A counterexample to conjecture "Catch 22"

TL;DR

This paper presents a finite deterministic graphical game without Nash equilibria, providing a counterexample to a longstanding conjecture and exploring conditions for Nash solvability in multi-player games.

Contribution

It constructs a counterexample to the 'Catch 22' conjecture and investigates conditions under which finite DG games have Nash equilibria, including a new proof for play-once games satisfying a specific condition.

Findings

Counterexample disproves 'Catch 22' conjecture.

Nash equilibrium exists for play-once games satisfying condition (C0).

Open questions remain on combined conditions for Nash solvability.

Abstract

We construct a finite deterministic graphical (DG) game without Nash equilibria in pure stationary strategies. This game has 3 players $I=\{1,2,3\}$ and 5 outcomes: 2 terminal $a_1$ and $a_2$ and 3 cyclic. Furthermore, for 2 players a terminal outcome is the best: $a_1$ for player 3 and $a_2$ for player 1. Hence, the rank vector $r$ is at most $(1,2,1)$. Here $r_i$ is the number of terminal outcomes that are worse than some cyclic outcome for the player $i \in I$. This is a counterexample to conjecture ``Catch 22" from the paper ``On Nash-solvability of finite $n$-person DG games, Catch 22" (2021) arXiv:2111.06278, according to which, at least 2 entries of $r$ are at least 2 for any NE-free game. However, Catch 22 remains still open for the games with a unique cyclic outcome, not to mention a weaker (and more important) conjecture claiming that an $n$-person finite DG game has a Nash equilibrium (in pure stationary strategies) when $r = (0^n)$, that is, all $n$ entries of $r$ are 0; in other words, when the following condition holds: $\qquad\bullet$ ($C_0$) any terminal outcome is better than every cyclic one for each player. A game is play-once if each player controls a unique position. It is known that any play-once game satisfying ($C_0$) has a Nash equilibrium. We give a new and very short proof of this statement. Yet, not only conjunction but already disjunction of the above two conditions may be sufficient for Nash-solvability. This is still open.