I have to find an example of a game that does not admit (mixed strategy) Nash equilibria.

Consider a game in normal form. Let $N=\{1,2\}$ be set of players and $S_i=\mathbb{R}$ a set of possible strategies for each player (uncountably many), $i=1,2$.

The rules of this game are as follows: each player chooses one real number (his strategy). Player who chose a bigger number than the other player wins $1$, player who chose smaller number gets $0$. If both players choose the same number, they both get $0$.

Formally, payoff functions of each player one looks like this: $$v_1(s_1,s_2)=\left\{\begin{matrix} 1 \ \mathrm{if}\ s_1>s_2 \\ 0 \ \mathrm{if}\ s_1\leq s_2 \end{matrix}\right. $$ $$v_2(s_1,s_2)=\left\{\begin{matrix} 1 \ \mathrm{if}\ s_2>s_1 \\ 0 \ \mathrm{if}\ s_2\leq s_1 \end{matrix}\right. $$

Do you think it's a good example?

  • $\begingroup$ It looks ok, but – just my two cents – I would choose $(0,1)$ instead of $\mathbb{R}$ to emphasize that the set of strategies of a player can indeed be uncountably infinite, but it has to be compact. $\endgroup$
    – Kolmin
    Jul 20, 2015 at 10:18
  • $\begingroup$ But $(0,1)$ is not compact as far as I know... $\endgroup$
    – luka5z
    Jul 20, 2015 at 10:19
  • 2
    $\begingroup$ Indeed, that's the point. If you choose $[0,1]$ (that is compact), you have uncountably many strategies for every player and a NE (think about Cournot's duopoly). $\endgroup$
    – Kolmin
    Jul 20, 2015 at 10:20
  • $\begingroup$ In my game also - if set of strategies was $[0,1]$, then $(1,1)$ would be pure strategy Nash equilibrium. Ok, thanks. $\endgroup$
    – luka5z
    Jul 20, 2015 at 10:25
  • $\begingroup$ You just wrote the example. You did not prove that it has no mixed strategy Nash equilibrium. $\endgroup$ Jul 23, 2015 at 20:55

1 Answer 1


There is none, unless the problem is wrongly planned. By definition, every game has at least one equilibrium in mixed strategies.

  • $\begingroup$ Please post this as a comment. $\endgroup$
    – Max0815
    Apr 29, 2019 at 22:15
  • $\begingroup$ This is absolutely WRONG !! By application of Kakutani's fixed point theorem, if the players have infinite strategies and this set is not compact, then there is no guarantee that there will be a NE in the game. $\endgroup$
    – Oniropolo
    Sep 22, 2020 at 0:12

You must log in to answer this question.

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