This paper introduces a new quantum game called Quantum Tapsilou that is inspired from the classical traditional Greek coin tossing game tapsilou. The new quantum game, despite its increased complexity and scope, retains the crucial characteristic of the traditional game, namely that of fairness. In the classical game, both players have $\frac { 1 } { 4 }$ probability to win. In its quantum version, both players have equal chances to win too, but now the probability to win varies considerably, depending on previous choices. The two most important novelties of Quantum Tapsilou can be attributed to its implementation of entanglement via the use of rotation gates instead of Hadamard gates, which generates Bell-like states with unequal probability amplitudes, and the integral use of groups. In Quantum Tapsilou both players agree on a specific cyclic rotation group of order $n$, for some sufficiently large $n$, which is the group upon which the game will be based, in the sense both players will pick rotations from this group to realize their actions using the corresponding $R_{ y }$ rotation gates. This fact is in accordance with a previous result in the literature showing that symmetric quantum games, where both players draw their moves from the same group, are fair.

Keywords:: Classical games, quantum games, quantum coin tossing games, quantum strategies.

§ INTRODUCTION

Undoubtedly, Game theory has established itself by now as a multifaceted framework that can readily and easily provide a multitude of tools that facilitate the analysis of complex situations, especially those concerning conflict, competition, or cooperation. The actors or agents involved in these situations are called players. The fundamental assumption of Game theory is that players are always rational, and strive to maximize their profit or minimize their loss. Each player possesses her own set of objectives, and her actions can disrupt the plans of other players. At the conclusion of the game, each player assesses her gains or losses by employing a payoff function. The premise of rationality plays a crucial role in guaranteeing that each agent is motivated to maximize their own payoff. Game theory is particularly suited for tackling serious and difficult problems from a plethora of different fields, that range ranging from economics, political and social sciences, to computer science and biology and psychology. In this work we shall be employing only a few elementary concepts from Game theory that can be readily found in standard textbooks such as [1], [2] or [3].

The realm of quantum computation holds the promise of ushering us in a new era of computational capabilities. Despite its origins dating back to the early 1980s, it took some time for this field to fully develop and reach its current state. Presently, there are commercial quantum computers in existence, albeit with a rather modest number of qubits. However, it is anticipated that this situation will undergo significant transformation in the years ahead. If these predictions prove accurate, there will be a substantial increase in the number of available qubits. Consequently, this quantum leap will lead to a remarkable enhancement of our computational power, enabling us to address complex decision-making and optimization challenges effectively.

In this broader context, it appears inevitable that quantum games would emerge as a natural consequence. The emerging field of quantum game theory explores the integration of quantum principles and concepts into classical games, such as coin flipping, the Prisoner's Dilemma, and various others.

§.§ Related work

It would be historically accurate to say that the inception of the field of quantum games occurred in 1999 when Meyer introduced the PQ penny flip game in his influential paper [4]. This game serves as the quantum counterpart to the classical penny flip game. One of the notable features of Meyer's formulation was the incorporation of two well-known fictional characters, Picard and Q, from the renowned television series Star Trek. In this game, Picard embodies the classical player, while Q embodies the quantum player. This signifies that Picard follows the rules of the classical penny flip game, while Q can utilize arbitrary unitary operators, a hallmark of the quantum realm. Meyer demonstrated that in this game, Q can consistently emerge victorious by employing the venerable Hadamard transform. Another influential work, also from the same year, was the one by Eisert et al. in [5]. There they demonstrated for the first time the so called Eisert-Wilkens-Lewenstein protocol. This protocol has gained wide acceptance and is now extensively used in the literature. Using their protocol, they defined a variant of the famous Prisoners' Dilemma and demonstrated that there exists a quantum strategy that is better than any classical strategy. Subsequently, many researchers followed that line of research, obtaining interesting results. For more recent developments we refer to [6] and [7], where the Eisert–Wilkens–Lewenstein protocol was generalized.

Subsequently, many researchers, inspired by Meyer's concepts, extended this game pattern to $n$-dimensional quantum systems. This avenue of exploration was pursued in [8, 9] and [10]. Later, a comprehensive mapping of every finite variation of the PQ penny flip game to easily constructed finite automata was provided in [11]. Recently, this same game underwent extensive analysis from the perspective of group theory, with its relation to dihedral groups being established in [12]. A general analysis of quantum games under the assumption that both players draw their move from groups and not just unstructured sets, coming to some interesting conclusions, can be found in [13]. While in many works, the quantum player appears to possess a distinct advantage over the classical player, the situation is not as straightforward, as demonstrated in [14]. In that study, the authors cleverly altered the rules of the PQ penny flip game, allowing the classical player to secure a victory.

Additionally, a broader and intriguing problem concerning quantum gambling based on Nash equilibrium has been explored recently in [15]. In a similar vein, quantum coin flipping has served as a crucial component in numerous quantum cryptographic protocols, where familiar figures like Alice and Bob take on the roles of remote parties striving to reach a consensus on a random bit. For a comprehensive presentation, the seminal work in [16] is invaluable, and for some state-of-the-art cryprographic protocols we refer to [17, 18, 19, 20, 21]). Some of these ideas have been extended in [22] to quantum dice rolling in scenarios involving multiple outcomes and more than two parties. Most of the papers mentioned above, rely in an integral way on the phenomenon of quantum entanglement, a trend that is also followed by the current work.

Let us also point out that the inclination to employ unconventional approaches for improving outcomes in classical games extends beyond the realm of quantum mechanics. Even in the domain of well-known classical games, such as the Prisoners' Dilemma, efforts have been made to interpret them in terms of biological concepts (for a comprehensive overview, refer to [23]). In fact, many classical games can be reinterpreted within the context of biological and bio-inspired processes, as evidenced by numerous references (for more sources, see [23, 24, 25]).

Contribution. This paper introduces Quantum Tapsilou, which is a novel quantum game inspired by a traditional Greek coin tossing game called tapsilou. While the classical game has a simple and fair probability distribution, with both players having a $\frac { 1 } { 4 }$ probability of winning, Quantum Tapsilou exhibit additional complexity and a new level of fairness using quantum mechanics. In Quantum Tapsilou, both players still have equal chances of winning, but these probabilities are now influenced by their previous choices and the use of quantum principles. There are two key innovations in Quantum Tapsilou:

* Entanglement through Rotation Gates. Instead of using Hadamard gates, which are commonly associated with creating entangled states with equal probability amplitudes, Quantum Tapsilou employs rotation gates, specifically $R_y$ rotation gates. These gates generate Bell-like states with unequal probability amplitudes, which adds a layer of complexity to the game dynamics.

* Integral Use of Groups. In Quantum Tapsilou, both players agree on a specific cyclic rotation group of order $n$, where $n$ should be sufficiently large integer in order to provide the players with additional choices and increase the suspense of the game. This group forms the basis of the game, and both players select rotations from this group to perform their actions using the corresponding $R_y$ rotation gates. The choice of this group and the rotations within it significantly impacts the game's outcome, introducing a strategic element based on group theory.

By incorporating these quantum principles and group theory, Quantum Tapsilou retains the fairness aspect of the classical tapsilou game while adding complexity and variability to the probabilities of winning. Players must consider their choices of rotations within the agreed-upon group, as well as the entangled states generated by the $R_y$ rotation gates, to maximize their chances of winning in this intriguing quantum adaptation of a traditional Greek game. This is another example of the critical role of groups in ensuring fairness in quantum games, a fact that is in accordance with a previous result in [13] establishing that symmetric quantum games, where both players draw their moves from the same group, are fair.

§.§ Organization

This paper is organized as follows: Section <ref> gives a concise introduction to the field of quantum games, along with some of the most relevant references. Section <ref> presents the traditional Greek game tapsilou, while section <ref> explains why the original game is meaningful and well-conceived.
Sections <ref> and <ref> give an easygoing account of the intuition behind the quantization of the classical game, and the formal presentation of the new quantum game, respectively. Section <ref> contains an extensive and rigorous analysis of the main properties of the proposed game. Finally, Section <ref> summarizes the main conclusions of this work and contains some possible research directions for future work.

§ THE TRADITIONAL GAME “TAPSILOU”

The Greek gambling game “tapsilou” is a coin tossing game that originates from the Greek island of Corfu. The name “tapsilou” is a local idiom meaning “up high.” This is a traditional game played in New Year’s Eve and New Year’s Day in Mantouki, Corfu. It is also known as “crosses and marks” (“stavri or markoi”) and goes back in the history of the island. It was extremely popular in the 60s, 70s and 80s among the young people and it was tradition to be played around the New Year celebration. Although it is considered a game, for many of the participants meant their economic ruin [26].

Tapsilou needs two coins. They used mostly Greek currency (drachmas), the so called 5-drachmas (“talara or talira”), like those depicted in Figure <ref>. It was the biggest in diameter coin of the days. The coin depicts a figure in one side, the heads side, and a number on the other, the tails side. Thus, the alternative name “crosses and marks.” As a terrain, the participants typically chose a sandlot. Two coins are placed in one of the player’s two-or-three fingers, as shown in Figure <ref>. This player is the tosser and from now on will also be referred to as player 1. The initial position of the coins on the fingers is one coin with the face up and one with the cross. The other players, usually more than one, are the gamblers, and, henceforth, will also be collectively referred to as player 2. The gamblers draw a circle around the tosser, who throws the coins high to the sky, thus the origin of the name “tapsilou.”  After the flip, the coins end up on the ground. If the coins end up with two faces, the tosser wins. If the coins end up with crosses, the gamblers win, and in case the coins end up with one face and one cross, the tossing is repeated until there is a winner.

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A typical example of coins from the 60s [27].

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The initial placement of the two coins, just before the tossing [28].

Although tapsilou requires at least 2 players, the tosser and at least one gambler, there is no limit as to what the maximum number of players is. It would not be unusual to have fifteen or more players, as long as there is an amount of money, large enough to bet. Before the tossing, the gamblers must announce their bet. If the two coins come up faces up, the tosser wins and takes all the bet money, whereas if they both come up crosses, the tosser pays each gambler twice the agreed bet. If the result is one coin face up and one coin cross, then the gamblers choose, individually, whether they will keep the same bet or raise it, and the tosser throws the coins again. This process is repeated, until the tosser or the gamblers win. The tosser loses when he has no more money left. If the tossers budget is running low, he may decide to take no more bets, or up to a certain amount, so as to not end up in debt. When the tosser loses all his money, another player takes his place.

Tapsilou is a forbidden game, so in many cases there may be lookout men. There are two common cheating strategies in this game. The most tricky one is to crack the coin just a little bit, so as to always drop from one side. This, of course, makes the coins biased, but the crack has to be unnoticeable. The other cheating strategy is during the tossing to employ a technique where the two coins seem to flip in the air before they end up in the ground, but in fact only one flips, while the other does not flip at all. This is just an illusion that only the best trained cheaters can achieve. If a gambler recognizes a certain cheat strategy, he has the right to call it (using Greek idioms such as“den paei” or “ta kovo”) and the bet is canceled.

§ TWO COINS SEEM TO BE THE PROPER CHOICE

In this section, we give a precise quantitative description of the game, and examine its possible generalizations.

Assuming that no cheating takes place, we see that the classical version of tapsilou is characterized by the following properties, where we use $H$ and $T$ to denote heads and tails, respectively.

* The number of players is $2$. Player $1$ is the tosser, and player $2$ is the gambler. Despite the fact that in real life many persons may play the gambler simultaneously, they can all be identified as one player in a theoretical analysis.

* The number of coins is $2$.

* Viewing the two coins as a system, with no particular ordering between them, the initial state of the system can taken to be $(H, T)$ or, equivalently, $(T, H)$. So, without loss of generality, we may assume from now on that the initial state is $(H, T)$.

* If the final state is $(H, H)$, then player $1$ wins. If the final state is $(T, T)$, then player $2$ wins. In any other case, the game goes on to the next round.

* The probability that player $1$ wins is equal to the probability that player $2$ wins, namely $\frac { 1 } { 4 }$, so the game is fair.

* The probability that neither player wins is $\frac { 1 } { 2 }$.

A question that comes to mind is whether it would be possible to generalize this game by tossing more than two coins. Under the plausible assumption that in such a generalization player $1$ wins if all coins end up heads and player $2$ wins if all coins end up tails, it is easy to calculate the probability for each player to win. If $n$ coins are used, the condition for player $1$ to win is that all $n$ coins end up heads, an event that may happen with probability $\frac { 1 } { 2^{ n } }$. Symmetrically, for player $2$ the probability that all $n$ coins come up tails is also $\frac { 1 } { 2^{ n } }$. The next Table <ref> gives the numerical values of these probabilities when the number of coins is $3$, $4$ and $5$. A clear pattern that emerges. If tapsilou were to use more than two coins, the game would not be very interesting because the probability of neither player winning and just spending most of the gaming tossing coins would be far greater, than having a winner. Therefore, using exactly two coins seems to be the best choice in order to have a game with suspense, symmetry and fairness.

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This table gives the probabilities of players $1$ and $2$ to win in a possible generalization of the game tapsilou.

[HTML]000000   Coins & Probability

[HTML]000000   # of Coins

$P( \text{ player 1 wins } )$

$P( \text{ player 2 wins } )$

$P( \text{ neither wins } )$

[HTML]000000   $3$

Events ( m ) := 2^ m ;
Probability ( m ) := 1 / Events ( m );

$\xintfloateval{ Probability ( 3 ) }$

Events ( m ) := 2^ m ;
Probability ( m ) := 1 / Events ( m );

$\xintfloateval{ Probability ( 3 ) }$

Events ( m ) := 2^m;
Probability ( m ) := 1 - ( 2 / Events ( m ) );
Probability ( 3 )

[HTML]000000   $4$

Events ( m ) := 2^ m ;
Probability ( m ) := 1 / Events ( m );

$\xintfloateval{ Probability ( 4 ) }$

Events ( m ) := 2^ m ;
Probability ( m ) := 1 / Events ( m );

$\xintfloateval{ Probability ( 4 ) }$

Events ( m ) := 2^m;
Probability ( m ) := 1 - ( 2 / Events ( m ) );
Probability ( 4 )

[HTML]000000   $5$

Events ( m ) := 2^ m ;
Probability ( m ) := 1 / Events ( m );

$\xintfloateval{ Probability ( 5 ) }$

Events ( m ) := 2^ m ;
Probability ( m ) := 1 / Events ( m );

$\xintfloateval{ Probability ( 5 ) }$

Events ( m ) := 2^m;
Probability ( m ) := 1 - ( 2 / Events ( m ) );
Probability ( 5 )

In view of our previous analysis, it would seem that the classical world cannot offer a meaningful generalization of tapsilou. If one hopes to achieve an interesting variation of the classical game, turning to the quantum realm seems to be the most promising option.

§ THE INTUITION BEHIND THE QUANTUM TAPSILOU

In this section we explain the rationale behind the proposed Quantum Tapsilou game (QTG from now), which, we believe, that captures the spirit of the classical tapsilou game. In the QTG, kets $\ket{ 0 }$ and $\ket{ 1 }$ correspond to the classical heads and tails states, respectively. Consequently, the initial state of the QTG is $\ket{ 0 } \ket{ 1 }$, and player $1$ (player $2$) wins if the final state of the system (after measurement) is $\ket{ 0 } \ket{ 0 }$ ($\ket{ 1 } \ket{ 1 }$). The following Table <ref> summarizes these facts.

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This table compares the characteristics of the classical tapsilou against the corresponding characteristics of the quantum tapsilou.

[HTML]000000   Classical vs quantum tapsilou

Initial State

Winning State for Player $1$

Winning State for Player $2$

[HTML]000000   Classical

$H$ (Heads) $T$ (Tails)

$(H, T)$

$(H, H)$

$(T, T)$

[HTML]000000   Quantum

$\ket{ 0 }$         $\ket{ 1 }$

$\ket{ 0 } \ket{ 1 }$

$\ket{ 0 } \ket{ 0 }$

$\ket{ 1 } \ket{ 1 }$

Let us first assume that player $1$ (the tosser) can employ quantum actions to his advantage. In order for him to win, both coins must end up heads. This is easily achievable in the quantum realm using entanglement. Thus, entanglement is the key for player $1$ to ensure that both coins end up in the same state. The typical quantum circuit used to generate entangled pairs, which is based on the Hadamard transform, is shown in Figure <ref>. The quantum circuit of Figure <ref>, and all forthcoming quantum circuits, are designed using the IBM Quantum Composer [29]. The same tool is used for the visualization of the state vector of the quantum system. This particular circuit will drive the system of the two coins in the $\ket{ \Phi^{ - } }$ state, which is characterized by the state vector depicted in Figure <ref>. However, if the coins are measured in this state, the tosser does not have any real advantage, as both players have equal probability $\frac { 1 } { 2 }$ to win.

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This figure shows a typical quantum circuit that generates $\ket{ \Phi^{ - } }$ pairs, given that the initial state is $\ket{ 0 } \ket{ 1 }$.

This figure depicts the state vector of the system of the two coins entangled in the $\ket{ \Phi^{ - } }$ state.

Additionally, player $1$ can drive the system to the $\ket{ \Psi^{ - } }$, via the circuit of Figure <ref>. As shown in Figure <ref>, if the coins are measured in this state, it is certain that nobody wins, and the game keeps going on ad infinitum. This is another option, albeit less likely, that the tosser may pursue as a bluff.

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Assuming that the initial state is $\ket{ 0 } \ket{ 1 }$, the above figure depicts a typical quantum circuit that generates $\ket{ \Psi^{ - } }$ pairs.

The above figure visualizes the state vector of the two coin system in the $\ket{ \Psi^{ - } }$ state.

Therefore, from the perspective of player $1$, it becomes evident that he must entangle the two coins in a more subtle way. Logic suggests that player $1$ should devise a kind of entanglement that satisfies the following requirements.

* Both coins end up in the same state with certainty, i.e., with probability $1$.

* While it may still be possible that both coins may come up tails, the probability that both coins may end up heads must be significantly higher. In other words, the odds should favor player $1$ decisively.

One way that player $1$ may achieve such an effect is to use rotation gates in place of the Hadamard gate. Towards this end, the simplest choice would be to use the quantum gate $R_{ y } ( \theta )$ with an  appropriate parameter $\theta \in \mathbb{R}$. $R_{ y } ( \theta )$, which is defined as

\begin{align} \label{eq:Ry Rotation Operator}
R_{ y } ( \theta )
\begin{bNiceMatrix}[ cell-space-limits = 2 pt ] % margin, hvlines, rules/width = 0.0025 mm ]
\cos \frac { \theta } { 2 } & - \sin \frac { \theta } { 2 }
\\
\sin \frac { \theta } { 2 } & \phantom{-} \cos \frac { \theta } { 2 }
\end{bNiceMatrix}
\ ,
\end{align}

rotates the state of the coin around the $y$ axis of the Bloch sphere through an angle $\frac { \theta } { 2 }$ counterclockwise. Figure <ref> illustrates a quantum circuit that employs the rotation gate $R_{ y } (  \frac { 12 \pi } { 16 } )$ to send the two coins to the state described by Figure <ref>.

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Given that the initial state is $\ket{ 0 } \ket{ 1 }$, the above quantum circuit applies the rotation gate $R_{ y } (  \frac { 12 \pi } { 16 } )$ to drive the system to the state visualized in Figure <ref>.

This circuit of Figure <ref> guarantees that both coins will end up in the same state, but state $\ket{ 0 } \ket{ 0 }$ is much more probable.

Clearly, if the final state of the wo coins is determined by measuring the two coins in the computational basis $B = \{ \ket{ 0 }, \ket{ 1 } \}$, the most probable outcome, by a significant margin, will be that player $1$ wins and player $2$ looses. The game is no longer fair, and player $2$, being rational, has no incentive whatsoever to participate in such a game. Nonetheless, it is quite possible to make the game fair again by letting player $2$ choose the basis with respect to which the coins are measured.

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The effect of the application of the two rotation gates prior to measurement is that the measuring basis is no longer the computational basis.

The circuit of Figure <ref> drives the system to the state described by the above state vector. Now the odds favor player $2$.

As demonstrated by the above Figures <ref> and <ref>, if player $2$ chooses the proper measurement basis, despite the rotation used by player $1$, he can turn the tables and make sure that the odds of winning are now in his favor.

§ THE FORMAL PRESENTATION OF THE QTG

This section contains the precise description of the QTG, together with its comprehensive mathematical analysis. In our vision of the QTG, rotations play a decisive role, and, since rotations form groups, and group are a prevalent theme in quantum mechanics, we shall employ some elementary concepts from the theory of groups, adhering to the notation and definitions found in standard textbooks such as [30, 31] and [32]. So, without further ado, we proceed to introduce the required machinery.

* The general cyclic rotation group of order $n$, denoted by $\langle r \rangle$, contains the following $n$ elements

\begin{align} \label{def:General Cyclic Rotation Group Elements}
\langle r \rangle
\{ \mathds{ 1 }, r, r^{ 2 }, \dots, r^{ n - 1 } \}
\ ,
\end{align}

where $r$ is the rotation on the plane by $\frac { 2 \pi } { n }$. Its elements $\mathds{ 1 }, r, r^{ 2 }, \dots, r^{ n - 1 }$ are plane rotations. Specifically, $r^{ k }$ is the rotation by $\frac { 2 \pi k } { n }$, $0 \leq k \leq n - 1$, and $\mathds{ 1 } = r^{ 0 }$ is the identity element.

* To each group element $r^{ k }$, which represents the plane rotation by $\frac { 2 \pi k } { n }$, we associate the rotation gate $R_{ y } ( \frac { 2 \pi k } { n } )$

\begin{align} \label{eq:Ry Rotation Operator Corresponding to Rotakion r^k}
R_{ y } \left( \frac { 2 \pi k } { n } \right)
\begin{bNiceMatrix}[ cell-space-limits = 2 pt ] % margin, hvlines, rules/width = 0.0025 mm ]
\cos \frac { \pi k } { n } & - \sin \frac { \pi k } { n }
\\
\sin \frac { \pi k } { n } & \phantom{-} \cos \frac { \pi k } { n }
\end{bNiceMatrix}
\ ,
\end{align}

the two orthonormal kets

\begin{align} \label{eq:Rotation of Ket 0 By k Pi}
\ket{ 0 }_{ \frac { 2 \pi k } { n } }
\begin{bmatrix}
\cos \frac { \pi k } { n }
\\
\sin \frac { \pi k } { n }
\end{bmatrix}
\end{align}

\begin{align} \label{eq:Rotation of Ket 1 By k Pi}
\ket{ 1 }_{ \frac { 2 \pi k } { n } }
\begin{bmatrix*}[r]
- \sin \frac{ \pi k } { n }
\\
\phantom{-} \cos \frac { \pi k } { n }
\end{bmatrix*}
\ ,
\end{align}

and the orthonomal basis

\begin{align} \label{eq:Rotation of Computational Basis By k Pi}
B_{ \frac { 2 \pi k } { n } }
\begin{bNiceMatrix}[ cell-space-limits = 2 pt ] % margin, hvlines, rules/width = 0.0025 mm ]
\cos \frac { \pi k } { n } & - \sin \frac{ \pi k } { n }
\\
\sin \frac { \pi k } { n } & \phantom{-} \cos \frac{ \pi k } { n }
\end{bNiceMatrix}
\ .
\end{align}

We formally state the rules of the QTG below.

* The initial state of the two coin system is $\ket{ 0 } \ket{ 1 }$. The winning state for player $1$ (the tosser) is $\ket{ 0 } \ket{ 0 }$, and the winning state for player $2$ (the gambler) is $\ket{ 1 } \ket{ 1 }$.

* Both players agree on a specific cyclic rotation group $\langle r \rangle$ of order $n$, for some sufficiently large $n$. This will be the group upon which the QTG will be based. Both players will pick elements from this group to realize their actions.

* Player $1$ chooses an element $r^{ k } \in \langle r \rangle$, $0 \leq k \leq n - 1$, and uses the rotation operator $R_{ y } ( \frac { 2 \pi k } { n } )$ (recall equation (<ref>)) to entangle the two coins. Formally, player $1$'s action is described by

\begin{align} \label{eq:General Action of Player 1}
CNOT \ \cdot \ \left( I \otimes R_{ y } \left( \frac { 2 \pi k } { n } \right) \right)
\ ,
\end{align}

where $CNOT$ and $I$ are the controlled NOT and identity operators, respectively.

* Player $2$ chooses an element $r^{ l } \in \langle r \rangle$, $0 \leq l \leq n - 1$, and measures the two coins with respect to the orthonormal basis $B_{ \frac { 2 \pi l } { n } }$ (recall formula (<ref>)). Formally, player $2$'s action is described by

\begin{align} \label{eq:General Action of Player 2}
R_{ y } \left( \frac { 2 \pi l } { n } \right) \otimes R_{ y } \left( \frac { 2 \pi l } { n } \right)
\ .
\end{align}

For concreteness, we have set as $\ket{ 0 } \ket{ 1 }$ the initial state of the game. It is noteworthy to mention that the choice of the initial state does not really matter. Picking a different initial state, would not change the nature of the QTG and the forthcoming analysis.

The QTG is captured by the abstract quantum circuit shown below in Figure <ref>.

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nobit AUX_C_0_0;
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nobit AUX_C_1_0;
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barrier ( - ) ;

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hspace 0.250 cm Coin_0;
cnot Coin_1 | Coin_0;
hspace 0.250 cm Coin_0;

[ name = Ph1, WordBlueVeryLight, line width = 0.50 mm, label = Phase $1$ ]
barrier ( - ) ;

hspace 0.50 cm Coin_0;
[ draw = GreenLighter2!50, fill = GreenLighter2!50, radius = 0.5 cm ] box RY$( \frac { 2 \pi l } { n } )$ Coin_0;
hspace 0.250 cm Coin_0;
hspace 0.50 cm Coin_1;
[ draw = GreenLighter2!50, fill = GreenLighter2!50, radius = 0.5 cm ] box RY$( \frac { 2 \pi l } { n } )$ Coin_1;
hspace 0.250 cm Coin_1;

[ name = Ph2, WordBlueVeryLight, line width = 0.50 mm, label = Phase $2$ ]
barrier ( - ) ;

[ draw = white, fill = MagentaDark ] [ radius = 0.5 cm ] measure Coin_0;
[ draw = white, fill = MagentaDark ] [ radius = 0.5 cm ] measure Coin_1;

hspace 0.5 cm Coin_0;
output $\ket{ ? }$ Coin_0;
output $\ket{ ? }$ Coin_1;

scale = 1.25,
anchor = center,
above right = 1.50 cm and 1.750 cm of Ph0,
label =  [ label distance = 0.00 cm ] north:  Tosser
( )   Entangles the coins ;
scale = 1.25,
anchor = center,
above right = 1.50 cm and 1.50 cm of Ph1,
label =  [ label distance = 0.00 cm ] north:  Gambler
( )   Changes the basis ;
[ below = 2.50 cm ] at (Ph0)  $\ket{ \psi_{ 0 } }$ ;
[ below = 2.50 cm ] at (Ph1)  $\ket{ \psi_{ 1 } }$ ;
[ below = 2.50 cm ] at (Ph2)  $\ket{ \psi_{ 2 } }$ ;

This figure shows the quantum circuit that corresponds to the QRG.

§ THE MATHEMATICAL ANALYSIS OF THE QTG

In this section we shall analyze the QTG from a rigorous mathematical viewpoint, in order to draw conclusions about the options of the two players and their odds of winning. We use the abstract quantum circuit of Figure <ref> as a guide. In our exposition, we follow the Qiskit [33] way of ordering the qubits, which means that the bottom qubit is the most significant and the top the least significant.

By rule (R$_{ 1 }$), the initial state of the system, denoted by $\ket{ \psi_{ 0 } }$, is given by

\begin{align} \label{eq:TQG Initial State}
\ket{ \psi_0 }
\ket{ 0 }
\ket{ 1 }
\begin{bmatrix}
\\
\\
\\
\end{bmatrix}
\ .
\end{align}

Player $1$, the tosser, acts first on the system, using the action descried by equation (<ref>) in rule (R$_{ 3 }$). In view of formula (<ref>), the matrix representation of the tensor product $( I \otimes R_{ y } ( \frac { 2 \pi k } { n } ) )$ is

\begin{align} \label{eq:I Tensor Product Ry}
% \hspace{-1.00 cm}
I \otimes R_{ y } \left( \frac { 2 \pi k } { n } \right)
\overset { \eqref{eq:Ry Rotation Operator Corresponding to Rotakion r^k} } { = }
\begin{bNiceMatrix}[ cell-space-limits = 2 pt ] % margin, hvlines, rules/width = 0.0025 mm ]
1 & 0
\\
0 & 1
\end{bNiceMatrix}
\otimes
\begin{bNiceMatrix}[ cell-space-limits = 2 pt ] % margin, hvlines, rules/width = 0.0025 mm ]
\cos \frac { \pi k } { n } & - \sin \frac { \pi k } { n } \\
\sin \frac { \pi k } { n } & \phantom{-} \cos \frac { \pi k } { n }
\end{bNiceMatrix}
% \nonumber \\
\begin{bNiceMatrix}[ cell-space-limits = 2 pt ] % margin, hvlines, rules/width = 0.0025 mm ]
\cos \frac { \pi k } { n } & - \sin \frac { \pi k } { n } & 0 & 0
\\
\sin \frac { \pi k } { n } & \phantom{-} \cos \frac { \pi k } { n } & 0 & 0
\\
0 & 0 & \cos \frac { \pi k } { n } & - \sin \frac { \pi k } { n }
\\
0 & 0 & \sin \frac { \pi k } { n } & \phantom{-} \cos \frac { \pi k } { n }
\end{bNiceMatrix}
\ .
\end{align}

Since, we follow the Qiskit conventions in the numbering of qubits, the matrix representation of the $CNOT$, as applied in the circuit of Figure <ref>, is the one given below, according to the Qiskit documentation [34].

\begin{align} \label{eq:CNOT Operator}
\begin{bNiceMatrix}[ cell-space-limits = 2 pt ] % margin, hvlines, rules/width = 0.0025 mm ]
1 & 0 & 0 & 0
\\
0 & 0 & 0 & 1
\\
0 & 0 & 1 & 0
\\
0 & 1 & 0 & 0
\end{bNiceMatrix}
\
\end{align}

Using (<ref>) and (<ref>), we may derive the matrix representation of the action of player $1$.

\begin{align} \label{eq:Matrix Form of the Action of Player 1}
CNOT \ \cdot \ \left( I \otimes R_{ y } \left( \frac { 2 \pi k } { n } \right) \right)
&\overset { \eqref{eq:I Tensor Product Ry}, \eqref{eq:CNOT Operator} } { = }
\begin{bNiceMatrix}[ cell-space-limits = 2 pt ] % margin, hvlines, rules/width = 0.0025 mm ]
1 & 0 & 0 & 0
\\
0 & 0 & 0 & 1
\\
0 & 0 & 1 & 0
\\
0 & 1 & 0 & 0
\end{bNiceMatrix}
\
\begin{bNiceMatrix}[ cell-space-limits = 2 pt ] % margin, hvlines, rules/width = 0.0025 mm ]
\cos \frac { \pi k } { n } & - \sin \frac { \pi k } { n } & 0 & 0
\\
\sin \frac { \pi k } { n } & \phantom{-} \cos \frac { \pi k } { n } & 0 & 0
\\
0 & 0 & \cos \frac { \pi k } { n } & - \sin \frac { \pi k } { n }
\\
0 & 0 & \sin \frac { \pi k } { n } & \phantom{-} \cos \frac { \pi k } { n }
\end{bNiceMatrix}
\nonumber \\
\begin{bNiceMatrix}[ cell-space-limits = 2 pt ] % margin, hvlines, rules/width = 0.0025 mm ]
\cos \frac { \pi k } { n } & - \sin \frac { \pi k } { n } & 0 & 0
\\
0 & 0 & \sin \frac { \pi k } { n } & \phantom{-} \cos \frac { \pi k } { n }
\\
0 & 0 & \cos \frac { \pi k } { n } & - \sin \frac { \pi k } { n }
\\
\sin \frac { \pi k } { n } & \phantom{-} \cos \frac { \pi k } { n } & 0 & 0
\end{bNiceMatrix}
\
\end{align}

By combining (<ref>) and (<ref>), we conclude that the state of the two coins at the end of Phase $1$, i.e., after the tosser's action, which is denoted by $\ket{ \psi_{ 1 } }$, is given by

\begin{align} \label{eq:TQG Phase 1}
\ket{ \psi_{ 1 } }
\begin{bmatrix}
- \sin \frac { \pi k }{ n }
\\
\\
\\
\phantom{-} \cos \frac { \pi k } { n }
\end{bmatrix}
\ .
\end{align}

[Playing in the group $\langle \frac { 2 \pi } { 16 } \rangle$]
This example examines how the QTG evolves up to Phase $1$ (see Figure <ref>), when the game is played in the cyclic rotation group $\langle \frac { 2 \pi } { 16 } \rangle$. Practically, this implies that both players may choose for their action one the $16$ rotations of this group.

The general form of the action of Player $1$ is given by (<ref>) and the state of the two coins by (<ref>). If the game were to end at this point in time, measuring the two coins would find the system in state $\ket{ 0 } \ket{ 0 }$ with probability $\sin^{ 2 } \frac { \pi k }{ n }$, and in state $\ket{ 1 } \ket{ 1 }$ with probability $\cos^{ 2 } \frac { \pi k }{ n }$. Hence, the probability that the tosser would win ranges from $0$, in case the tosser uses the identity element of the group, namely the $0$ radians rotation, to $1$, in case he uses the $\frac { \pi }{ 2 }$ radians rotation. For every possible choice of rotation from the $\langle \frac { 2 \pi } { 16 } \rangle$ group, the corresponding probability for player $1$ is given by the red line of Figure <ref>. Symmetrically, for every possible choice of rotation from $\langle \frac { 2 \pi } { 16 } \rangle$, the corresponding probability for player $2$ is given by the green line of Figure <ref>. The specific numerical values of the players' probabilities are contained in Table <ref>.

grow to left by = 1.00 cm,
grow to right by = 1.00 cm,
colback = MagentaVeryLight!03,
enhanced jigsaw,

sharp corners,

toprule = 1.0 pt,
bottomrule = 1.0 pt,
leftrule = 0.1 pt,
rightrule = 0.1 pt,
sharp corners,
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compat = 1.16
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width = 7.00 cm, height = 5.50 cm,
axis line style =  WordBlueDarker, line width = 0.25 mm ,
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axis y line = middle,
axis line style =  ->, > =  latex [ length = 7pt ]  ,
xmin = - 0.1, xmax = 3.40,
ymin = - 0.1, ymax = 1.1,
xtick =

pi * 1 / 16, pi * 2 / 16, pi * 3 / 16, pi * 4 / 16,
pi * 5 / 16, pi * 6 / 16, pi * 7 / 16, pi * 8 / 16,
pi * 9 / 16, pi * 10 / 16, pi * 11 / 16, pi * 12 / 16,
pi * 13 / 16, pi * 14 / 16, pi * 15 / 16, pi * 16 / 16
xticklabels =

$\frac { \pi } { 16 }$, $\frac { 2 \pi } { 16 }$,
$\frac { 3 \pi } { 16 }$, $\frac { \pi } { 4 }$,
$\frac { 5 \pi } { 16 }$, $\frac { 6 \pi } { 16 }$,
$\frac { 7 \pi } { 16 }$, $\frac { \pi } { 2 }$,
$\frac { 9 \pi } { 16 }$, $\frac { 10 \pi } { 16 }$,
$\frac { 11 \pi } { 16 }$, $\frac { 3 \pi } { 4 }$,
$\frac { 13 \pi } { 16 }$, $\frac { 14 \pi } { 16 }$,
$\frac { 15 \pi } { 16 }$, $\pi$
ytick =  0, 0.125, 0.25, 0.375, 0.50, 0.625, 0.75, 0.875, 1.00 ,
yticklabels =  0, 0.125, 0.250, 0.375, 0.500, 0.625, 0.750, 0.875, 1 ,
x label style =

at =  ( axis description cs: 0.5, -0.075 ) , fill = WordVeryLightTeal, anchor = north
xlabel =  Rotation angle $\frac { \pi k } { 16 } \ (0 \leq k \leq 15)$ ,
y label style =

at =  ( axis description cs: 0.025, 1.040 ) , fill = WordVeryLightTeal, anchor = south
ylabel =  Probabilities ,
x post scale = 2.0,
y post scale = 1.5,
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trig format plots = rad,
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declare function =  t() = sin( )^2; g() = cos( )^2; ,

legend style =  at =  ( 0.5, 1.2 ) , anchor = south

empty legend
Phase $1$ in $\langle \frac { 2 \pi } { 16 } \rangle$

[ RedPurple, line width = 0.4 mm ]  t() ;

[ pin = 30 :   Tosser  ] at ( 10 * pi / 16, 0.855 ) ;
Probability Player $1$ wins

[ GreenLighter2, line width = 0.4 mm ]  g ( ) ;
[ pin = 30 :   Gambler  ] at ( pi / 8, 0.855 ) ;
Probability Player $2$ wins

[ color = RedPurple, draw = none, mark size = 3 pt, mark = * ] coordinates

( 0,  t ( 0 )  )
( pi * 1 / 16,  t ( pi * 1 / 16 )  )
( pi * 2 / 16,  t ( pi * 2 / 16 )  )
( pi * 3 / 16,  t ( pi * 3 / 16 )  )
( pi * 4 / 16,  t ( pi * 4 / 16 )  )
( pi * 5 / 16,  t ( pi * 5 / 16 )  )
( pi * 6 / 16,  t ( pi * 6 / 16 )  )
( pi * 7 / 16,  t ( pi * 7 / 16 )  )
( pi * 8 / 16,  t ( pi * 8 / 16 )  )
( pi * 9 / 16,  t ( pi * 9 / 16 )  )
( pi * 10 / 16,  t ( pi * 10 / 16 )  )
( pi * 11 / 16,  t ( pi * 11 / 16 )  )
( pi * 12 / 16,  t ( pi * 12 / 16 )  )
( pi * 13 / 16,  t ( pi * 13 / 16 )  )
( pi * 14 / 16,  t ( pi * 14 / 16 )  )
( pi * 15 / 16,  t ( pi * 15 / 16 )  )
( pi * 16 / 16,  t ( pi * 16 / 16 )  )

[ color = GreenLighter2, draw = none, mark size = 3 pt, mark = square* ] coordinates

( 0,  g ( 0 )  )
( pi * 1 / 16,  g ( pi * 1 / 16 )  )
( pi * 2 / 16,  g ( pi * 2 / 16 )  )
( pi * 3 / 16,  g ( pi * 3 / 16 )  )
( pi * 4 / 16,  g ( pi * 4 / 16 )  )
( pi * 5 / 16,  g ( pi * 5 / 16 )  )
( pi * 6 / 16,  g ( pi * 6 / 16 )  )
( pi * 7 / 16,  g ( pi * 7 / 16 )  )
( pi * 8 / 16,  g ( pi * 8 / 16 )  )
( pi * 9 / 16,  g ( pi * 9 / 16 )  )
( pi * 10 / 16,  g ( pi * 10 / 16 )  )
( pi * 11 / 16,  g ( pi * 11 / 16 )  )
( pi * 12 / 16,  g ( pi * 12 / 16 )  )
( pi * 13 / 16,  g ( pi * 13 / 16 )  )
( pi * 14 / 16,  g ( pi * 14 / 16 )  )
( pi * 15 / 16,  g ( pi * 15 / 16 )  )
( pi * 16 / 16,  g ( pi * 16 / 16 )  )

This figure shows both players probabilities to win at the end of Phase $1$ when the QTG is played in $\langle \frac { 2 \pi } { 16 } \rangle$. The red and green lines depict the tosser and the gambler's, respectively, probabilities to win.

grow to left by = 1.00 cm,
grow to right by = 1.00 cm,
colback = WordIceBlue!25,
enhanced jigsaw,

sharp corners,
boxrule = 0.1 pt,
toprule = 0.1 pt,
bottomrule = 0.1 pt

This table gives the probabilities for each player at the end of Phase $1$ when QTG is played in $\langle \frac { 2 \pi } { 16 } \rangle$.

[HTML]000000   Probabilities at the end of Phase $1$ in $\langle \frac { 2 \pi } { 16 } \rangle$

[HTML]000000   $k$

[HTML]000000   $0$

[HTML]000000   $1, 15$

[HTML]000000   $2, 14$

[HTML]000000   $3, 13$

[HTML]000000   $4, 12$

[HTML]000000   $5, 11$

[HTML]000000   $6, 10$

[HTML]000000   $7, 9$

[HTML]000000   $8$

[HTML]000000   Tosser

[HTML]000000   Gambler

Continuing with our analysis, let us now proceed to Phase $2$ (see Figure <ref>). During Phase $2$, player $2$, the gambler, acts by changing the measurement basis. This action is expressed by the tensor product $R_{ y } ( \frac { 2 \pi l } { n } ) \otimes R_{ y } ( \frac { 2 \pi l } { n } )$ of equation (<ref>), which is given below in explicit form.

\begin{align} \label{eq:Ry Tensor Product Ry}
% \hspace{ - 1 cm}
R_{ y } \left( \frac { 2 \pi l } { n } \right) \otimes R_{ y } \left( \frac { 2 \pi l } { n } \right)
&\overset { \eqref{eq:General Action of Player 2} } { = }
\begin{bNiceMatrix}[ margin, cell-space-limits = 2 pt ] % , hvlines, rules/width = 0.0025 mm ]
\cos \frac { \pi l } { n } & - \sin \frac { \pi l } { n } \\
\sin \frac { \pi l } { n } & \cos \frac { \pi l } { n }
\end{bNiceMatrix}
\otimes
\begin{bNiceMatrix}[ margin, cell-space-limits = 2 pt ] % , hvlines, rules/width = 0.0025 mm ]
\cos \frac { \pi l } { n } & - \sin \frac { \pi l } { n } \\
\sin \frac { \pi l } { n } & \cos \frac { \pi l } { n }
\end{bNiceMatrix}
\nonumber \\
\begin{bNiceMatrix}[ margin, cell-space-limits = 2 pt ] % , hvlines, rules/width = 0.0025 mm ]
\cos \frac { \pi l } { n } \cos \frac { \pi l } { n } &
- \cos \frac { \pi l } { n } \sin \frac { \pi l } { n } &
- \sin \frac{ \pi l } { n } \cos \frac { \pi l } { n } &
\phantom{-} \sin \frac { \pi l } { n } \sin \frac { \pi l } { n }
\\
\cos \frac { \pi l } { n } \sin \frac { \pi l } { n } &
\phantom{-} \cos \frac { \pi l } { n } \cos \frac { \pi l } { n } &
- \sin \frac{ \pi l }{ n } \sin \frac { \pi l } { n } &
- \sin \frac { \pi l } { n } \cos \frac { \pi l } { n }
\\
\sin \frac { \pi l } { n } \cos \frac { \pi l } { n } &
- \sin \frac { \pi l } { n } \sin \frac{ \pi l } { n } &
\phantom{-} \cos \frac { \pi l } { n } \cos \frac { \pi l } { n } &
- \cos \frac { \pi l } { n } \sin \frac { \pi l } { n }
\\
\sin \frac { \pi l } { n } \sin \frac { \pi l } { n } &
\phantom{-} \sin \frac { \pi l } { n } \cos \frac { \pi l } { n } &
\phantom{-} \cos \frac { \pi l } { n } \sin \frac { \pi l } { n } &
\phantom{-} \cos \frac { \pi l } { n } \cos \frac { \pi l } { n }
\end{bNiceMatrix}
\ .
\end{align}

Therefore, as a result of the gambler's action, the system of the two coins at the end of Phase $2$ will be in the following $\ket{ \psi_{ 2 } }$ state.

\begin{align} \label{eq:TQG Phase 2}
\ket{ \psi_{ 2 } }
&\overset { \eqref{eq:TQG Phase 1}, \eqref{eq:Ry Tensor Product Ry} } { = }
\begin{bNiceMatrix}[ margin, cell-space-limits = 2 pt ] % , hvlines, rules/width = 0.0025 mm ]
\cos \frac { \pi l } { n } \cos \frac { \pi l } { n } &
- \cos \frac { \pi l } { n } \sin \frac { \pi l } { n } &
- \sin \frac{ \pi l } { n } \cos \frac { \pi l } { n } &
\phantom{-} \sin \frac { \pi l } { n } \sin \frac { \pi l } { n }
\\
\cos \frac { \pi l } { n } \sin \frac { \pi l } { n } &
\phantom{-} \cos \frac { \pi l } { n } \cos \frac { \pi l } { n } &
- \sin \frac{ \pi l }{ n } \sin \frac { \pi l } { n } &
- \sin \frac { \pi l } { n } \cos \frac { \pi l } { n }
\\
\sin \frac { \pi l } { n } \cos \frac { \pi l } { n } &
- \sin \frac { \pi l } { n } \sin \frac{ \pi l } { n } &
\phantom{-} \cos \frac { \pi l } { n } \cos \frac { \pi l } { n } &
- \cos \frac { \pi l } { n } \sin \frac { \pi l } { n }
\\
\sin \frac { \pi l } { n } \sin \frac { \pi l } { n } &
\phantom{-} \sin \frac { \pi l } { n } \cos \frac { \pi l } { n } &
\phantom{-} \cos \frac { \pi l } { n } \sin \frac { \pi l } { n } &
\phantom{-} \cos \frac { \pi l } { n } \cos \frac { \pi l } { n }
\end{bNiceMatrix}
\
\begin{bmatrix}
- \sin \frac { \pi k } { n }
\\
\\
\\
\cos \frac { \pi k } { n }
\end{bmatrix}
\nonumber \\
\begin{bNiceMatrix}[ margin, cell-space-limits = 2 pt ]
- \sin \frac { \pi k } { n } \cos^{ 2 } \frac { \pi l } { n } +
\cos \frac { \pi k } { n } \sin^{ 2 } \frac { \pi l } { n }
\\
- \sin \frac { \pi k } { n } \cos \frac { \pi l } { n } \sin \frac { \pi l } { n } -
\cos \frac { \pi k } { n } \cos \frac { \pi l } { n } \sin \frac { \pi l } { n }
\\
- \sin \frac { \pi k } { n } \cos \frac { \pi l } { n } \sin \frac { \pi l } { n } -
\cos \frac { \pi k } { n } \cos \frac { \pi l } { n } \sin \frac { \pi l } { n }
\\
- \sin \frac { \pi k } { n } \sin^{ 2 } \frac { \pi l } { n } +
\cos \frac { \pi k } { n } \cos^{ 2 } \frac { \pi l } { n }
\end{bNiceMatrix}
\ .
\end{align}

After the completion of Phase $2$, the measurement takes place. As formula (<ref>) implies, the probability $P_{ T }$ of player $1$ winning, that is of observing the system in state $\ket{ 0 } \ket{ 0 }$, is given by the next equation

\begin{align} \label{eq:Probability Tosser Wins}
P_{ T } ( k, l )
\left(
\cos \frac { \pi k } { n } \sin^{ 2 } \frac { \pi l } { n } -
\sin \frac { \pi k } { n } \cos^{ 2 } \frac { \pi l } { n }
\right)^{ 2 }
\ .
\end{align}

In a symmetric way, the probability $P_{ G }$ of player $2$ winning, i.e, finding the system in state $\ket{ 1 } \ket{ 1 }$ is

\begin{align} \label{eq:Probability Gambler Wins}
P_{ G } ( k, l )
\left(
\cos \frac { \pi k } { n } \cos^{ 2 } \frac { \pi l } { n } -
\sin \frac { \pi k } { n } \sin^{ 2 } \frac { \pi l } { n }
\right)^{ 2 }
\ .
\end{align}

Another important remark is that the probability $P_{ \text{draw} }$ that neither player wins, namely that the system ends up in the state $\ket{ 0 } \ket{ 1 }$ or $\ket{ 1 } \ket{ 0 }$, is expressed by

\begin{align} \label{eq:Draw Probability}
P_{ \text{draw} } ( k, l )
2 \
\left(
\cos \frac { \pi k } { n } + \sin \frac { \pi k } { n }
\right)^{ 2 }
\cos^{ 2 } \frac { \pi l } { n } \sin^{ 2 } \frac { \pi l } { n }
\ ,
\end{align}

and is, in general, nonzero.

A close examination of equations (<ref>) and (<ref>) reveals the characteristic property of the QTG: the game is fair for both players. In other words, irrespective of the tosser's choice of rotation, the subsequent action of the gambler balances the situation and puts both players on a equal footing. For any given $k$, $0 \leq k \leq n - 1$, chosen by the tosser, the gambler picks some orthonomal basis $B_{ \frac { 2 \pi l } { n } }$, $0 \leq l \leq n - 1$, which leads to the probabilities $P_{ T } ( k, l )$ and $P_{ G } ( k, l )$ that express the chances of the former and the latter, respectively, to win the game. The crucial observation is that for every basis $B_{ \frac { 2 \pi l } { n } }$, there exists a dual basis $B_{ \frac { 2 \pi l^{ \ast } } { n } }$ that exchanges the aforementioned probabilities, so that $P_{ T } ( k, l^{ \ast } ) = P_{ G } ( k, l )$ and $P_{ G } ( k, l^{ \ast } ) = P_{ T } ( k, l )$. For instance, say that for some basis $B_{ \frac { 2 \pi l } { n } }$, $P_{ T } ( k, l ) > P_{ G } ( k, l )$, i.e., player $1$ is more likely to win than player $2$. Then for the dual basis $B_{ \frac { 2 \pi l^{ \ast } } { n } }$, $P_{ G } ( k, l^{ \ast } ) = P_{ T } ( k, l ) > P_{ G } ( k, l ) = P_{ T } ( k, l^{ \ast } )$, which shows that the odds are perfectly reserved and player $2$ is now more likely to win than player $1$ by exactly the same amount. This fact is formally proved in the next Theorem <ref>.

The QTG is fair for both players. Formally, the characteristic property of the QTG is that for each $k$, $0 \leq k \leq n - 1$, to every orthonomal basis $B_{ \frac { 2 \pi l } { n } }$, $0 \leq l \leq n - 1$, corresponds a dual basis $B_{ \frac { 2 \pi l^{ \ast } } { n } }$ such that:

\begin{align} \label{eq:The Characteristic Property of the QTG}
P_{ T } ( k, l^{ \ast } ) &= P_{ G } ( k, l )
\\
\nonumber \\
P_{ G } ( k, l^{ \ast } ) &= P_{ T } ( k, l )
\ .
\end{align}

The proof is quite straightforward. First, let us recall some well-known trigonometric identities (see any standard textbook, such as [35]).

\begin{align}
\cos ( \theta \pm \frac { \pi } { 2 } ) &= \mp \sin \theta
\sin ( \theta \pm \frac { \pi } { 2 } ) &= \pm \cos \theta
\tag{$\star$}
\end{align}

Let us fix a specific $k$, $0 \leq k \leq n - 1$, representing the plane rotation $r^{ k }$ by $\frac { 2 \pi k } { n }$ radians, which is employed by the tosser during Phase $1$. Let us further assume that during Phase $2$ the gambler picks the orthonomal basis $B_{ \frac { 2 \pi l } { n } }$ for some $l$, $0 \leq l \leq n - 1$. According to formulas (<ref>) and (<ref>), the corresponding probabilities that player $1$ and $2$ wins are

\begin{align}
P_{ T } ( k, l )
\left(
\cos \frac { \pi k } { n } \sin^{ 2 } \frac { \pi l } { n } -
\sin \frac { \pi k } { n } \cos^{ 2 } \frac { \pi l } { n }
\right)^{ 2 }
\tag{\ref{eq:Probability Tosser Wins}}
\\
\nonumber \\
P_{ G } ( k, l )
\left(
\cos \frac { \pi k } { n } \cos^{ 2 } \frac { \pi l } { n } -
\sin \frac { \pi k } { n } \sin^{ 2 } \frac { \pi l } { n }
\right)^{ 2 }
\ .
\tag{\ref{eq:Probability Gambler Wins}}
\end{align}

Given $l$, we define its dual, denoted by $l^{ \ast }$, as follows:

\begin{align} \label{eq:Dual l^* D}
l^{ \ast }
\equiv
\left( l + \frac { n } { 2 } \right) \bmod n
\ .
\end{align}

The above compact definition, can be cast in the more expedient form shown below.

\begin{align} \label{eq:Explicit Dual l^*}
l^{ \ast }
\left\{
\begin{matrix*}[l]
l + \frac { n } { 2 } & \text{ if } l + \frac { n } { 2 } < n \\
l - \frac { n } { 2 } & \text{ if } l + \frac { n } { 2 } \geq n
\end{matrix*}
\right.
\ , \ 0 \leq l \leq n - 1 \ .
\end{align}

We now proceed with the proof by distinguishing the following cases.

* $l + \frac { n } { 2 } < n$. In this case, according to (<ref>), $l^{ \ast } = l + \frac { n } { 2 }$. Therefore,

\begin{align}
P_{ T } ( k, l^{ \ast } )
&\overset{\eqref{eq:Probability Tosser Wins}}{=}
\left(
\cos \frac { \pi k } { n } \sin^{ 2 } \frac { \pi l^{ \ast } } { n } -
\sin \frac { \pi k } { n } \cos^{ 2 } \frac { \pi l^{ \ast } } { n }
\right)^{ 2 }
\nonumber
\\
\left(
\cos \frac { \pi k } { n } \sin^{ 2 } \left( \frac { 2 \pi l + \pi n} { 2 n } \right) -
\sin \frac { \pi k } { n } \cos^{ 2 } \left( \frac { 2 \pi l + \pi n} { 2 n } \right)
\right)^{ 2 }
\nonumber
\\
\left(
\cos \frac { \pi k } { n } \sin^{ 2 } \left( \frac { \pi l } { n } + \frac { \pi } { 2 } \right) -
\sin \frac { \pi k } { n } \cos^{ 2 } \left( \frac { \pi l } { n } + \frac { \pi } { 2 } \right)
\right)^{ 2 }
\nonumber
\\
\left(
\cos \frac { \pi k } { n } \cos^{ 2 } \frac { \pi l } { n } -
\sin \frac { \pi k } { n } \sin^{ 2 } \frac { \pi l } { n }
\right)^{ 2 }
\nonumber
\\
&\overset{\eqref{eq:Probability Gambler Wins}}{=}
P_{ G } ( k, l )
\\
\nonumber
\\
P_{ G } ( k, l^{ \ast } )
&\overset{\eqref{eq:Probability Gambler Wins}}{=}
\left(
\cos \frac { \pi k } { n } \cos^{ 2 } \frac { \pi l^{ \ast } } { n } -
\sin \frac { \pi k } { n } \sin^{ 2 } \frac { \pi l^{ \ast } } { n }
\right)^{ 2 }
\nonumber
\\
\left(
\cos \frac { \pi k } { n } \cos^{ 2 } \left( \frac { 2 \pi l + \pi n} { 2 n } \right) -
\sin \frac { \pi k } { n } \sin^{ 2 } \left( \frac { 2 \pi l + \pi n} { 2 n } \right)
\right)^{ 2 }
\nonumber
\\
\left(
\cos \frac { \pi k } { n } \cos^{ 2 } \left( \frac { \pi l } { n } + \frac { \pi } { 2 } \right) -
\sin \frac { \pi k } { n } \sin^{ 2 } \left( \frac { \pi l } { n } + \frac { \pi } { 2 } \right)
\right)^{ 2 }
\nonumber
\\
\left(
\cos \frac { \pi k } { n } \sin^{ 2 } \frac { \pi l } { n } -
\sin \frac { \pi k } { n } \cos^{ 2 } \frac { \pi l } { n }
\right)^{ 2 }
\nonumber
\\
&\overset{\eqref{eq:Probability Tosser Wins}}{=}
P_{ T } ( k, l )
\ .
\end{align}

* $l + \frac { n } { 2 } \geq n$. In this case, according to (<ref>), $l^{ \ast } = l - \frac { n } { 2 }$. Therefore,

\begin{align}
P_{ T } ( k, l^{ \ast } )
&\overset{\eqref{eq:Probability Tosser Wins}}{=}
\left(
\cos \frac { \pi k } { n } \sin^{ 2 } \frac { \pi l^{ \ast } } { n } -
\sin \frac { \pi k } { n } \cos^{ 2 } \frac { \pi l^{ \ast } } { n }
\right)^{ 2 }
\nonumber
\\
\left(
\cos \frac { \pi k } { n } \sin^{ 2 } \left( \frac { 2 \pi l - \pi n} { 2 n } \right) -
\sin \frac { \pi k } { n } \cos^{ 2 } \left( \frac { 2 \pi l - \pi n} { 2 n } \right)
\right)^{ 2 }
\nonumber
\\
\left(
\cos \frac { \pi k } { n } \sin^{ 2 } \left( \frac { \pi l } { n } - \frac { \pi } { 2 } \right) -
\sin \frac { \pi k } { n } \cos^{ 2 } \left( \frac { \pi l } { n } - \frac { \pi } { 2 } \right)
\right)^{ 2 }
\nonumber
\\
\left(
\cos \frac { \pi k } { n } \cos^{ 2 } \frac { \pi l } { n } -
\sin \frac { \pi k } { n } \sin^{ 2 } \frac { \pi l } { n }
\right)^{ 2 }
\nonumber
\\
&\overset{\eqref{eq:Probability Gambler Wins}}{=}
P_{ G } ( k, l )
\\
\nonumber
\\
P_{ G } ( k, l^{ \ast } )
&\overset{\eqref{eq:Probability Gambler Wins}}{=}
\left(
\cos \frac { \pi k } { n } \cos^{ 2 } \frac { \pi l^{ \ast } } { n } -
\sin \frac { \pi k } { n } \sin^{ 2 } \frac { \pi l^{ \ast } } { n }
\right)^{ 2 }
\nonumber
\\
\left(
\cos \frac { \pi k } { n } \cos^{ 2 } \left( \frac { 2 \pi l - \pi n} { 2 n } \right) -
\sin \frac { \pi k } { n } \sin^{ 2 } \left( \frac { 2 \pi l - \pi n} { 2 n } \right)
\right)^{ 2 }
\nonumber
\\
\left(
\cos \frac { \pi k } { n } \cos^{ 2 } \left( \frac { \pi l } { n } - \frac { \pi } { 2 } \right) -
\sin \frac { \pi k } { n } \sin^{ 2 } \left( \frac { \pi l } { n } - \frac { \pi } { 2 } \right)
\right)^{ 2 }
\nonumber
\\
\left(
\cos \frac { \pi k } { n } \sin^{ 2 } \frac { \pi l } { n } -
\sin \frac { \pi k } { n } \cos^{ 2 } \frac { \pi l } { n }
\right)^{ 2 }
\nonumber
\\
&\overset{\eqref{eq:Probability Tosser Wins}}{=}
P_{ T } ( k, l )
\ .
\end{align}

Ergo, for every rotation $r^{ k }$, $0 \leq k \leq n - 1$, chosen by the tosser during Phase $1$, both players end up with the same probability distribution.

[Playing in the group $\langle \frac { 2 \pi } { 16 } \rangle$ continued]
This example continues and concludes Example <ref> by examining Phase $2$ (see Figure <ref>) of the QTG under the same assumption that the game is played in the cyclic rotation group $\langle \frac { 2 \pi } { 16 } \rangle$.

Recall that Figure <ref> visualizes the tosser's and the gambler's probabilities to win at the end of Phase $1$ as a function of the rotation $r^{ k }$, $0 \leq k \leq n - 1$, chosen by the tosser, and that the numerical values of the players' probabilities are contained in Table <ref>.

Let us consider the scenario in which the tosser chose the rotation $r^{ 6 } = \frac { 12 \pi } { 16 }$ in Phase $1$. This choice was implemented via the use of the rotation gate $R_{ y } ( \frac { 12 \pi } { 16 } )$, which, in reality, causes a rotation through an angle $\frac { 6 \pi } { 16 }$ around the $y$ axis of the Bloch sphere, and gave player $1$ probability $0.854$ to win at the end of Phase $1$, as shown in Table <ref>. The motivation for such a choice could be that the tosser, being somewhat greedy, wanted a clear advantage over the gambler, while still allowing the latter a nonzero, but very slim, probability to win so as to keep appearances. However, player $2$ can still turn the tables and gain the upper hand by picking an appropriate measurement basis. This fact is visualized in Figure <ref>, which depicts both players' probabilities as a function of the measurement basis. The exact numerical values for each possible choice of player $2$ are given in Table <ref>. With that in mind, it becomes evident that, if player $2$ picks $B_{ \frac { 16 \pi } { 12 } }$ as measurement basis, his probability to win becomes $0.854$, equal to the tosser's probability at the end of Phase $1$, which is another verification of Theorem <ref>.

A final observation is that the tossers choice has affected the range of probabilities of both players. Although both have the same probability distribution as dictated by Theorem <ref>, the maximum probability for each player at the end of Phase $2$ is $0.854$, as determined by tosser's initial choice of $R_{ y } ( \frac { 12 \pi } { 16 } )$ during Phase $1$.

grow to left by = 1.00 cm,
grow to right by = 1.00 cm,
colback = MagentaVeryLight!03,
enhanced jigsaw,

sharp corners,

toprule = 1.0 pt,
bottomrule = 1.0 pt,
leftrule = 0.1 pt,
rightrule = 0.1 pt,
sharp corners,
center title,
fonttitle = ]

compat = 1.16
[ scale = 1.00 ]

width = 7.00 cm, height = 5.50 cm,
axis line style =  WordBlueDarker, line width = 0.25 mm ,
axis x line = center,
axis y line = middle,
axis line style =  ->, > =  latex [ length = 7pt ]  ,
xmin = - 0.1, xmax = 3.40,
ymin = - 0.1, ymax = 1.1,
xtick =

pi * 1 / 16, pi * 2 / 16, pi * 3 / 16, pi * 4 / 16,
pi * 5 / 16, pi * 6 / 16, pi * 7 / 16, pi * 8 / 16,
pi * 9 / 16, pi * 10 / 16, pi * 11 / 16, pi * 12 / 16,
pi * 13 / 16, pi * 14 / 16, pi * 15 / 16, pi * 16 / 16
xticklabels =

$\frac { \pi } { 16 }$, $\frac { 2 \pi } { 16 }$,
$\frac { 3 \pi } { 16 }$, $\frac { \pi } { 4 }$,
$\frac { 5 \pi } { 16 }$, $\frac { 6 \pi } { 16 }$,
$\frac { 7 \pi } { 16 }$, $\frac { \pi } { 2 }$,
$\frac { 9 \pi } { 16 }$, $\frac { 10 \pi } { 16 }$,
$\frac { 11 \pi } { 16 }$, $\frac { 3 \pi } { 4 }$,
$\frac { 13 \pi } { 16 }$, $\frac { 14 \pi } { 16 }$,
$\frac { 15 \pi } { 16 }$, $\pi$
ytick =  0, 0.125, 0.25, 0.375, 0.50, 0.625, 0.75, 0.875, 1.00 ,
yticklabels =  0, 0.125, 0.250, 0.375, 0.500, 0.625, 0.750, 0.875, 1 ,
x label style =

at =  ( axis description cs: 0.5, -0.075 ) , fill = WordVeryLightTeal, anchor = north
xlabel =  Rotation angle $\frac { \pi l } { 16 } \ (0 \leq l \leq 15)$ ,
y label style =

at =  ( axis description cs: 0.025, 1.040 ) , fill = WordVeryLightTeal, anchor = south
ylabel =  Probabilities ,
x post scale = 2.0,
y post scale = 1.5,
grid = minor,
trig format plots = rad,
domain = 0 * pi : pi,
samples = 800,
declare function =

t6() = ( - sin( 6 * pi / ) * cos( )^2 + cos( 6 * pi / ) * sin( )^2 )^2;
g6() = ( - sin( 6 * pi / ) * sin( )^2 + cos( 6 * pi / ) * cos( )^2 )^2;

legend style =  at =  ( 0.5, 1.2 ) , anchor = south

empty legend
Phase $2$ in $\langle \frac { 2 \pi } { 16 } \rangle$ for $R_{ y } ( \frac { 12 \pi } { 16 } )$

[ RedPurple, line width = 0.4 mm ]  t6() ;

Probability Player $1$ wins

[ GreenLighter2, line width = 0.4 mm ]  g6() ;
Probability Player $2$ wins

[ color = RedPurple, draw = none, mark size = 3 pt, mark = * ] coordinates

( 0,  t6 ( 0 )  )
( pi * 1 / 16,  t6 ( pi * 1 / 16 )  )
( pi * 2 / 16,  t6 ( pi * 2 / 16 )  )
( pi * 3 / 16,  t6 ( pi * 3 / 16 )  )
( pi * 4 / 16,  t6 ( pi * 4 / 16 )  )
( pi * 5 / 16,  t6 ( pi * 5 / 16 )  )
( pi * 6 / 16,  t6 ( pi * 6 / 16 )  )
( pi * 7 / 16,  t6 ( pi * 7 / 16 )  )
( pi * 8 / 16,  t6 ( pi * 8 / 16 )  )
( pi * 9 / 16,  t6 ( pi * 9 / 16 )  )
( pi * 10 / 16,  t6 ( pi * 10 / 16 )  )
( pi * 11 / 16,  t6 ( pi * 11 / 16 )  )
( pi * 12 / 16,  t6 ( pi * 12 / 16 )  )
( pi * 13 / 16,  t6 ( pi * 13 / 16 )  )
( pi * 14 / 16,  t6 ( pi * 14 / 16 )  )
( pi * 15 / 16,  t6 ( pi * 15 / 16 )  )
( pi * 16 / 16,  t6 ( pi * 16 / 16 )  )

[ color = GreenLighter2, draw = none, mark size = 3 pt, mark = square* ] coordinates

( 0,  g6 ( 0 )  )
( pi * 1 / 16,  g6 ( pi * 1 / 16 )  )
( pi * 2 / 16,  g6 ( pi * 2 / 16 )  )
( pi * 3 / 16,  g6 ( pi * 3 / 16 )  )
( pi * 4 / 16,  g6 ( pi * 4 / 16 )  )
( pi * 5 / 16,  g6 ( pi * 5 / 16 )  )
( pi * 6 / 16,  g6 ( pi * 6 / 16 )  )
( pi * 7 / 16,  g6 ( pi * 7 / 16 )  )
( pi * 8 / 16,  g6 ( pi * 8 / 16 )  )
( pi * 9 / 16,  g6 ( pi * 9 / 16 )  )
( pi * 10 / 16,  g6 ( pi * 10 / 16 )  )
( pi * 11 / 16,  g6 ( pi * 11 / 16 )  )
( pi * 12 / 16,  g6 ( pi * 12 / 16 )  )
( pi * 13 / 16,  g6 ( pi * 13 / 16 )  )
( pi * 14 / 16,  g6 ( pi * 14 / 16 )  )
( pi * 15 / 16,  g6 ( pi * 15 / 16 )  )
( pi * 16 / 16,  g6 ( pi * 16 / 16 )  )

This figure shows the probabilities at the end of Phase $2$, assuming the QTG is played in $\langle \frac { 2 \pi } { 16 } \rangle$ and player $1$ has utilized the rotation gate $R_{ y } ( \frac { 12 \pi } { 16 } )$ during Phase $1$.

grow to left by = 1.00 cm,
grow to right by = 1.00 cm,
colback = WordIceBlue!25,
enhanced jigsaw,

sharp corners,
boxrule = 0.1 pt,
toprule = 0.1 pt,
bottomrule = 0.1 pt

This table gives the probabilities for each player at the end of Phase $2$ when QTG is played in $\langle \frac { 2 \pi } { 16 } \rangle$ and player $1$ has utilized the rotation gate $R_{ y } ( \frac { 12 \pi } { 16 } )$ during Phase $1$.

[HTML]000000   Probabilities at the end of Phase $2$ in $\langle \frac { 2 \pi } { 16 } \rangle$

[HTML]000000   $k$

[HTML]000000   $0$

[HTML]000000   $1, 15$

[HTML]000000   $2, 14$

[HTML]000000   $3, 13$

[HTML]000000   $4, 12$

[HTML]000000   $5, 11$

[HTML]000000   $6, 10$

[HTML]000000   $7, 9$

[HTML]000000   $8$

[HTML]000000   Tosser

[HTML]000000   Gambler

The last observation of Example <ref>, can be cast in the form of the next Corollary <ref>.

The tosser's choice of the rotation gate $R_{ y } ( \frac { 2 \pi k } { n } )$ during Phase $1$, determines the maximum probability at the end of Phase $2$ for both players.

§ DISCUSSION AND CONCLUSIONS

This paper introduces Quantum Tapsilou, a new quantum game inspired by the traditional Greek coin-tossing game known as tapsilou. While the classical version of the game features a straightforward and equal probability distribution, where both players have an equal $\frac { 1 } { 4 }$ probability of winning, Quantum Tapsilou introduces a new level of complexity and fairness by leveraging principles from quantum mechanics.

In Quantum Tapsilou, both players still maintain an equal chance of winning, but these probabilities are now influenced by their prior choices and the utilization of quantum principles. The following two key innovations are at the core of Quantum Tapsilou.

Entanglement via rotation gates instead of relying on the commonly used Hadamard gates for creating entangled states with equal probability amplitudes. Quantum Tapsilou employs rotation gates, specifically $R_y$ rotation gates in order to generate Bell-like states with unequal probability amplitudes, introducing a layer of complexity into the game dynamics. Quantum Tapsilou makes integral use of Groups in the sense that both players agree on a specific cyclic rotation group of order $n$, where $n$ should be a sufficiently large integer to provide players with additional choices and enhance the suspense of the game. This group serves as the foundation of the game, and both players select rotations from this group to execute their actions using the corresponding $R_y$ rotation gates. The selection of this group and the specific rotations within it significantly affect the game's outcome, introducing a strategic element based on group theory.

By incorporating these quantum principles and group theory, Quantum Tapsilou preserves the fairness aspect of the classical tapsilou game while introducing complexity and variability into the probabilities of winning. Players must carefully consider their choices of rotations within the agreed-upon group and the entangled states generated by the $R_y$ rotation gates to maximize their chances of winning in this intriguing quantum adaptation of a traditional Greek game. This example highlights the crucial role of groups in ensuring fairness in quantum games, which aligns with the findings in a previous result presented in [13], establishing that symmetric quantum games, where both players draw their moves from the same group, are inherently fair.

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