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Consider the following 2x2 game:

\begin{array}{|c|c|c|} \hline &C&R\\\hline M&0,0&3,5\\\hline D&4,4&0,3\\\hline \end{array}

This game has two distinct pure-strategy Nash equilibria.

I recall that in cases like this there is almost always a third mixed NE, and this had to do with the index theorem and having an odd number of equilibria.

Can someone provide the idea behind the index theorem and why having two distinct pure strategy NE in 2x2, 2 person game, static, complete information results into a third mixed NE with probability 1?

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What about \begin{array}{|c|c|c|} \hline &C&R\\\hline M&0,0&0,0\\\hline D&0,0&1,1\\\hline \end{array} ?

This game has two pure equilibria, $(M,C)$ and $(D,R)$ but no mixed equilibra.

You are probably referring to Wilson's Oddness Theorem, which guarantess your claim for generic games. See Oddness of equilibrium points.

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  • $\begingroup$ I just got what you meant by probability 1. I think the word order of that sentence is a little off. $\endgroup$
    – Giskard
    Nov 9 '17 at 13:38

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