Tripartite Dynamic Zero-Sum Quantum Games

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.

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 Here, we assume that Alice h the ball into the box ⋄B or ⋄C 1, with P ⋄B , P ⋄C ∈ [0, 1].
From winning rules W1-W winning probability P Rice 1 f finds the ball after opening ⋄ P Rice 1 = P Moreover, the winning proba Bob finds the ball after open P Bob 1 = P 2 P ⋄C P ⋄B For Alice, there are three sub winning. It follows that P Alice = 1 − P 1 + P ⋄B P ⋄C +P ⋄B P ⋄C P 1 + P The winning probability for B The same result holds for R P Bob 2 .  The classical game tree. The rectangle with color represents the black box with the ball while the rectangle without color represents an entity, i.e., Alice, Bob and Rice. The diamond represents an operation. N means no ball being found, and Y means the ball being found. Now, for convenience, we define the following probabilities.
P1. Denote P 1 as the probability that Alice puts the ball into A. P2. Denote P 2 as the probability that Alice puts the ball into B or C. Here, we assume that Alice has equal probability to put the ball into B or C. It follows that P 1 + 2P 2 = 1, with P B , P C ∈ [0, 1].
From winning rules W1-W7 of G, it is easy to get the winning probability P Rice 1 for Rice from W3 (i.e., Rice finds the ball after opening C) is given by Moreover, the winning probability for Bob from W6 (i.e., Rice chooses to open C and does not find the ball, but Bob finds the ball after opening B) is given by For Alice, there are three subcases W2, W5 and W7 for winning. It follows that The winning probability for Bob from W1 and W4 is given by The same result holds for Rice from W1 and W4, i.e., P Rice 2 = P Bob 2 .
Here, we analyze players' strategies for the present game 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 B . So, he should consider a known parameter P 1 given by Alice and an unknown parameter P 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.

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 G 1 with the gain setting given in Table 1. Our goal is to show the no-fairness of this game with classical resources. Table 1. The gain settings of players in the first classical game G 1 . Here, g c(b,a) denotes the gain of Rice (or Bob, or Alice). R succ ( C) means that Rice finds the ball after opening box C. B succ ( B) means that Bob finds the ball after opening box B. R/B f ail (·) means that neither Rice or Bob finds the ball. R/B succ ( A) means that either Rice or Bob successes by opening box A. η ≥ 1 in the first classical game.

Cases\ Gains
From Equation (6), we get that the partial derivative of G Rice with respect to P C is given by From Equation (6), if ∂G Rice ∂P C < 0, G Rice is a deceasing function. In this case, Rice has to set P C = 0 to maximize her gains. If ∂G Rice ∂P C > 0, i.e., G Rice is increasing function, Rice will choose P C = 1 to maximize her gains. Moreover, when ∂G Rice ∂P C = 0, i.e., G Rice is a constant, P C can be any probability.

The Average Gain of Bob
Similar to Equation (6), we get that the average gain of Bob is given by 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.
All the results are given in Table 2. Table 2. The values of P B and P C . P B( C) depends on the different cases in the game, where . Here, ∆ i denotes the intervals given by 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.
For the case of 0 ≤ P 1 < 3 2η+5 , it follows from Table 2 that P C = 1 and P B = 1. Equation (9) is then rewritten into where P 2 = 1−P 1 2 . It is easy to prove that G Alice achieves the maximum when P 1 = 3 2η+5 .
where is a small constant satisfying > 0. It follows from Equation (10) that By using the same method for the rest of cases (see Appendix B for details), we can get Table 3. From Table 3, we get that Thus, Alice will choose P 1 = 3 2η+5 − to maximize her gain when 1 ≤ η < 2, while P 1 = 3 2η+5 when η ≥ 2. Table 3. G m Alice denotes the maximum of G Alice , and P max denotes the corresponding point of P 1 .

Fair Zero-Sum Game
From Section 2.2.3, we get that P 1 = 3 2η+5 − for 1 ≤ η < 2.. By using induction we know that P B = 1 and P C = 1 from Table 2. From Equations (6), (8) and (9), the expressions of G Alice , G Bob and G Rice with respect to η are shown in Table 4.
Result 1 The tripartite game G 1 is unfair for any η ≥ 1.
Proof. The numeric evaluations of G Alice , G Bob and G Rice are shown in Figure 3 for η ≤ 100 and = 10 −5 . It shows that the tripartite classical game G 1 is unfair. Formally, since the tripartite game 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 ≤ η < 2, from Table 4, we get that C2. For η ≥ 2, from Table 4, we obtain that if is very small. This completes the proof. The gains of players Figure 3. The average gains of three parties in the first classical game G 1 . In this simulation, we assume η ≤ 100 and = 10 −5 . The gain of Alice is strictly larger than the gains of Bob and Rice for any η ≥ 1.

The Second Tripartite Classical Game
In this section, we present the second game G 2 with different settings shown in Table 5. 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 G 1 , our goal is to prove its unfair. Table 5. The gain settings of the second game G 2 . Here, g c(b,a) denotes gain of Rice (or Bob, or Alice). We assume that µ ≥ 2 in this second game.

Cases\ Gains
Similar to Section 2.2, we can evaluate the gains of all parties, as shown in Table 6. The details are shown in Appendixes C-E. From Table 6, we can prove the following theorem.
Result 2 The tripartite classical game G 2 is unfair for any µ ≥ 2.
Proof. The numeric evaluations of G Alice , G Bob and G Rice are shown in Figure 4 for µ ≤ 100. It shows that the tripartite classical game G 2 is unfair. This can be proved formally as follows. From the assumption, the second classical game G 2 is zero-sum. It is sufficient to prove that there is no µ such that G Alice , G Bob and G Rice equal to zero. The proof is completed by two cases. , from Table 6, we obtain that This completes the proof.  Here, µ ≤ 100. G Alice and G Bob and G Rice have no common intersection. Moreover, when µ > µ 1 , G Bob and G Rice coincide, but do not intersect with G Alice .

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 Entropy 2021, 23, 154 9 of 40 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 A to prevent Alice from putting the ball into the box 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, A, B and C used to store a photon. The state of the photon in the boxes are denoted |a , |b and |c . 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, A, B and C. Alice sends the box C to Rice, and sends B to Bob. S2. Bob gives his own strategy, and he opens box B. S3. Rice chooses her own strategy and takes action, i.e., she opens 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 A to verify whether the state of photon prepared by Alice is the same as the committed state . In the stages of G, the wining rules of the game is given by W1. Rice win if Rice finds the particle after opening ⋄C.
W2. Bob win if Bob finds the particle after opening ⋄B.
W3. Bob and Rice win if neither Rice nor Bob finds the particle but Alice is not honest.
W4. Alice wins if neither Rice nor Bob finds the particle and Alice is honest.
FIG. 4: The schematic model of tripartite dynamic zero-sum quantum game. Here, three parties use a single particle state to complete the game while the classical box is used in the game given in Fig.1.
We consider the dynamic zero-sum quantum games with the same setting parameters given in Tables I and V, but the difference is that each symbol has a slightly different 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 G, the wining rules of the game is given by W1. Rice wins if Rice finds the photon after opening C. W2. Bob wins if Bob finds the photon after opening 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. Figure 6. The quantum game tree. The rectangle with color represents the black box with the photon while the rectangle without color represents an entity Alice. The diamond represents an operation. N means no photon being found, and Y means the photon being found.
We consider the dynamic zero-sum quantum game with the same setting parameters given in Tables 1 and 5, but the difference is that each symbol has a slightly different meaning, i.e., in quantum games, R succ ( C) means that Rice finds the photon after opening the box C. B succ ( B) means that Bob finds the photon after opening the box B. R/B f ail (·) means that neither Rice nor Bob finds the photon and Alice is honest. R/B succ ( A) means that neither Rice nor Bob finds the photon but Alice is not honest.

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 α 1 as the probability that Alice puts the photon into the box A. P2. Denote α 2 as the probability that Alice puts the photon into box B or box C. Similar to classical game shown in Figure 1, we have α 1 + 2α 2 = 1 and γ, β ∈ [0, 1]. In quantum scenarios, the box of A, B, or C is realized by a quantum state |a , |b or |c . The statement of one party finding the photon by opening one box ( A for example) means that one party find the photon in the state |a after projection measurement under the basis {|a , |b , |c }. With these assumption, we get an experimental quantum game as follows.
Alice's preparation. Alice prepares the single photon in the following supposition state where |a , |b and |c can be realized by using different paths, α 1 + 2α 2 = 1 and α 1 ∈ [0, 1] is a parameter controlled by Alice. Bob's operation. Bob splits the box B into box B and box B according the following transformation where β ∈ [0, 1] is a parameter controlled by Bob. Rice's operation. Rice splits the box C into two parts C and C according to the following transformation where γ ∈ [0, 1] is a parameter controlled by Rice. Similar to classical games, Alice may choose a large α 1 to increase the probability of the photon appearing in box 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 using path encoding. When Bob (Rice) receives its box B( C) and splits it into two parts, the final state of the photon is given by From Equation (20), the probability that Rice finds the photon in C (using single photon detector) is Moreover, the probability that Bob finds the photon in B is given by If neither Rice nor Bob finds the photon, the state in Equation (20) will collapse into If Alice did prepare the photon in the committed state |φ = √ ω 1 |a + √ ω 2 |b + √ ω 2 |c initially, it is easy to prove that the state at this stage should be where ω 1 + 2ω 2 = 1. By performing a projection measurement on |ψ 1 , Rice or Bob gets an ideal state |ψ C with the probability where The probability that neither Rice nor Bob finds the photon when Alice did prepare the photon in the committed state |φ is give by 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 Rice 2 or P Bob 2 , which is given by with P Rice 2 = P Bob 2 from winning rule W3.

The First Tripartite Quantum Game
In this subsection, we introduce the quantum implementation of the first game 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, R succ ( C) means that Rice finds the photon after opening the box C. B succ ( B) means that Bob finds the photon after opening the box B. R/B f ail (·) means that neither Rice or Bob finds the photon and Alice is honest. R/B succ ( A) means that neither Rice or Bob finds the photon but Alice is not honest.

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 The partial derivative of G Rice with respect to the variable γ is given by 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, If Alice chooses α 1 < ω 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 α 1 < ω 1 . It follows that α 1 ≥ ω 1 and α 2 ≤ ω 2 . Hence, from Equation (30) we get (30) and (31), G Rice is a decreasing function in x when x ≤ x * 1 and increasing when

The Average Gain of Bob
Denote G Bob as the average gain of Bob. From Table 1, we get G Bob as The partial derivative of G Bob with respect to β is given by Similar to Equations (30) and (31), we can get (33) and (34), Since 0 ≤ x ≤ 2, G Bob gets the maximum value at x = 0 for for x * 4 ≤ 0, or gets the maximum value at x = 2 for x * 4 ≥ 2, or gets the maximum value at x = x * 4 for 0 < x * 4 < 2. From η ≥ 1, we get x * 2 ≥ x * 4 . Similarly, we can get the detailed analysis of Rice, shown in Appendix F. The values of β * and γ * which depend on x * i in the first quantum game such that G Bob achieves the maximum are given in Table 7. Table 7. The values of β * and γ * depend on x * i in the first quantum game such that G Rice and G Bob achieve the maximums, where β * and γ * are the probability of Bob opening box B and Rice opening box C respectively. Here, x * 1 and x * 2 are given in Equation (31), and x * 3 and x * 4 are given in Equation (34).

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 Table 7, we discuss Alice's gain in five cases. Here, we only discuss one of them. Other cases are shown in Appendix G.
It follows that ω 1 ≤ α 1 ≤ D. In this case, we get β = γ = 0. From Equation (36), we get that the average gain of Alice is given by Note that dG Alice dα 1 = 3 from Equation (37). G Alice increases with α 1 when ω 1 ≤ α 1 ≤ D. So, G Alice gets the maximum value at α 1 = D, which is denoted by G Alice 1 given by where the corresponding point is denoted by p 1 . By using the same method for other cases (see Appendix G), we get Table 8.

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 If G Alice 1 = max{G Alice 1 ,G Alice 2 ,G Alice 3 , G Alice 4 , G Alice 5 }, from Table 8 we obtain that α 1 = p 1 , and β = γ = 0 from Table 7. The average gains of three players are evaluated using η and ω 1 , as shown in Figure 7. Here, for each η ≥ 1, the gains of Alice, Bob and Rice can tend to zero by changing ω 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 η → ∞.
To sum up, we get the following theorem.
Result 3 The first quantum game G 1 is asymptotically fair.

The Second Tripartite Quantum Game
In this section, we give the quantum realization of the second game G 2 . According to Table 5, similar to the first quantum game, the gain of Rice is given by where x = β + γ. The detailed maximizing G Rice is shown in Appendix H. Similarly, the gain of Bob is given by Its maximization is shown in Appendix I. The gain of Alice is given by Its maximization is shown in Appendix J.
The numeric evaluations of average gains are shown in Figure 9 in terms of µ and ω 1 . For each µ, we can make the gains of Alice, Bob and Rice equal to zero as much as possible by adjusting ω 1 . The deviation degree is shown in Figure 10a. The relationship between the appropriate ω 1 and µ 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 µ → ∞. To sum up, we get the following result.  Result 4 The second quantum game G 2 is asymptotically fair.
Proof. Note that the second quantum game G 2 is zero-sum, i.e., the summation of the average gains of all players is zero. From Equation (46), Alice will make α 2 = 0 when µ → ∞, i.e., α 1 = 1, in order to maximize her own gain, while Bob and Rice will choose β = γ = 1 accordingly. Hence, we get three gains as follows We get ω 1 = 1 2 and ω 2 = 1 4 from G Alice = G Bob = G Rice = 0. Thus, when µ → ∞, the quantum game G 2 converges a fair game, i.e., asymptotically fair. This completes the proof.

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.

Experimental Measurement Error
In the case of measurement error, from Equation (20), we can get that the probability of Rice opens box C and finds the photon becomes where ε is measurement error which may be very small. From Equation (20), we get the probability that Bob opens box B and finds the photon is given by Rice or Bob makes a projection measurement on |ψ C for the verification the final state with success probability where x = β + γ. 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 where λ i are given by Since λ 3 ≤ 1 and λ 4 ≤ 1, and ε is very small, denotes O(ε) = −2ελ 4 + ελ 3 − 2ε 2 , which can be treated as measurement error. Thus Equation (49) can be rewritten as The probability that neither Rice nor Bob finds the photon but Rice or Bob detects the forage preparation of Alice is denoted by P Rice 2 or P Bob 2 which given by where P Rice 2 = P Bob 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.

White Noises
In this subsection, we consider that Alice prepares a noisy photon in the state where I = |a a| + |b b| + |c c| denotes the identity operator, |ψ is given in Equation (16), and v ∈ [0, 1]. After Bob's and Rice's splitting operation, the state of the photon becomes where |ψ 0 is given by Denote P Rice 1 as the probability that Rice finds the photon after opening the box C. From Equation (53) it is given by Denote P Bob 1 as the probability that Bob finds the photon after opening the box B.

From Equation (53) it is given by
For the case that neither Bob nor Rice detects the photon, the density operator for the photon is given by Now, Rice or Bob makes a projection measurement with positive operator {|ψ C ψ C |, I − |ψ C ψ C |} on the photon for verifying the committed state of Alice with success probability 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 Denote P Rice 2 or P Bob 2 as the probability that neither Rice nor Bob finds the photon but Rice or Bob detects the forage preparation. We obtain that where P Rice 2 = P Bob 2 from the winning rule W4. Take the first quantum game as an example. The Rice's average gain G Rice is given by The partial derivative of G Rice with respect to γ is Assume that v is very close to one, i.e., 1−v 3 ≈ 0. It follows that is bounded. Thus Equation (62) can be rewritten as The partial derivative of G Bob with respect to β is given by The Alice's average gain G Alice is given by 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.

Conclusions
It has shown that two present quantum games are asymptotically fair. Interestingly, these games can be easily changed to biased versions from Figures 5 and 8, by choosing different η and ω 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.

Conflicts of Interest:
The authors declare no conflict of interest.

Appendix A. The Average Gain of Bob in the First Classical Game
In order to calculate Bob's average gain, we need to consider whether ∂G Rice ∂P C is greater than zero or not. And the case of ∂G Rice ∂P C ≤ 0 has given in Section 2.2.2 The case of ∂G Rice ∂P C > 0 is as follows.
For ∂G Rice ∂P C > 0, we have 0 ≤ P B < 3P 2 −P 1 P 2 (η+1) P 1 (η+1)(1−P 2 ) . In this case, P C = 1. From Equation (8) we get The partial derivative of G Bob with respect to P B is given by Two cases will be considered as follows.

Appendix B. The Average Gain of Alice in the First Classical Game
For the case of 0 ≤ P 1 < 3 2η+5 , we have given in Section 2.2.3. The remaining three cases are as follows.

Appendix C. The Average Gain of RICE in the Second Classical Game
It is straightforward to calculate the average gain of Rice according to Table 5 as follows The partial derivative of G Rice with respect to P C is given by , G Rice increases with P C , Rice will choose P C = 1 to maximize her gains.
However, if ∂G Rice ∂P C ≤ 0, i.e., G Rice decreases with P C , Rice will set P C = 0 in order to maximize her gains.

Appendix D. The Average Gain of Bob in Classical Second Game
Similar to G Rice , we can get Bob's average gain G Bob as follows According to the sign of ∂G Rice ∂P C , we can divide it into two cases.