# On the Nash Equilibria of a Duel with Terminal Payoffs

## Abstract

**:**

## 1. Introduction

## 2. Game Description

- The game stays in state 11 ad infinitum (no player is ever killed);
- At some ${t}^{\prime}$ the game moves to a state $\mathbf{s}\left({t}^{\prime}\right)\in \left\{10,01,00\right\}$ (one or both players are killed). These are terminal states, i.e., as soon as they are reached, the game terminates.

## 3. Stationary Equilibria

- The transition to state 10 gives payoff ${a}_{1}$ and takes place with probability ${x}_{1}{p}_{1}$ (${P}_{1}$ shot and hit ${P}_{2}$) multiplied by $\left({x}_{2}{\overline{p}}_{2}+{\overline{x}}_{2}\right)$ (${P}_{2}$ either shot and missed or did not shoot);
- The transition to state 01 gives payoff $-{b}_{1}$ and takes place with probability ${x}_{2}{p}_{2}$ (${P}_{2}$ shot and hit ${P}_{1}$) multiplied by $\left({x}_{1}{\overline{p}}_{1}+{\overline{x}}_{1}\right)$ (${P}_{1}$ either shot and missed or did not shoot);
- The transition to state 00 gives payoff ${a}_{1}-{b}_{1}$ and takes place with probability ${x}_{1}{p}_{1}$ (${P}_{1}$ shot and hit ${P}_{2}$) multiplied by ${x}_{2}{p}_{2}$ (${P}_{2}$ shot and hit ${P}_{1}$);
- The transition to state 11 gives payoff ${V}_{1}$ (it is as if the game starts from the beginning) and takes place with probability $\left({\overline{x}}_{1}+{x}_{1}{\overline{p}}_{1}\right)$ (${P}_{1}$ either shot and missed or did not shoot) multiplied by $\left({x}_{2}{\overline{p}}_{2}+{\overline{x}}_{2}\right)$ (${P}_{2}$ either shot and missed or did not shoot).

**Proposition**

**1.**

**Proof.**

## 4. Connection to the Iterated Prisoner’s Dilemma

- The IPD is a deterministic game, while $\Gamma \left(\mathbf{p},\mathbf{a},\mathbf{b}\right)$ involves randomness;
- In the IPD, each player receives a payoff in every turn and the total payoff is the discounted (by a discount factor $\gamma $) sum of turn payoffs, while in $\Gamma \left(\mathbf{p},\mathbf{a},\mathbf{b}\right)$, payoff is obtained only at the final turn and is undiscounted;
- The IPD will last an infinite number of turns, while $\Gamma \left(\mathbf{p},\mathbf{a},\mathbf{b}\right)$ may (depending on the p values and the strategy used) terminate in a finite number of turns (in fact, it may be the case that it will terminate in a finite number of terms with probability one).

## 5. Non-Stationary Equilibria

**Proposition**

**2.**

**Proof.**

**Remark**

**1.**

**Proposition**

**3.**

**Proof.**

- They can continue shooting in all subsequent turns, in which case, so will ${P}_{2}$;
- They can revert to not shooting, in which case, in the next turn, they are in the same situation as at the start of the game.

- Consider first the case in which ${P}_{1}$ adopts the strategy ${\sigma}^{D}$ of shooting in each turn. Then we have$${Q}_{n}\left(11,{\sigma}^{D},{\sigma}^{TfT}\right)={Q}_{1}\left(11,\left({\sigma}^{D},{\sigma}^{G}\right)\right)={V}_{1}^{R}$$$${p}_{2}\left(1-{p}_{1}\right)\left({p}_{1}{a}_{1}-{b}_{1}\right)+{p}_{1}{a}_{1}<0$$$$\frac{{b}_{1}}{{a}_{1}}>\frac{1+{p}_{2}\left(1-{p}_{1}\right)}{1-{p}_{1}}\xb7\frac{{p}_{1}}{{p}_{2}}\phantom{\rule{4.pt}{0ex}}.$$
- Next consider the case in which ${P}_{1}$ alternates between shooting and not shooting. Then their payoff will be$${V}_{1}^{S}={p}_{1}{a}_{1}+\left(1-{p}_{1}\right)\left(0+{p}_{2}\left(-{b}_{1}\right)+\left(1-{p}_{2}\right){V}_{1}^{S}\right).$$The above equation holds because the expected payoff ${V}_{1}^{S}$ is computed by summing the following possibilities. ${P}_{1}$ will certainly shoot and then:
- (a)
- With probability ${p}_{1}$, ${P}_{2}$ will kill ${P}_{2}$ and hence, receive payoff ${a}_{1}$;
- (b)
- With probability $1-{p}_{1}$, ${P}_{2}$ will miss (and receive zero payoff) and in the next turn ${P}_{2}$ will shoot and kill ${P}_{1}$; this combination has probability $\left(1-{p}_{1}\right){p}_{2}$ and gives to ${P}_{1}$ payoff $-{b}_{1}$;
- (c)
- With probability $1-{p}_{1}$, ${P}_{1}$ will miss and in the next turn ${P}_{2}$ will shoot and miss ${P}_{1}$; this combination has probability $\left(1-{p}_{1}\right)\left(1-{p}_{2}\right)$ and returns the game to the original state, in which ${P}_{1}$ receives payoff ${V}_{1}^{S}$.

Simplifying the above equation and solving we obtain$${V}_{1}^{S}=\frac{{p}_{1}{a}_{1}-{p}_{2}{b}_{1}+{p}_{2}{b}_{1}{p}_{1}}{{p}_{1}+{p}_{2}-{p}_{2}{p}_{1}}.$$For an NE we must have ${V}_{1}^{C}-{V}_{1}^{S}>0$ and this will hold when$$0>{p}_{1}{a}_{1}-{p}_{2}{b}_{1}+{p}_{2}{b}_{1}{p}_{1}\iff {p}_{2}{b}_{1}\left(1-{p}_{1}\right)>{p}_{1}{a}_{1}\iff \frac{{b}_{1}}{{a}_{1}}>\frac{1}{1-{p}_{1}}\xb7\frac{{p}_{1}}{{p}_{2}}.$$However, from our assumption (10), we have$$\frac{{b}_{1}}{{a}_{1}}>\frac{1+{p}_{2}\left(1-{p}_{1}\right)}{1-{p}_{1}}\xb7\frac{{p}_{1}}{{p}_{2}}\phantom{\rule{4.pt}{0ex}}>\frac{1}{1-{p}_{1}}\xb7\frac{{p}_{1}}{{p}_{2}}\phantom{\rule{4.pt}{0ex}}.$$Hence ${V}_{1}^{C}-{V}_{1}^{S}>0$.

**Corollary**

**1.**

## 6. Conclusions

- sequential play, in which a single player is allowed to shoot in each turn;
- random play, in which the player allowed to shoot in each turn is chosen randomly and equi-probably.

## Funding

## Data Availability Statement

## Conflicts of Interest

## Notes

1 | |

2 | In the sequel we use the standard game theoretic notation by which ${s}_{-1}={s}_{2}$, ${s}_{-2}={s}_{1}$. The same notation is used for players, actions etc. |

3 | Several parts of this paper require rather involved algebraic calculations. We have always performed these using the computer algebra system Maple and afterwards verified the results by hand. |

4 | We should clarify at this point that, despite the use of the terms “cooperation” and “cooperative”, the duel is not a cooperative game in Shapley’s sense [34]. In other words, it does not involve external enforcement of cooperative behavior. Instead, the duel is a non-cooperative game and “cooperation” is used in the same sense as in the Prisoner’s Dilemma literature; i.e., “cooperation” is understood as a spontaneous emergence of coordinated moves due to the players’ selfish behavior, rather than due to an explicit alliance mechanism. |

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Kehagias, A.
On the Nash Equilibria of a Duel with Terminal Payoffs. *Games* **2023**, *14*, 62.
https://doi.org/10.3390/g14050062

**AMA Style**

Kehagias A.
On the Nash Equilibria of a Duel with Terminal Payoffs. *Games*. 2023; 14(5):62.
https://doi.org/10.3390/g14050062

**Chicago/Turabian Style**

Kehagias, Athanasios.
2023. "On the Nash Equilibria of a Duel with Terminal Payoffs" *Games* 14, no. 5: 62.
https://doi.org/10.3390/g14050062