Consider a 3 player game (e.g. poker for 3 people).

Nash equilibrium is a set of strategies $s_1, s_2, s_3$, such that the expected profit value for each player $s_i$

$E_i(s_1, ..., s_i', ..., s_N)$

is maximized when $s_i'= s_i$

If I understand correctly, this applies when all other players are playing according to the equilibrium strategy.

However, suppose that the 3-rd player deviates from the equilibrium strategy $s_3' \neq s_3$. Could the first player choose a strategy $s_1'$ to exploit both the 3-th and the 2nd player, such that

$E_2(s_1', s_2, s_3')$ < $E_2(s_1, s_2, s_3)$ for the second player?


1 Answer 1


Here's an example from the book Naive Decision Making by Tom Körner. We have a 3 player game in which each player has two strategies A and B. The payoffs are that if everyone plays the same strategy they each get 2, otherwise the majority players get 3 and the minority 0. "All choose A" is a Nash equilibrium, since any individual player loses out by switching to B (they would get 0 instead of 2). On the other hand if player three deviates then player 1 does best by also deviating, assuming player 2 stays with A.

  • $\begingroup$ That's a great example. A following question: can player 2 always change his strategy to not be exploited, or is there a pair of strategies s1 and s3 (not necessary optimal for p1 or p3), such that p2 is always exploited? Feel free to choose other game as an example $\endgroup$ Jun 17, 2022 at 18:02
  • $\begingroup$ I'm not sure I quite understand what you mean by p2 being exploited. It's certainly possible p3 picks a strategy for which every outcome for p2 is worse than before. If all three players get to change their strategies,the original Nash equilibrium isn't relevant any more. OTOH once p3 picks their strategy, assuming p1 and p2 know what it was, p1 and p2 are playing one another in a 2 person game which will have its own Nash equilibria. $\endgroup$ Jun 17, 2022 at 18:21
  • $\begingroup$ So in this example, each player pays 2 cents to play a game, and in the NE case, they win 2 cents back. Hence, EV for each player is 0. Being exploitable in this game means that your EV is negative. So I'm wondering if p1 and p3 can choose a strategy, such that p2 is exploited, i.e. EV < 0 for p2 no matter how he plays. But you already told that the answer is yes. $\endgroup$ Jun 18, 2022 at 7:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .