I am trying to construct a normal-form game satisfying all of the following conditions:

  • there are two players;
  • each player has a finite number of strategies;
  • there is a unique Nash equilibrium, in which
  • one of the players uses a pure strategy; and
  • the other player uses a non-degenerate mixed strategy.

I have put considerable effort into trying to construct an example, but I keep finding that if a pure–mixed Nash equilibrium exists, then there exist other Nash equilibria as well. I am actually wondering if it is possible to construct an example with a unique such Nash equilibrium.

If this is not possible, I am wondering how to prove the following implication of this:

Let $\Gamma$ be a normal-form game with two players. The strategy sets are $S_1$ and $S_2$, each of which is a non-empty finite set. If $(\sigma_1^*,\sigma_2^*)\in\Delta(S_1)\times\Delta(S_2)$ is a Nash equilibrium of this game such that

  • $\sigma_1^*(s_1)=1$ for some $s_1\in S_1$; and

  • $\sigma_2^*(s_2)<1$ for all $s_2\in S_2$,

then there exist other Nash equilibria.

Any feedback is much appreciated.

  • 1
    $\begingroup$ Intuitively, the player with the mixed strategy knows what the other player is going to play so they should pick the best response to it. Either that's one particular move, or if there are several best responses which yield the same results. The only reason their best response could be mixed is if there are several, in which case they can choose them with any probabilities, so it is not unique $\endgroup$ – spaceisdarkgreen Feb 2 '18 at 1:07
  • $\begingroup$ @spaceisdarkgreen I see your point, but it is possible that for the player that is playing the pure strategy, this pure strategy is a best response only if the other player uses a particular mixed strategy. For example, consider the game: $$\begin{array}{cc}(4,0)&(0,1)\\(0,1)&(4,0)\\(2,1)&(2,1)\end{array}$$ In this case, the row player’s bottom action is a best reply only if the column player mixes with probabilities $1/2-1/2$. (Yet, there are other Nash equilibria in this game, so uniqueness breaks down after all, but for a different reason.) $\endgroup$ – triple_sec Feb 2 '18 at 1:14
  • $\begingroup$ Yes you are right, it is not that simple. $\endgroup$ – spaceisdarkgreen Feb 2 '18 at 1:42
  • $\begingroup$ It is not possible, so you should focus on proving that. Sorry don't have time to write a full proof. $\endgroup$ – Rahul Savani Feb 2 '18 at 8:21
  • 1
    $\begingroup$ @triple_sec: Sorry, I was too quick to comment. I actually tried to work through my proof idea, but it ends with needing a complicated case analysis. It's certainly not easy to construct an example, and I would still conjecture it is probably not possible (though in general a unique equilibrium with unequal size supports may be, but it seems that one being pure is a big constraint). Nice question - upvoted! Will return if I learn more. BTW, you may find my online bimatrix solver useful for exploring this: cgi.csc.liv.ac.uk/~rahul/bimatrix_solver. $\endgroup$ – Rahul Savani Feb 4 '18 at 12:57

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