In game theory, are there easy ways to rule out the existence of either (i) all pure Nash, (ii) all completely mixed Nash?

Question

For example,

1. in a rock paper scissors game, there are no pure Nash equilibrium and only a completely mixed Nash equilibrium.

2. in prisoner's dilemma, there is no mixed Nash equilibriumand only a pure Nash equilibrium.

Is it established in game theory how you can know if a game only has completely mixed Nash equilibrium or only pure Nash equilibrium?

For example, games with only mixed nash have a cyclical payoff structure: there are no fixed points in the players pure best responses.

Whereas for a game to have no mixed strategy, the pure strategy must have a strict dominance property: if it is obviously the best strategy that the players can use, then there is no need for the players to mix. But I am not certain if any game with a strict dominating pure strategy will necessarily not have a mixed strategy.

Has someone thought about these before?

Answer 1

By definition, if there is a strict dominant strategy then it will be strictly preferred to any mixed strategy. A necessary condition for existence of a NE in mixed strategis is that there is no strictly dominant strategy. Otherwise, you will want to play that strategy vs any other one.