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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?

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  • $\begingroup$ 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 '15 at 10:18
  • $\begingroup$ But $(0,1)$ is not compact as far as I know... $\endgroup$ – luka5z Jul 20 '15 at 10:19
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    $\begingroup$ 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 '15 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 '15 at 10:25
  • $\begingroup$ You just wrote the example. You did not prove that it has no mixed strategy Nash equilibrium. $\endgroup$ – Sergio Parreiras Jul 23 '15 at 20:55
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There is none, unless the problem is wrongly planned. By definition, every game has at least one equilibrium in mixed strategies.

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  • $\begingroup$ Please post this as a comment. $\endgroup$ – Max0815 Apr 29 at 22:15

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