Let's say I want to calculate Nash equilibrium with mixed strategies for a two-players game, in which there is no Nash equilibrium with pure strategies (no dominant strategy for any of the two players), for example, take the Matching Pennies game with the following payoffs:
\begin{bmatrix} & H & T\\ H & 1,-1 & -1,1\\ T & -1,1 & 1,-1 \end{bmatrix}
Now to calculate the Nash equilibrium, let player 2 (columns player) play $H$ with probability $p$, and $T$ with probability $1-p$.
If player 1 (rows player) best responds with mixed strategy, then player 2 must make him indifferent between choosing $H$ or $T$.
This gives us the constraint:
$u_1(H)=u_1(T)$
$\Rightarrow 1\cdot p-1(1-p)=-1\cdot p+1(1-p)$
$\Rightarrow p=0.5$
No surprise here.
What interests me is when trying to apply the same logic in a game where we do have a Nash equilibrium (and dominant strategy to the players), specifically, the Prisoner Dillema game:
\begin{bmatrix} & C & D\\ C & -1,-1 & -4,0\\ D & 0,-4 & -3,-3 \end{bmatrix}
Since intuitively you can't make player 1 (rows player) indifferent between choosing $C$ or $D$ (he is always better-off choosing $D$), we should expect $p=1$ - meaning we should expect that player 2 need to play $D$ with probability $1$, and play $C$ with probability $0$, but that's not the case:
$u_1(C)=u_1(D)$
$\Rightarrow -1\cdot p-4(1-p)=0\cdot p-3(1-p)$
$\Rightarrow -3=-4$
No such $p$.
Why is that?