# Quantum Tapsilou—A Quantum Game Inspired by the Traditional Greek Coin Tossing Game Tapsilou

^{*}

^{†}

## Abstract

**:**

## 1. Introduction

#### 1.1. Related Work

**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 probability distribution, with both players having a $\frac{1}{4}$ probability of winning, Quantum Tapsilou exhibits an additional level of complexity 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 new 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 a sufficiently large integer in order to provide the players with extra choices and increase the suspense of the game. This group forms the basis of the game because both players select rotations from this group to perform their actions using the corresponding ${R}_{y}$ rotation gates. Agreeing on this group, and, subsequently, choosing the precise rotations within the group, significantly impact the outcome of the game, introducing a strategic element based on group theory.

#### 1.2. Organization

## 2. The Traditional Game “Tapsilou”

## 3. Two Coins Seem to Be the Proper Choice

- (
**C**_{1}) - 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;
- (
**C**_{2}) - The number of coins is 2;
- (
**C**_{3}) - Viewing the two coins as comprising a system, we assume from now on that the initial state of this system is $(H,T)$;
- (
**C**_{4}) - 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;
- (
**C**_{5}) - The probability that player 1 wins is equal to the probability that player 2 wins, namely $\frac{1}{4}$;
- (
**C**_{6}) - The probability that neither player wins is $\frac{1}{2}$.

## 4. The Intuition behind the Quantum Tapsilou

- 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.

## 5. The Formal Presentation of the QTG

- (
**N**_{1}) - The general cyclic rotation group of order n, denoted by $\langle r\rangle $, contains the following n elements$$\begin{array}{c}\hfill \langle r\rangle =\{\mathbb{1},r,{r}^{2},\dots ,{r}^{n-1}\}\phantom{\rule{4pt}{0ex}},\end{array}$$
- (
**N**_{2}) - To each group element ${r}^{k}$, which represents the plane rotation by $\frac{2\pi k}{n}$, we associate the rotation gate ${R}_{y}\left(\frac{2\pi k}{n}\right)$$$\begin{array}{c}\hfill {R}_{y}\left(\frac{2\pi k}{n}\right)=\left[\begin{array}{cc}\hfill cos\frac{\pi k}{n}& \hfill -sin\frac{\pi k}{n}\\ \hfill sin\frac{\pi k}{n}& \hfill \phantom{-}cos\frac{\pi k}{n}\end{array}\right]\phantom{\rule{4pt}{0ex}},\end{array}$$$$\begin{array}{c}\hfill {|0\rangle}_{\frac{2\pi k}{n}}=\left[\begin{array}{c}cos\frac{\pi k}{n}\\ sin\frac{\pi k}{n}\end{array}\right]\end{array}$$$$\begin{array}{c}\hfill {|1\rangle}_{\frac{2\pi k}{n}}=\left[\begin{array}{c}\hfill -sin\frac{\pi k}{n}\\ \hfill \phantom{-}cos\frac{\pi k}{n}\end{array}\right]\phantom{\rule{4pt}{0ex}},\end{array}$$$$\begin{array}{c}\hfill {B}_{\frac{2\pi k}{n}}=\left[\begin{array}{cc}\hfill \phantom{-}cos\frac{\pi k}{n}& \hfill sin\frac{\pi k}{n}\\ \hfill -sin\frac{\pi k}{n}& \hfill cos\frac{\pi k}{n}\end{array}\right]\phantom{\rule{4pt}{0ex}},\end{array}$$

- (
**R**_{1}) - The initial state of the two coin system is $|0\rangle |1\rangle $. The winning state for player 1 (the tosser) is $|0\rangle |0\rangle $, and the winning state for player 2 (the gambler) is $|1\rangle |1\rangle $;
- (
**R**_{2}) - 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;
- (
**R**_{3}) - Player 1 chooses an element ${r}^{k}\in \langle r\rangle $, $0\le k\le n-1$, and uses the rotation operator ${R}_{y}\left(\frac{2\pi k}{n}\right)$ (recall Equation (7)) to entangle the two coins. Formally, player 1’s action is described by$$\begin{array}{c}\hfill CNOT\phantom{\rule{4pt}{0ex}}\xb7\phantom{\rule{4pt}{0ex}}\left(I\otimes {R}_{y}\left(\frac{2\pi k}{n}\right)\right)\phantom{\rule{4pt}{0ex}},\end{array}$$
- (
**R**_{4}) - Player 2 chooses an element ${r}^{l}\in \langle r\rangle $, $0\le l\le n-1$, and measures the two coins with respect to the orthonormal basis ${B}_{\frac{2\pi l}{n}}$ (recall Formula (10)). Formally, player 2’s action is described by$$\begin{array}{c}\hfill {R}_{y}\left(\frac{2\pi l}{n}\right)\otimes {R}_{y}\left(\frac{2\pi l}{n}\right)\phantom{\rule{4pt}{0ex}}.\end{array}$$

## 6. The Mathematical Analysis of the QTG

**R**${}_{1}$), the initial state of the system, denoted by $|{\psi}_{0}\rangle $, is given by

**R**${}_{3}$). In view of Formula (7), the matrix representation of the tensor product $(I\otimes {R}_{y}\left(\frac{2\pi k}{n}\right))$ is

**Example**

**1**

**.**This example examines how the QTG evolves up to Phase 1 (see Figure 11), 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.

**Theorem**

**1.**

**Proof.**

- $l+\frac{n}{2}<n$. In this case, according to (27), ${l}^{*}=l+\frac{n}{2}$. Therefore,$$\begin{array}{cc}\hfill {P}_{T}(k,{l}^{*})\phantom{\rule{1.em}{0ex}}& \stackrel{\left(20\right)}{=}{\left(cos\frac{\pi k}{n}{sin}^{2}\frac{\pi {l}^{*}}{n}-sin\frac{\pi k}{n}{cos}^{2}\frac{\pi {l}^{*}}{n}\right)}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& ={\left(cos\frac{\pi k}{n}{sin}^{2}\left(\frac{2\pi l+\pi n}{2n}\right)-sin\frac{\pi k}{n}{cos}^{2}\left(\frac{2\pi l+\pi n}{2n}\right)\right)}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& ={\left(cos\frac{\pi k}{n}{sin}^{2}\left(\frac{\pi l}{n}+\frac{\pi}{2}\right)-sin\frac{\pi k}{n}{cos}^{2}\left(\frac{\pi l}{n}+\frac{\pi}{2}\right)\right)}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \stackrel{\left(25\right)}{=}{\left(cos\frac{\pi k}{n}{cos}^{2}\frac{\pi l}{n}-sin\frac{\pi k}{n}{sin}^{2}\frac{\pi l}{n}\right)}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \stackrel{\left(21\right)}{=}{P}_{G}(k,l)\hfill \end{array}$$$$\begin{array}{cc}\hfill \phantom{\rule{1.em}{0ex}}& \mathrm{and}\hfill \\ \hfill {P}_{G}(k,{l}^{*})\phantom{\rule{1.em}{0ex}}& \stackrel{\left(21\right)}{=}{\left(cos\frac{\pi k}{n}{cos}^{2}\frac{\pi {l}^{*}}{n}-sin\frac{\pi k}{n}{sin}^{2}\frac{\pi {l}^{*}}{n}\right)}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& ={\left(cos\frac{\pi k}{n}{cos}^{2}\left(\frac{2\pi l+\pi n}{2n}\right)-sin\frac{\pi k}{n}{sin}^{2}\left(\frac{2\pi l+\pi n}{2n}\right)\right)}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& ={\left(cos\frac{\pi k}{n}{cos}^{2}\left(\frac{\pi l}{n}+\frac{\pi}{2}\right)-sin\frac{\pi k}{n}{sin}^{2}\left(\frac{\pi l}{n}+\frac{\pi}{2}\right)\right)}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \stackrel{\left(25\right)}{=}{\left(cos\frac{\pi k}{n}{sin}^{2}\frac{\pi l}{n}-sin\frac{\pi k}{n}{cos}^{2}\frac{\pi l}{n}\right)}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \stackrel{\left(20\right)}{=}{P}_{T}(k,l)\phantom{\rule{4pt}{0ex}}.\hfill \end{array}$$
- $l+\frac{n}{2}\ge n$. In this case, according to (27), ${l}^{*}=l-\frac{n}{2}$. Therefore,$$\begin{array}{cc}\hfill {P}_{T}(k,{l}^{*})\phantom{\rule{1.em}{0ex}}& \stackrel{\left(21\right)}{=}{\left(cos\frac{\pi k}{n}{sin}^{2}\frac{\pi {l}^{*}}{n}-sin\frac{\pi k}{n}{cos}^{2}\frac{\pi {l}^{*}}{n}\right)}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& ={\left(cos\frac{\pi k}{n}{sin}^{2}\left(\frac{2\pi l-\pi n}{2n}\right)-sin\frac{\pi k}{n}{cos}^{2}\left(\frac{2\pi l-\pi n}{2n}\right)\right)}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& ={\left(cos\frac{\pi k}{n}{sin}^{2}\left(\frac{\pi l}{n}-\frac{\pi}{2}\right)-sin\frac{\pi k}{n}{cos}^{2}\left(\frac{\pi l}{n}-\frac{\pi}{2}\right)\right)}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \stackrel{\left(25\right)}{=}{\left(cos\frac{\pi k}{n}{cos}^{2}\frac{\pi l}{n}-sin\frac{\pi k}{n}{sin}^{2}\frac{\pi l}{n}\right)}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \stackrel{\left(21\right)}{=}{P}_{G}(k,l)\hfill \end{array}$$$$\begin{array}{cc}\hfill \phantom{\rule{1.em}{0ex}}& \mathrm{and}\hfill \\ \hfill {P}_{G}(k,{l}^{*})\phantom{\rule{1.em}{0ex}}& \stackrel{\left(21\right)}{=}{\left(cos\frac{\pi k}{n}{cos}^{2}\frac{\pi {l}^{*}}{n}-sin\frac{\pi k}{n}{sin}^{2}\frac{\pi {l}^{*}}{n}\right)}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& ={\left(cos\frac{\pi k}{n}{cos}^{2}\left(\frac{2\pi l-\pi n}{2n}\right)-sin\frac{\pi k}{n}{sin}^{2}\left(\frac{2\pi l-\pi n}{2n}\right)\right)}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& ={\left(cos\frac{\pi k}{n}{cos}^{2}\left(\frac{\pi l}{n}-\frac{\pi}{2}\right)-sin\frac{\pi k}{n}{sin}^{2}\left(\frac{\pi l}{n}-\frac{\pi}{2}\right)\right)}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \stackrel{\left(25\right)}{=}{\left(cos\frac{\pi k}{n}{sin}^{2}\frac{\pi l}{n}-sin\frac{\pi k}{n}{cos}^{2}\frac{\pi l}{n}\right)}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \stackrel{\left(20\right)}{=}{P}_{T}(k,l)\phantom{\rule{4pt}{0ex}}.\hfill \end{array}$$

**Example**

**2**

**.**This example continues and concludes Example 1 by examining Phase 2 (see Figure 11) of the QTG under the same assumption that the game is played in the cyclic rotation group $\langle \frac{2\pi}{16}\rangle $.

**Corollary**

**1**

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

## 7. Discussion and Conclusions

#### Benefits and Future Applications

## Author Contributions

## Funding

## Institutional Review Board Statement

## Informed Consent Statement

## Data Availability Statement

## Acknowledgments

## Conflicts of Interest

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**Figure 1.**A typical example of coins from the 1960s (see also Ref. [37]).

**Figure 2.**The initial placement of the two coins, just before the tossing (see also Ref. [38]).

**Figure 3.**This figure shows a typical quantum circuit that generates $|{\Phi}^{-}\rangle $ pairs, given that the initial state is $|0\rangle |0\rangle $.

**Figure 4.**This figure depicts the state vector of the system of the two coins entangled in the $|{\Phi}^{-}\rangle $ state.

**Figure 5.**Assuming that the initial state is $|0\rangle |0\rangle $, the above figure depicts a typical quantum circuit that generates $|{\Psi}^{-}\rangle $ pairs.

**Figure 6.**The above figure visualizes the state vector of the two coin system in the $|{\Psi}^{-}\rangle $ state.

**Figure 7.**Given that the initial state is $|0\rangle |0\rangle $, the above quantum circuit applies the rotation gate ${R}_{y}\left(\frac{12\pi}{16}\right)$ to drive the system to the state visualized in Figure 8.

**Figure 8.**This circuit of Figure 7 guarantees that both coins will end up in the same state, but state $|0\rangle |0\rangle $ is much more probable.

**Figure 9.**The effect of the application of the two rotation gates prior to measurement is that the measuring basis is no longer the computational basis.

**Figure 10.**The circuit of Figure 9 drives the system to the state described by the above state vector. Now the odds favor player 2.

**Figure 12.**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.

**Figure 13.**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 ${R}_{y}\left(\frac{12\pi}{16}\right)$ during Phase 1.

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

Coins & Probability | |||
---|---|---|---|

# of Coins | P (Player 1 Wins) | P (Player 2 Wins) | P (Neither Wins) |

3 | 0.125 | 0.125 | 0.750 |

4 | 0.0625 | 0.0625 | 0.875 |

5 | 0.0313 | 0.0313 | 0.938 |

**Table 2.**This table compares the characteristics of the classical tapsilou against the corresponding characteristics of the quantum tapsilou.

Classical vs. Quantum Tapsilou | ||||
---|---|---|---|---|

State | Initial State | Winning State for Player 1 | Winning State for Player 2 | |

Classical | H (Heads) T (Tails) | $(H,T)$ | $(H,H)$ | $(T,T)$ |

Quantum | |0〉 |1〉 | $|0\rangle |1\rangle $ | $|0\rangle |0\rangle $ | $|1\rangle |1\rangle $ |

**Table 3.**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 $.

Probabilities at the End of Phase 1 in $\langle \frac{2\mathit{\pi}}{16}\rangle $ | |||||||||
---|---|---|---|---|---|---|---|---|---|

$k$ | 0 | $\mathbf{1},\mathbf{15}$ | $\mathbf{2},\mathbf{14}$ | $\mathbf{3},\mathbf{13}$ | $\mathbf{4},\mathbf{12}$ | $\mathbf{5},\mathbf{11}$ | $\mathbf{6},\mathbf{10}$ | $\mathbf{7},\mathbf{9}$ | 8 |

Tosser | 0 | $0.038$ | $0.146$ | $0.309$ | $0.5$ | $0.691$ | $0.854$ | $0.962$ | 1 |

Gambler | 1 | $0.962$ | $0.854$ | $0.691$ | $0.5$ | $0.309$ | $0.146$ | $0.038$ | 0 |

**Table 4.**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 ${R}_{y}\left(\frac{12\pi}{16}\right)$ during Phase 1.

Probabilities at the End of Phase 2 in $\langle \frac{2\mathit{\pi}}{16}\rangle $ | |||||||||
---|---|---|---|---|---|---|---|---|---|

$k$ | 0 | $\mathbf{1},\mathbf{15}$ | $\mathbf{2},\mathbf{14}$ | $\mathbf{3},\mathbf{13}$ | $\mathbf{4},\mathbf{12}$ | $\mathbf{5},\mathbf{11}$ | $\mathbf{6},\mathbf{10}$ | $\mathbf{7},\mathbf{9}$ | 8 |

Tosser | $0.854$ | $0.764$ | $0.537$ | $0.271$ | $0.073$ | 0 | $0.037$ | $0.111$ | $0.146$ |

Gambler | $0.146$ | $0.111$ | $0.037$ | 0 | $0.073$ | $0.271$ | $0.537$ | $0.764$ | $0.854$ |

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## Share and Cite

**MDPI and ACS Style**

Kastampolidou, K.; Andronikos, T.
Quantum Tapsilou—A Quantum Game Inspired by the Traditional Greek Coin Tossing Game Tapsilou. *Games* **2023**, *14*, 72.
https://doi.org/10.3390/g14060072

**AMA Style**

Kastampolidou K, Andronikos T.
Quantum Tapsilou—A Quantum Game Inspired by the Traditional Greek Coin Tossing Game Tapsilou. *Games*. 2023; 14(6):72.
https://doi.org/10.3390/g14060072

**Chicago/Turabian Style**

Kastampolidou, Kalliopi, and Theodore Andronikos.
2023. "Quantum Tapsilou—A Quantum Game Inspired by the Traditional Greek Coin Tossing Game Tapsilou" *Games* 14, no. 6: 72.
https://doi.org/10.3390/g14060072