# if $G$ has no Nash equilibrium in pure strategies then $G$ has single Nash equilibrium in mixed strategies.

Let $G=(S,T,\pi _1 ,\pi_2)$ be a 2 player game with strategies $T$ for player 1 and $S$ for player 2 such that $|T|=|S|=2$, and payoff functions $\pi _1 ,\pi_2$.
prove that if $G$ has no Nash equilibrium in pure strategies then $G$ has single Nash equilibrium in mixed strategies.

I noticed that in this case the payoff for the Nash equilibrium for each player is always between the second worst case and the second best case for them, yet I don't know how continue.

Let $T=\{t_1,t_2\}$ and $S=\{s_1,s_2\}$. Let payoffs be as follows:
$$\begin{array} \hline & s_1 & s_2 \\ t_1 & a,w & b,y \\ t_2 & c,x & d,z \\ \end{array}$$
We know that a Nash Equilibrium must exist (Nash proved this!), but that it does not exist in pure strategies (given by the problem). For there to exist a NE in strictly mixed actions, any player mixing must be indifferent between his two actions. If we denote $p_{t_1}$ as the probability weight P1 puts on $t_1$, and similarly for other actions, it must be that: $$p_{s_1}a + (1-p_{s_1})b = p_{s_1}c + (1-p_{s_1})d$$ $$p_{t_1}w + (1-p_{t_1})x = p_{t_1}y + (1-p_{t_1})z$$ where the first equation (on P2's mix) guarantees that P1 is indifferent between his two actions (and thus willing to mix), and where the second equation (on P1's mix) guarantees that P2 is indifferent between his two actions (and thus willing to mix).
Now the argument to make is that each of the above equations above solves for a unique $p_{s_1}$ and $p_{t_1}$ respectively, as this would then imply a unique equilibrium. Well, solve for$p_{t_1}$ and $p_{s_1}$, and : $$p_{s_1} = \frac{b-d}{b+c-a-d}$$ $$p_{t_1} = \frac{x-z}{x+y-w-z}$$
A detail needs to be added to the previous answer. We need to explain why the system of linear equations that one obtains when writing down the indifference conditions does not have more than one solution. Without loss of generality consider the first equation. It will have more than one solution if $b+c-a-d=0$ and $b-d=0$. This is equivalent to: $a-c=0$ and $b-d=0$. This means that player 1 is indifferent between both strategies, regardless of what player 2 does. But if this were the case, then the game would have pure strategy Nash equilibria. For example, $(t_1,s_1)$ (if $w\geq y$) or $(t_1,s_2)$ (if $w\leq y$) would be a pure strategy Nash equilibrium. Because by assumption the game has no pure strategy Nash equilibrium, we have obtained a contradiction.