Game Theory: Is there a Mixed Strategy Nash Equilibrium? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T09:50:14Z http://mathoverflow.net/feeds/question/89346 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89346/game-theory-is-there-a-mixed-strategy-nash-equilibrium Game Theory: Is there a Mixed Strategy Nash Equilibrium? unknown (google) 2012-02-24T00:13:02Z 2013-05-10T17:30:35Z <p>The game looks like this:</p> <pre><code> a b A [(-12, 1) (8, 8)] B [(15, 1), (8,-1)] </code></pre> <p>(15, 1) and (8,8) are Nash Equlibria. However, could you still mix between (8,8) and (15,1)? For example, for P2 (column player) to make P1 indifferent he could play b with a probability of 1. And player 1 could make player 2 indifferent with another probability mix. </p> <p>However, can mixed strategies include weakly dominated strategies? </p> http://mathoverflow.net/questions/89346/game-theory-is-there-a-mixed-strategy-nash-equilibrium/95548#95548 Answer by Nixon for Game Theory: Is there a Mixed Strategy Nash Equilibrium? Nixon 2012-04-30T04:27:55Z 2012-04-30T04:27:55Z <p>In the mixed strategy Nash equilibrium the column players will choose b with probability 1, thus there is never a mix that includes (15,1). However, the row player can mix with their weakly dominated strategy A. (probability A = 2/9, B = 7/9)</p> http://mathoverflow.net/questions/89346/game-theory-is-there-a-mixed-strategy-nash-equilibrium/98183#98183 Answer by Niki for Game Theory: Is there a Mixed Strategy Nash Equilibrium? Niki 2012-05-28T09:56:02Z 2012-05-28T09:56:02Z <p>The answer given by Nixon is correct. If you label p the probability that player 1 chooses A, and q the probability that player 2 chooses a, then you have: \$E(a)=E(b)\$, so \$1=8p-(1-p)\$, that is \$p=2/9\$. On the other hand, you can see that \$q=0\$.</p> <p>So, yes, mixed strategy can include weakly dominated strategy.</p>