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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?

enter image description here

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

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  • $\begingroup$ 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. $\endgroup$
    – Walrasian Auctioneer
    Feb 23, 2022 at 0:55

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Nash's theorem says that every finite game has a NE in mixed strategies, but here mixed, coming without the qualifier strictly, implies the weak version that includes pure strategies. So your statement is indeed false and your game is a counterexample.

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Your very example shows the statement is false.

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  • $\begingroup$ Could you explain it, please? Is p=1 and (1-p)=0 the strictly mix strategy? As it says the probability must be strictly positive, I suppose (1-p) can't be 0, so p can't be 1. I'm lost. $\endgroup$
    – Constanza Zepeda
    Feb 23, 2022 at 2:16
  • $\begingroup$ "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" is false, for the very reason you have stated. $\endgroup$
    – Walrasian Auctioneer
    Feb 23, 2022 at 3:20

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