Take the 2-minute tour ×
Computer Science Stack Exchange is a question and answer site for students, researchers and practitioners of computer science. It's 100% free, no registration required.

Is the following statement always true:

if there is a mixed-strategy Nash equilibria then it is unique.

I know that there can be several pure strategy Nash equilibrias.

share|improve this question

migrated from cstheory.stackexchange.com Jun 12 '12 at 18:05

This question came from our site for theoretical computer scientists and researchers in related fields.

1 Answer 1

Shameless Advertisement: Game Theory proposal on Area 51 that you should totally follow!

No. The easiest way to prove this is by counterexample. Expanding matching pennies, we can create a game with two mixed strategies (players either mix between A and B or between C and D):

Dual Matching Pennies

What might be true about that statement is for a mix of particular pure strategies for two players, there is a unique mixed strategy equilibrium. This presumes that there is a threshold after which you switch from using strategy $0$ to strategy $1$. At this threshold, your opponent is ambivalent between her strategies. As depicted below, the intersection (and hence mix) will be unique.

Reaction Correspondence Matching Pennies

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.