hfill \end{array}$$

(28)

where $x=\beta +\gamma$.

The partial derivative of $G_{Rice}$ with respect to the variable $\gamma$ is given by

$$\begin{array}{ccc}\hfill \frac{\partial G_{Rice}}{\partial \gamma }& =& \frac{1}{{({\omega }_1+{\omega }_{2}x)}^2}(-3{\omega }_2^2{\alpha }_2{x}^2-6{\omega }_1{\omega }_2{\alpha }_{2}x\hfill \\ & & +(\eta -2){\omega }_1^2{\alpha }_2+(\eta +1){\omega }_1{\omega }_2{\alpha }_1\hfill \\ & & -2{\omega }_1(\eta +1)\sqrt{{\alpha }_1{\alpha }_2{\omega }_1{\omega }_2})\hfill \end{array}$$

(29)

Each participant has the perfect knowledge before the game reaching this stage, i.e., each participant is exactly aware of what previous participant has done. Hence, ${\omega }_1+{\omega }_{2}x\ne 0$. Let $\frac{\partial G_{Rice}}{\partial \gamma }=0$. We get

$$\begin{array}{ccc}\hfill x_1^{*}& =& -\frac{\sqrt{{\omega }_1(\eta +1)}\left|\sqrt{{\alpha }_1{\omega }_2}-\sqrt{{\alpha }_2{\omega }_1}\right|}{{\omega }_2\sqrt{3{\alpha }_2}}-\frac{{\omega }_1}{{\omega }_2}\hfill \\ \hfill x_2^{*}& =& \frac{\sqrt{{\omega }_1(\eta +1)}\left|\sqrt{{\alpha }_1{\omega }_2}-\sqrt{{\alpha }_2{\omega }_1}\right|}{{\omega }_2\sqrt{3{\alpha }_2}}-\frac{{\omega }_1}{{\omega }_2}\hfill \end{array}$$

(30)

If Alice chooses ${\alpha }_1<{\omega }_1$, the probability that Alice finds the photon will decrease while the probability that Rice or Bob detects the difference between the prepared and the committed states will increase in the verification stage. This means that Alice has no benefit if she chooses ${\alpha }_1<{\omega }_1$. It follows that ${\alpha }_1\ge {\omega }_1$ and ${\alpha }_2\le {\omega }_2$. Hence, from Equation (30) we get

$$\begin{array}{ccc}\hfill x_1^{*}& =& -\frac{\sqrt{{\omega }_1(\eta +1)}(\sqrt{{\alpha }_1{\omega }_2}-\sqrt{{\alpha }_2{\omega }_1})}{\sqrt{3}\sqrt{{\alpha }_2}{\omega }_2}-\frac{{\omega }_1}{{\omega }_2}\hfill \\ \hfill x_2^{*}& =& \frac{\sqrt{{\omega }_1(\eta +1)}(\sqrt{{\alpha }_1{\omega }_2}-\sqrt{{\alpha }_2{\omega }_1})}{\sqrt{3}\sqrt{{\alpha }_2}{\omega }_2}-\frac{{\omega }_1}{{\omega }_2}\hfill \end{array}$$

(31)

From Equation (31), we get $x_1^{*}\le 0$ and $x_2^{*}\ge x_1^{*}$. From Equations (30) and (31), $G_{Rice}$ is a decreasing function in x when $x\le x_1^{*}$ and increasing when $x_1^{*}<x\le x_2^{*}$. Moreover, when $x>x_2^{*}$, $G_{Rice}$ is a decreasing function in x. Since $0\le x\le 2$, $G_{Rice}$ gets the maximum value at $x=0$ for $x_2^{*}\le 0$, or gets the maximum value at $x=2$ for $x_2^{*}\ge 2$, or gets the maximum value at $x=x_2^{*}$ for $0<x_2^{*}<2$.

3.2.2. The Average Gain of Bob

Denote $G_{Bob}$ as the average gain of Bob. From Table 1, we get $G_{Bob}$ as

$$\begin{array}{ccc}\hfill G_{Bob}& =& 2P_{Bo{b}_1}+P_{Bo{b}_2}-P_{Ric{e}_1}-P_{Alice}\hfill \\ & =& -\frac{2{(\sqrt{{\alpha }_1{\omega }_1}+x\sqrt{{\alpha }_2{\omega }_2})}^2}{({\omega }_1+{\omega }_{2}x)}+{\alpha }_1+{\alpha }_2\hfill \\ & & -{\alpha }_2\beta +2{\alpha }_2\gamma \hfill \end{array}$$

(32)

where $x=\beta +\gamma$.

The partial derivative of $G_{Bob}$ with respect to $\beta$ is given by

$$\begin{array}{ccc}\hfill \frac{\partial G_{Bob}}{\partial \beta }& =& \frac{1}{{({\omega }_1+{\omega }_{2}x)}^2}(-3{\omega }_2^2{\alpha }_2{x}^2-6{\omega }_1{\omega }_2{\alpha }_{2}x\hfill \\ & & -{\omega }_1^2{\alpha }_2-4{\omega }_1\sqrt{{\alpha }_1{\alpha }_2{\omega }_1{\omega }_2}+2{\omega }_1{\omega }_2{\alpha }_1)\hfill \end{array}$$

(33)

Similar to Equations (30) and (31), we can get

$$\begin{array}{ccc}\hfill x_3^{*}& =& -\frac{\sqrt{2{\omega }_1}(\sqrt{{\alpha }_1{\omega }_2}-\sqrt{{\alpha }_2{\omega }_1})}{{\omega }_2\sqrt{3{\alpha }_2}}-\frac{{\omega }_1}{{\omega }_2}\hfill \\ \hfill x_4^{*}& =& \frac{\sqrt{2{\omega }_1}(\sqrt{{\alpha }_1{\omega }_2}-\sqrt{{\alpha }_2{\omega }_1})}{{\omega }_2\sqrt{3{\alpha }_2}}-\frac{{\omega }_1}{{\omega }_2}\hfill \end{array}$$

(34)

from $\frac{\partial G_{Bob}}{\partial \beta }=0$.

From Equation (34), we obtain $x_3^{*}\le 0$ and $x_4^{*}\ge x_3^{*}$. From Equations (33) and (34), $G_{Bob}$ decreases with $x\le x_3^{*}$ and increases with $x_3^{*}<x\le x_4^{*}$. Moreover, $G_{Bob}$ decreases with $x>x_4^{*}$. Since $0\le x\le 2$, $G_{Bob}$ gets the maximum value at $x=0$ for for $x_4^{*}\le 0$, or gets the maximum value at $x=2$ for $x_4^{*}\ge 2$, or gets the maximum value at $x=x_4^{*}$ for $0<x_4^{*}<2$. From $\eta \ge 1$, we get $x_2^{*}\ge x_4^{*}$.

Similarly, we can get the detailed analysis of Rice, shown in Appendix F. The values of ${\beta }^{*}$ and ${\gamma }^{*}$ which depend on $x_i^{*}$ in the first quantum game such that $G_{Bob}$ achieves the maximum are given in

Table 7. The values of ${\beta }^{*}$ and ${\gamma }^{*}$ depend on $x_i^{*}$ in the first quantum game such that $G_{Rice}$ and $G_{Bob}$ achieve the maximums, where ${\beta }^{*}$ and ${\gamma }^{*}$ are the probability of Bob opening box $\diamond B^{\prime }$ and Rice opening box $\diamond C^{\prime }$ respectively. Here, $x_1^{*}$ and $x_2^{*}$ are given in Equation (31), and $x_3^{*}$ and $x_4^{*}$ are given in Equation (34).

Cases	${\mathit{x}}_{\mathbf{2}}^{*}\le \mathbf{0}$	$\mathbf{0}<{\mathit{x}}_{\mathbf{2}}^{*}\le \mathbf{1}$	${\mathit{x}}_{\mathbf{4}}^{*}\le \mathbf{1}<{\mathit{x}}_{\mathbf{2}}^{*}$	$\mathbf{1}<{\mathit{x}}_{\mathbf{4}}^{*}<\mathbf{2}$	${\mathit{x}}_{\mathbf{4}}^{*}\ge \mathbf{2}$
${\beta }^{*}$	0	0	0	$x_4^{*}-1$	1
${\gamma }^{*}$	0	$x_2^{*}$	1	1	1

.

3.2.3. The Average Gain of Alice

Denote $G_{Alice}$ as the average gain of Alice. From Table 1, it is easy to evaluate $G_{Alice}$ as follows

$$\begin{array}{ccc}\hfill G_{Alice}& =& 2P_{Alice}-(\eta +1)P_{Ric{e}_2}-P_{Bo{b}_1}-P_{Ric{e}_1}\hfill \\ & =& (\eta +3)\frac{{(\sqrt{{\alpha }_1{\omega }_1}+x\sqrt{{\alpha }_2{\omega }_2})}^2}{{\omega }_1+{\omega }_{2}x}\hfill \\ & & -\eta {\alpha }_{2}x-1-\eta {\alpha }_1\hfill \end{array}$$

(35)

where $x=\beta +\gamma$.

From Table 7, we discuss Alice’s gain in five cases. Here, we only discuss one of them. Other cases are shown in Appendix G.

If $x_2^{*}\le 0$, i.e., $\sqrt{\frac{{\alpha }_1{\omega }_1(\eta +1)}{3{\alpha }_2{\omega }_2}}-\frac{{\omega }_1}{{\omega }_2}(\sqrt{\frac{\eta +1}{3}}+1)\le 0$, we get that

$$\begin{array}{c}\hfill {\omega }_1\le {\alpha }_1\le \frac{\frac{{\omega }_1}{{\omega }_2}{(\sqrt{\frac{\eta +1}{3}}+1)}^2}{\frac{{\omega }_1}{{\omega }_2}{(\sqrt{\frac{\eta +1}{3}}+1)}^2+\frac{2}{3}(\eta +1)}\end{array}$$

(36)

Denote

$$\begin{array}{c}\hfill D=\frac{\frac{{\omega }_1}{{\omega }_2}{(\sqrt{\frac{\eta +1}{3}}+1)}^2}{\frac{{\omega }_1}{{\omega }_2}{(\sqrt{\frac{\eta +1}{3}}+1)}^2+\frac{2}{3}(\eta +1)}\end{array}$$

It follows that ${\omega }_1\le {\alpha }_1\le D$. In this case, we get $\beta =\gamma =0$. From Equation (36), we get that the average gain of Alice is given by

$$\begin{array}{c}\hfill G_{Alice}=3{\alpha }_1-1\end{array}$$

(37)

Note that $\frac{d{G}_{Alice}}{d{\alpha }_1}=3$ from Equation (37). $G_{Alice}$ increases with ${\alpha }_1$ when ${\omega }_1\le {\alpha }_1\le D$. So, $G_{Alice}$ gets the maximum value at ${\alpha }_1=D$, which is denoted by $G_{Alic{e}_1}$ given by

$$\begin{array}{c}\hfill G_{Alic{e}_1}=3D-1\end{array}$$

(38)

where the corresponding point is denoted by $p_1$.

By using the same method for other cases (see Appendix G), we get

Table 8. $G_{Alice}^m$ denotes the maximum of $G_{Alice}$ where the corresponding point is denoted by $p_{max}$.

Cases	${\mathit{x}}_{\mathbf{2}}^{*}\le \mathbf{0}$	$\mathbf{0}<{\mathit{x}}_{\mathbf{2}}^{*}\le \mathbf{1}$	${\mathit{x}}_{\mathbf{4}}^{*}\le \mathbf{1}<{\mathit{x}}_{\mathbf{2}}^{*}$	$\mathbf{1}<{\mathit{x}}_{\mathbf{4}}^{*}<\mathbf{2}$	${\mathit{x}}_{\mathbf{4}}^{*}\ge \mathbf{2}$
$p_{max}$	$p_1$	$p_2$	$p_3$	$p_4$	$p_5$
$G_{Alice}^m$	$G_{Alic{e}_1}$	$G_{Alic{e}_2}$	$G_{Alic{e}_3}$	$G_{Alic{e}_4}$	$G_{Alic{e}_5}$

.

3.2.4. Quantum Fair Game

In this subsection, we prove the proposed quantum game is fair. From Equations (29) and (32) and Table 8 we get that

$$\begin{array}{ccc}\hfill G_{Alice}& =& max\{G_{Alic{e}_1},G_{Alic{e}_2},G_{Alic{e}_3},G_{Alic{e}_4},G_{Alic{e}_5}\}\hfill \\ \hfill G_{Bob}& =& -\frac{2{(\sqrt{{\alpha }_1{\omega }_1}+x\sqrt{{\alpha }_2{\omega }_2})}^2}{{\omega }_1+{\omega }_{2}x}\hfill \\ & & +2{\alpha }_2\gamma -{\alpha }_2\beta -{\alpha }_2+1\hfill \\ \hfill G_{Rice}& =& -\frac{(\eta +1){\left(\sqrt{{\alpha }_1{\omega }_1}+x\sqrt{{\alpha }_2{\omega }_2}\right)}^2}{{\omega }_1+{\omega }_{2}x}\hfill \\ & & +(\eta -2){\alpha }_2\gamma +\eta {\alpha }_2\beta +\eta {\alpha }_1+2{\alpha }_2\hfill \end{array}$$

(39)

If $G_{Alic{e}_1}=max\{G_{Alic{e}_1}$,$G_{Alic{e}_2}$,$G_{Alic{e}_3},G_{Alic{e}_4},G_{Alic{e}_5}\}$, from Table 8 we obtain that ${\alpha }_1=p_1$, and $\beta =\gamma =0$ from Table 7. The average gains of three players are evaluated using $\eta$ and ${\omega }_1$, as shown in Figure 7. Here, for each $\eta \ge 1$, the gains of Alice, Bob and Rice can tend to zero by changing ${\omega }_1$.

From Figure 8a,b, we get that the degree of deviation from the fair game is very small even if the game is not completely fair. The game converges to the fair game when $\eta \to \infty$. To sum up, we get the following theorem.

Result 3 The first quantum game ${\mathsf{G}}_1$ is asymptotically fair.

Proof. 

Note that the first quantum game ${\mathsf{G}}_1$ is zero-sum, i.e., the summation of the average gains of all players is zero. From Equation (36), Alice will make $|\langle {\psi }_1|{\psi }_C\rangle |=0$ when $\eta \to \infty$, i.e., ${\alpha }_1={\omega }_1$, in order to maximize her own gain, while Bob and Rice will choose $\beta =\gamma =0$ accordingly. Hence, we get three gains as follows

$$\begin{array}{ccc}\hfill G_{Alice}& =& 2{\omega }_1-2{\omega }_2\hfill \\ \hfill G_{Bob}& =& {\omega }_2-{\omega }_1\hfill \\ \hfill G_{Rice}& =& {\omega }_2-{\omega }_1\hfill \end{array}$$

(40)

Combining with the assumption of ${\omega }_1+2{\omega }_2=1$, we get ${\omega }_2={\omega }_1=\frac{1}{3}$ from $G_{Alice}=G_{Bob}=G_{Rice}=0$. Thus, when $\eta \to \infty$, the quantum game ${\mathsf{G}}_1$ converges a fair game, i.e., asymptotically fair. This completes the proof. □

3.3. The Second Tripartite Quantum Game

In this section, we give the quantum realization of the second game ${\mathsf{G}}_2$. According to Table 5, similar to the first quantum game, the gain of Rice is given by

$$\begin{array}{ccc}\hfill G_{Rice}& =& -2\frac{{(\sqrt{{\alpha }_1{\omega }_1}+x\sqrt{{\alpha }_2{\omega }_2})}^2}{{\omega }_1+{\omega }_{2}x}\hfill \\ & & +(\mu -1){\alpha }_2(1-\gamma )+{\alpha }_1+2{\alpha }_2\beta \hfill \end{array}$$

(41)

where $x=\beta +\gamma$. The detailed maximizing $G_{Rice}$ is shown in Appendix H.

Similarly, the gain of Bob is given by

$$\begin{array}{ccc}\hfill G_{Bob}& =& -2\frac{{(\sqrt{{\alpha }_1{\omega }_1}+x\sqrt{{\alpha }_2{\omega }_2})}^2}{({\omega }_1+{\omega }_{2}x)}+{\alpha }_1\hfill \\ & & +{\alpha }_2-{\alpha }_2\beta +2{\alpha }_2\gamma \hfill \end{array}$$

(42)

Its maximization is shown in Appendix I.

The gain of Alice is given by

$$\begin{array}{ccc}\hfill G_{Alice}& =& \frac{4{(\sqrt{{\alpha }_1{\omega }_1}+x\sqrt{{\alpha }_2{\omega }_2})}^2}{{\omega }_1+{\omega }_{2}x}-\mu {\alpha }_2\hfill \\ & & +(\mu -3){\alpha }_2\gamma -2{\alpha }_1-{\alpha }_2\beta \hfill \end{array}$$

(43)

Its maximization is shown in Appendix J.

The numeric evaluations of average gains are shown in Figure 9 in terms of $\mu$ and ${\omega }_1$. For each $\mu$, we can make the gains of Alice, Bob and Rice equal to zero as much as possible by adjusting ${\omega }_1$. The deviation degree is shown in Figure 10a. The relationship between the appropriate ${\omega }_1$ and $\mu$ is shown in Figure 10b. From Figure 10a,b, the degree of deviation from the fair game is very small even if the game is not completely fair. The game converges to the fair game when $\mu \to \infty$. To sum up, we get the following result.

Result 4 The second quantum game ${\mathsf{G}}_2$ is asymptotically fair.

Proof. 

Note that the second quantum game ${\mathsf{G}}_2$ is zero-sum, i.e., the summation of the average gains of all players is zero. From Equation (46), Alice will make ${\alpha }_2=0$ when $\mu \to \infty$, i.e., ${\alpha }_1=1$, in order to maximize her own gain, while Bob and Rice will choose $\beta =\gamma =1$ accordingly. Hence, we get three gains as follows

$$\begin{array}{ccc}\hfill G_{Alice}& =& 4{\omega }_1-2\hfill \\ \hfill G_{Bob}& =& 1-2{\omega }_1\hfill \\ \hfill G_{Rice}& =& 1-2{\omega }_1\hfill \end{array}$$

(44)

We get ${\omega }_1=\frac{1}{2}$ and ${\omega }_2=\frac{1}{4}$ from $G_{Alice}=G_{Bob}=G_{Rice}=0$. Thus, when $\mu \to \infty$, the quantum game ${\mathsf{G}}_2$ converges a fair game, i.e., asymptotically fair. This completes the proof. □

4. Quantum Game with Noises

In this section, we consider quantum games with noises. One is from the experimental measurement. The other is from the preparation noise of resource state.

4.1. Experimental Measurement Error

In the case of measurement error, from Equation (20), we can get that the probability of Rice opens box $\diamond C$ and finds the photon becomes

$$\begin{array}{c}\hfill P_{Ric{e}_1}={\left|\langle c|{\psi }_0\rangle \right|}^2+\epsilon ={\alpha }_2(1-\gamma )+\epsilon \end{array}$$

(45)

where $\epsilon$ is measurement error which may be very small.

From Equation (20), we get the probability that Bob opens box $\diamond B$ and finds the photon is given by

$$\begin{array}{c}\hfill P_{Bo{b}_1}={\left|\langle b|{\psi }_0\rangle \right|}^2+\epsilon ={\alpha }_2(1-\beta )+\epsilon \end{array}$$

(46)

Rice or Bob makes a projection measurement on $|{\psi }_C\rangle$ for the verification the final state with success probability

$$\begin{array}{c}\hfill |\langle {\psi }_1|{\psi }_C\rangle {|}^2=\frac{(\sqrt{{\alpha }_1{\omega }_1}+x\sqrt{{\alpha }_2{\omega }_2})}{({\alpha }_1+{\alpha }_{2}x)({\omega }_1+{\omega }_{2}x)}+\epsilon \end{array}$$

(47)

where $x=\beta +\gamma$.

$P_{Alice}$ is the probability that neither Rice nor Bob finds the photon and Alice did prepare the photon in the committed state, is given by

$$\begin{array}{ccc}\hfill P_{Alice}& =& (1-P_{Ric{e}_1}-P_{Bo{b}_1}){\left|\langle {\psi }_1|{\psi }_C\rangle \right|}^2\hfill \\ & =& \frac{{(\sqrt{{\alpha }_1{\omega }_1}+x\sqrt{{\alpha }_2{\omega }_2})}^2}{{\omega }_1+{\omega }_{2}x}-2\epsilon {\lambda }_4\hfill \\ & & +\epsilon {\lambda }_3-2{\epsilon }^2\hfill \end{array}$$

(48)

where ${\lambda }_i$ are given by

$$\begin{array}{ccc}\hfill {\lambda }_3& =& {\alpha }_1+{\alpha }_{2}x\le 1\hfill \\ \hfill {\lambda }_4& =& \frac{{(\sqrt{{\alpha }_1{\omega }_1}+x\sqrt{{\alpha }_2{\omega }_2})}^2}{({\omega }_1+{\omega }_{2}x)({\alpha }_1+{\alpha }_{2}x)}\le 1\hfill \end{array}$$

Since ${\lambda }_3\le 1$ and ${\lambda }_4\le 1$, and $\epsilon$ is very small, denotes $O\left(\epsilon \right)=-2\epsilon {\lambda }_4+\epsilon {\lambda }_3-2{\epsilon }^2$, which can be treated as measurement error. Thus Equation (49) can be rewritten as

$$\begin{array}{c}\hfill P_{Alice}=\frac{{(\sqrt{{\alpha }_1{\omega }_1}+x\sqrt{{\alpha }_2{\omega }_2})}^2}{{\omega }_1+{\omega }_{2}x}+O\left(\epsilon \right)\end{array}$$

(49)

The probability that neither Rice nor Bob finds the photon but Rice or Bob detects the forage preparation of Alice is denoted by $P_{Ric{e}_2}$ or $P_{Bo{b}_2}$ which given by

$$\begin{array}{ccc}\hfill P_{Ric{e}_2}& =& (1-P_{Ric{e}_1}-P_{Bo{b}_1})(1-|\langle {\psi }_1|{\psi }_C\rangle {|}^2)\hfill \\ & =& \frac{-{(\sqrt{{\alpha }_1{\omega }_1}+x\sqrt{{\alpha }_2{\omega }_2})}^2}{{\omega }_1+{\omega }_{2}x}+{\alpha }_1+{\alpha }_{2}x-2\epsilon -O\left(\epsilon \right)\hfill \\ & =& \frac{-{(\sqrt{{\alpha }_1{\omega }_1}+x\sqrt{{\alpha }_2{\omega }_2})}^2}{{\omega }_1+{\omega }_{2}x}+{\alpha }_1+{\alpha }_{2}x+O\left(\epsilon \right)\hfill \end{array}$$

(50)

where $P_{Ric{e}_2}=P_{Bo{b}_2}$ from the winning rule W4.

From Equations (21) and (22), Equations (25)–(27), Equations (45)–(47) and Equations (49) and (50), we get the present quantum games are also asymptotically fair if the measurement error is small enough.

4.2. White Noises

In this subsection, we consider that Alice prepares a noisy photon in the state

$$\begin{array}{c}\hfill {\rho }_0=v|\psi \rangle \langle \psi |+\frac{1-v}{3}I\end{array}$$

(51)

where $I=|a\rangle \langle a|+|b\rangle \langle b|+|c\rangle \langle c|$ denotes the identity operator, $|\psi \rangle$ is given in Equation (16), and $v\in [0,1]$. After Bob’s and Rice’s splitting operation, the state of the photon becomes

$$\begin{array}{ccc}\hfill {\rho }_1& =& v|{\psi }_0\rangle \langle {\psi }_0|+\frac{1-v}{3}\left(\right|a\rangle \langle a|+\left(\sqrt{1-\beta }\right|b\rangle \hfill \\ & & +\sqrt{\beta }|b^{\prime }\rangle \left)\right(\sqrt{1-\beta }\langle b|+\sqrt{\beta }\langle b^{\prime }\left|\right)+\left(\sqrt{1-\gamma }\right|c\rangle \hfill \\ & & +\sqrt{\gamma }|c^{\prime }\rangle \left)\right(\sqrt{1-\gamma }\langle c|+\sqrt{\gamma }\langle c^{\prime }\left|\right))\hfill \end{array}$$

(52)

where $|{\psi }_0\rangle$ is given by

$$\begin{array}{ccc}\hfill |{\psi }_0\rangle & =& \sqrt{{\alpha }_1}|a\rangle +\sqrt{{\alpha }_2(1-\beta )}|b\rangle +\sqrt{{\alpha }_2\beta }|b^{\prime }\rangle \hfill \\ & & +\sqrt{{\alpha }_2(1-\gamma )|c\rangle }+\sqrt{{\alpha }_2\gamma }|c^{\prime }\rangle \hfill \end{array}$$

(53)

Denote $P_{Ric{e}_1}$ as the probability that Rice finds the photon after opening the box $\diamond C$. From Equation (53) it is given by

$$\begin{array}{c}\hfill P_{Ric{e}_1}=v{\alpha }_2(1-\gamma )+\frac{1-v}{3}(1-\gamma )\end{array}$$

(54)

Denote $P_{Bo{b}_1}$ as the probability that Bob finds the photon after opening the box $\diamond B$. From Equation (53) it is given by

$$\begin{array}{c}\hfill P_{Bo{b}_1}=v{\alpha }_2(1-\beta )+\frac{1-v}{3}(1-\beta )\end{array}$$

(55)

For the case that neither Bob nor Rice detects the photon, the density operator for the photon is given by

$$\begin{array}{ccc}\hfill {\rho }_2& =& \frac{v}{1-P_{Bo{b}_1}-P_{Ric{e}_1}}\left[\begin{array}{ccccc}{\alpha }_1& \sqrt{{\alpha }_1{\alpha }_2(1-\beta )}& \sqrt{{\alpha }_1{\alpha }_2\beta }& \sqrt{{\alpha }_1{\alpha }_2(1-\gamma )}& \sqrt{{\alpha }_1{\alpha }_2\gamma }\\ \sqrt{{\alpha }_1{\alpha }_2(1-\beta )}& 0& {\alpha }_2\sqrt{\beta (1-\beta )}& {\alpha }_2\sqrt{(1-\beta )(1-\gamma )}& {\alpha }_2\sqrt{(1-\beta )\gamma }\\ \sqrt{{\alpha }_1{\alpha }_2\beta }& {\alpha }_2\sqrt{\beta (1-\beta )}& {\alpha }_2\beta & {\alpha }_2\sqrt{\beta (1-\gamma )}& {\alpha }_2\sqrt{\beta \gamma }\\ \sqrt{{\alpha }_1{\alpha }_2(1-\gamma )}& {\alpha }_2\sqrt{(1-\beta )(1-\gamma )}& {\alpha }_2\sqrt{\beta (1-\gamma )}& 0& {\alpha }_2\sqrt{\gamma (1-\gamma )}\\ \sqrt{{\alpha }_1{\alpha }_2\gamma }& {\alpha }_2\sqrt{(1-\beta )\gamma }& {\alpha }_2\sqrt{\beta \gamma }& {\alpha }_2\sqrt{\gamma (1-\gamma )}& {\alpha }_2\gamma \end{array}\right]\hfill \\ & & +\frac{1-v}{3(1-P_{Bo{b}_1}-P_{Ric{e}_1})}\left[\begin{array}{ccccc}1& 0& 0& 0& 0\\ 0& 0& \sqrt{\beta (1-\beta )}& 0& 0\\ 0& \sqrt{\beta (1-\beta )}& \beta & 0& 0\\ 0& 0& 0& 0& \sqrt{\gamma (1-\gamma )}\\ 0& 0& 0& \sqrt{\gamma (1-\gamma )}& 1-\gamma \end{array}\right]\hfill \end{array}$$

(56)

Now, Rice or Bob makes a projection measurement with positive operator $\left\{\right|{\psi }_C\rangle \langle {\psi }_C|,I-|{\psi }_C\rangle \langle {\psi }_C\left|\right\}$ on the photon for verifying the committed state of Alice with success probability

$$\begin{array}{ccc}\hfill \langle {\psi }_C\left|{\rho }_2\right|{\psi }_C\rangle & =& \frac{v{\left(\sqrt{{\alpha }_1{\omega }_1}+x\sqrt{{\alpha }_2{\omega }_2}\right)}^2}{({\omega }_1+{\omega }_{2}x)(1-P_{Bo{b}_1}-P_{Ric{e}_1})}\hfill \\ & & +\frac{1-v}{3}\frac{{\omega }_1+{\omega }_2{\beta }^2+{\omega }_2{\gamma }^2}{({\omega }_1+{\omega }_{2}x)(1-P_{Bo{b}_1}-P_{Ric{e}_1})}\hfill \\ \hfill \end{array}$$

(57)

Denote $P_{Alice}$ as the probability that neither Rice nor Bob finds the photon, and Alice did prepare the photon in the committed state. It is easy to obtain that

$$\begin{array}{ccc}\hfill P_{Alice}& =& (1-P_{Ric{e}_1}-P_{Bo{b}_1})\langle {\psi }_C|{\rho }_2|{\psi }_C\rangle \hfill \\ & =& \frac{v{(\sqrt{{\alpha }_1{\omega }_1}+x\sqrt{{\alpha }_2{\omega }_2})}^2}{{\omega }_1+{\omega }_{2}x}\hfill \\ & & +\frac{(1-v)({\omega }_1+{\omega }_2{\beta }^2+{\omega }_2{\gamma }^2)}{3({\omega }_1+{\omega }_{2}x)}\hfill \end{array}$$

(58)

Denote $P_{Ric{e}_2}$ or $P_{Bo{b}_2}$ as the probability that neither Rice nor Bob finds the photon but Rice or Bob detects the forage preparation. We obtain that

$$\begin{array}{ccc}\hfill P_{Ric{e}_2}& =& (1-P_{Ric{e}_1}-P_{Bo{b}_1})(1-\langle {\psi }_C|{\rho }_2|{\psi }_C\rangle )\hfill \\ & =& 1-P_{Ric{e}_1}-P_{Bo{b}_1}-P_{Alice}\hfill \end{array}$$

(59)

where $P_{Ric{e}_2}=P_{Bo{b}_2}$ from the winning rule W4.

Take the first quantum game as an example. The Rice’s average gain $G_{Rice}$ is given by

$$\begin{array}{ccc}\hfill G_{Rice}& =& 2P_{Ric{e}_1}+\eta P_{Ric{e}_2}-P_{Bo{b}_1}-P_{Alice}\hfill \\ & =& (v{\alpha }_2+\frac{1-v}{3})(2-\eta )(1-\gamma )\hfill \\ & & -(v{\alpha }_2+\frac{1-v}{3})(\eta +1)(1-\beta )+\eta \hfill \\ & & -(\eta +1)(v\frac{{(\sqrt{{\alpha }_1{\omega }_1}+x\sqrt{{\alpha }_2{\omega }_2})}^2}{{\omega }_1+{\omega }_{2}x}\hfill \\ & & +\frac{1-v}{3}\frac{{\omega }_1+{\omega }_2{\beta }^2+{\omega }_2{\gamma }^2}{{\omega }_1+{\omega }_{2}x})\hfill \end{array}$$

(60)

The partial derivative of $G_{Rice}$ with respect to $\gamma$ is

$$\begin{array}{ccc}\hfill \frac{\partial G_{Rice}}{\partial \gamma }& =& \frac{v}{{({\omega }_1+{\omega }_{2}x)}^2}(-3{\omega }_2^2{\alpha }_2{x}^2-6{\omega }_1{\omega }_2{\alpha }_{2}x\hfill \\ & & +(\eta -2){\omega }_1^2{\alpha }_2+(\eta +1){\omega }_1{\omega }_2{\alpha }_1\hfill \\ & & -2{\omega }_1(\eta +1)\sqrt{{\alpha }_1{\alpha }_2{\omega }_1{\omega }_2})+\frac{1-v}{3}ϵ\hfill \end{array}$$

(61)

where $ϵ=-\frac{(\eta +1)(2{\omega }_1{\omega }_2\gamma +2{\omega }_2^2\beta \gamma +{\omega }_2^2{\gamma }^2-{\omega }_1{\omega }_2-{\omega }_2^2{\beta }^2)}{{({\omega }_1+{\omega }_{2}x)}^2}+\eta -2$.

Assume that v is very close to one, i.e., $\frac{1-v}{3}\approx 0$. It follows that $ϵ$ is bounded. Thus Equation (62) can be rewritten as

$$\begin{array}{ccc}\hfill \frac{\partial G_{Rice}}{\partial \gamma }& =& \frac{v}{{({\omega }_1+{\omega }_{2}x)}^2}(-3{\omega }_2^2{\alpha }_2{x}^2-6{\omega }_1{\omega }_2{\alpha }_{2}x\hfill \\ & & +(\eta -2){\omega }_1^2{\alpha }_2+(\eta +1){\omega }_1{\omega }_2{\alpha }_1\hfill \\ & & -2{\omega }_1(\eta +1)\sqrt{{\alpha }_1{\alpha }_2{\omega }_1{\omega }_2})+O\left(ϵ\right)\hfill \end{array}$$

(62)

Similarly, we get

$$\begin{array}{ccc}\hfill G_{Bob}& =& (v{\alpha }_2+\frac{1-v}{3})(1-\beta )\hfill \\ & & -2(v{\alpha }_2+\frac{1-v}{3})(1-\gamma )+1\hfill \\ & & -\frac{2v{(\sqrt{{\alpha }_1{\omega }_1}+x\sqrt{{\alpha }_2{\omega }_2})}^2}{{\omega }_1+{\omega }_{2}x}\hfill \\ & & -\frac{2(1-v)({\omega }_1+{\omega }_2{\beta }^2+{\omega }_2{\gamma }^2)}{3({\omega }_1+{\omega }_{2}x)}\hfill \end{array}$$

(63)

The partial derivative of $G_{Bob}$ with respect to $\beta$ is given by

$$\begin{array}{ccc}\hfill \frac{\partial G_{Bob}}{\partial \beta }& =& \frac{v}{{({\omega }_1+{\omega }_{2}x)}^2}(-3{\omega }_2^2{\alpha }_2{x}^2-6{\omega }_1{\omega }_2{\alpha }_{2}x\hfill \\ & & -{\omega }_1^2{\alpha }_2-4{\omega }_1\sqrt{{\alpha }_1{\alpha }_2{\omega }_1{\omega }_2}+2{\omega }_1{\omega }_2{\alpha }_1)\hfill \\ & & +\frac{1-v}{3}ϵ\hfill \\ & =& \frac{v}{{({\omega }_1+{\omega }_{2}x)}^2}(-3{\omega }_2^2{\alpha }_2{x}^2-6{\omega }_1{\omega }_2{\alpha }_{2}x\hfill \\ & & -{\omega }_1^2{\alpha }_2-4{\omega }_1\sqrt{{\alpha }_1{\alpha }_2{\omega }_1{\omega }_2}\hfill \\ & & +2{\omega }_1{\omega }_2{\alpha }_1)+O\left(ϵ\right)\hfill \end{array}$$

(64)

where $ϵ=-\frac{4{\omega }_1{\omega }_2\gamma +4{\omega }_2^2\beta \gamma +2{\omega }_2^2{\gamma }^2-2{\omega }_1{\omega }_2-2{\omega }_2^2{\beta }^2}{{({\omega }_1+{\omega }_{2}x)}^2}-1$.

The Alice’s average gain $G_{Alice}$ is given by

$$\begin{array}{ccc}\hfill G_{Alice}& =& \frac{v(\eta +3){(\sqrt{{\alpha }_1{\omega }_1}+x\sqrt{{\alpha }_2{\omega }_2})}^2}{{\omega }_1+{\omega }_{2}x}\hfill \\ & & +\eta v{\alpha }_2(2-x)-\eta -1\hfill \\ & & +\frac{1-v}{3}(\frac{(\eta +3)({\omega }_1+{\omega }_2{\beta }^2+{\omega }_2{\gamma }^2)}{{\omega }_1+{\omega }_{2}x}\hfill \\ & & +\eta (2-x))\hfill \end{array}$$

(65)

$$\begin{array}{ccc}& & =\frac{v(\eta +3){(\sqrt{{\alpha }_1{\omega }_1}+x\sqrt{{\alpha }_2{\omega }_2})}^2}{{\omega }_1+{\omega }_{2}x}+v\eta -\eta \hfill \\ & & -v\eta {\alpha }_{2}x-v\eta {\alpha }_1-1+O\left(\epsilon \right)\hfill \end{array}$$

(66)

From Equations (30), (33), (36), (63), (65) and (66), we get that the first quantum game is asymptotically fair if white noisy is small enough, i.e., v is close to 1.

5. Conclusions

It has shown that two present quantum games are asymptotically fair. Interestingly, these games can be easily changed to biased versions from Figure 5 and Figure 8, by choosing different $\eta$ and ${\omega }_1$. These kind of schemes may be applicable in gambling theory. Similar to bipartite scheme [27], a proof-of-principle optical demonstration may be followed for each scheme.

In this paper, we present one tripartite zero-sum game with different settings. This game is unfair if all parties use of classical resources. Interestingly, this can be resolved by using only pure state in similar quantum games. Comparing with the classical games, the present quantum games provide asymptotically fair. Moreover, these quantum games are robust against the measurement errors and preparation noises. This kind of protocols provide interesting features of pure state in resolving specific tasks. The present examples may be extended for multipartite games in theory. Unfortunately, these extensions should be nontrivial because of high complexity depending on lots of free parameters.