0
$\begingroup$

The game looks like this:

         a       b
   A [(-12, 1) (8, 8)]
   B [(15, 1), (8,-1)]

(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.

However, can mixed strategies include weakly dominated strategies?

$\endgroup$
2
  • $\begingroup$ In the usual formulation of the definition, yes. There are other equilibrium concepts in which they cannot, then $(B,b)$ is the unique equilibrium. $\endgroup$
    – Will Sawin
    Feb 24, 2012 at 2:21
  • 1
    $\begingroup$ (B, b) can never be an equilibrium... $\endgroup$
    – user21641
    Feb 25, 2012 at 22:27

2 Answers 2

1
$\begingroup$

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)

$\endgroup$
0
$\begingroup$

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 $q=0$.

So, yes, mixed strategy can include weakly dominated strategy.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.