# Reason for non-existence of pure Nash in Matching Pennies versus Prisoner's Dilemma

I am reading some stuff on normal form game.

It is claimed that Matching Pennies do not have pure nash strategy because the strategy space $S$ is "non-convex", whereas Prisoner's Dilemma has a pure nash strategy

(Advanced Workshop And Tutorials On Operations Research (AWTOR-2012))

The strategy set for each case is:

$S = \{H, T\}$ for MP, and $S = \{\text{defect}, \text{cooperate}\}$ in PD.

Let $H, \text{defect} = 1, T, \text{cooperate} = 0$, then $S = \{0,1\}$ in each case...both of these sets are non-convex and more over every matrix game would have strategy set that is nonconvex. What am I not understanding and where does the non-convexity come in?

• Your quote does not say that the Prisoner's Dilemma has a "convex" strategy space. (It does not.) – mlc Aug 18 '16 at 16:14

The choices $\{0,1\}$ are called pure strategies. The set of pure strategies is not convex.
This set can be convexified by introducing randomised strategies, viewed as probability distributions over the pure strategies. In your simple example, this is equivalent to assuming that $T$ is played with probability $p$ and $H$ is played with probability $1-p$. The strategy set for $p$ is the closed interval $[0,1]$, which is convex.