# Is there an equilibrium for this non-zero-sum game?

The game $$G(N,M)$$ is played:

$$N$$ ($$N\geq 2$$) is the number of players, labeled $$1$$~$$N$$. In the beginning they have a pot with some chips in it. Players move alternatively in the order from $$1$$ to $$N$$. In their move, a player announces an integer $$C$$ and toss a fair coin: if head, $$C$$ more chips are added to the pot; if tail, $$C$$ chips are removed from the pot, where $$1\leq C\leq$$ the current number of chips in the pot.

The game ends on two conditions:

1. The pot is empty after a player's move, in which case that player loses the game, and everyone else wins.
2. There're $$M$$ or more chips in the pot, in which case everyone wins.

Communication is not allowed, and we assume a player's choice of integer $$C$$ is a function only of the current number of chips in the pot. Formally, in game $$G(N,M)$$ a player's strategy is a function $$f: \{1,2,...,M-1\} \longmapsto \{1,2,...,M-1\}$$, with the restriction $$f(x)\leq x, \forall x$$.

Question: Is there always an equilibrium for $$G(N,M)$$? If so, what can we say about the equilibria? Is it feasible to search for an equilibrium of, say $$G(3,1000)$$?

Edit: Notice that the strategy of always betting all but one chips can't be an equilibrium for many games. For example in $$G(3,5)$$, if the other 2 players stick to that strategy and there are 4 chips, you're better off betting 1 rather than 3.

• What if after the move of player N there are 0<x<M chips left in the pot? Do they start another round from player 1? – Pietro Majer Jul 21 at 15:26
• @PietroMajer Yes, they do. – Eric Jul 21 at 15:27
• The obvious play is to bet all but one of the chips if there's more than one. You can't lose immediately, and if you throw tails you have 50% odds that the player after you loses. If everyone does this (and by symmetry the best strategy is the same for everyone) then with large $N$ it's increasingly likely that someone will lose before it comes back to you. – Peter Taylor Jul 21 at 15:38
• @PeterTaylor That sounds reasonable. But if $M\gg N$, will it be better to bet $M-K$ chips if the current number of chips $K$ is very close to $M$? – Eric Jul 21 at 15:57
• I would like to comment on the part "by symmetry the best strategy is the same for everyone". It is true that this is a symmetric game, hence there is a symmetric equilibrium. But a player's equilibrium strategy depends on the current number of chips in the pot, and it is not clear that for K (where K is the number of chips in the pot) the symmetric equilibrium strategy will dictate the same behavior (same is whatever meaning you would want). – Eilon Jul 22 at 5:29