# Nash equilibrium in strictly mixed strategies

I have the following statement which I have been said it is false, but I don't understand why:

"All finite games have at least one Nash equilibrium in strictly mixed strategies, as long as there is no player with a single pure strategy".

It confuses me because the Nash theorem (1950) says that every finite game has a mixed strategy equilibrium. But if I make a simple game 2x2 as in the image, I don't find any mix strategy. Or does playing "a" with p=1 is the mix strategy?

P.S. A strictly mixed strategy is a mixed strategy that assigns strictly positive probability to at least 2 pure strategies.

• Do you know how to solve a game? Try making (or look up) some simple 2x2 examples and see if the above statement is true or not. Feb 23 at 0:55