Nash Equilibrium for the prisoners dilemma when using mixed strategies Consider the following game matrix
$$
 \begin{array}{l|c|c}
    & \textbf{S}       & \textbf{G} \\ \hline
    \textbf{S} & (-2,-2) & (-6, -1) \\ \hline
    \textbf{G} & (-1,-6) & (-4, -4)
 \end{array}
$$
which corresponds to the well-known prisonder's dilemma. Now a Nash Equilibrium by using pure strategies would be (G,G) cause by choosing them neither can improve his outcome by unilaterally changing his strategy.
Now I wanted to calculate a Nash Equilibrium for mixed strategies using this payoff-matrix.
(I am using an algorithm proposed at the German wikipedia Article on Nash-Equilibrium). So, I am looking for a mixed strategy for player II which makes player I indifferent regarding his strategy choices, and vice versa.
Let $q$ be the probability that player II chooses $S$ and accordingly $(1-q)$ that he chooses $G$, then the expected values for player I are
\begin{align*}
  EV(I | S) & = (-2)q + (-6)(1-q) \qquad \textrm{if he chooses $S$} \\
  EV(I | G) & = (-1)q + (-4)(1-q) \qquad \textrm{if he chooses $G$}
\end{align*}
(I use the notation $EV(I | S)$ to mean, expected value of player $I$ when he chooses $S$ and so on).
Equating them, to find the probablities for player II which make player I indifferent, I have to solve
$$
 -2q - 6(1-q) = -q - 4(1-q)
$$
which has the solution $q = 2$, which is impossible since q should be a probablity, so with the additional restriction $0 \le q \le 1$ it has no solution? But I heard that Nash-Equilibrium for mixed strategies always exists in finite games, so what did I wrong?

What you've established is that, in this game, no player will ever be indifferent, no matter which (mixed or pure) strategy her opponent chooses.
The NE you already found is the only NE in this game. It's a pure-strategy equilibrium, which is a degenerate case of the mixed-strategy equilibria. The algorithm that you mention only works for non-degenerate mixed-strategy equilibria.
When it is stated, like above, that a mixed-strategy equilibrium always exists, invariably one implies that this might be a degenerate case.
