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