Determine a finite game game with only one pure Nash Equilibrium and Unaccountably many mixed strategy Nash Equilibria

Question

I have this problem in my Game Theory problem set:

My intuition tells me that there must be some sort of indifference in order for the game to support uncountably many mixes strategy NE. So one could support the equilibria with different probabilities and so on.

My approach to that has been trying to draw best-responses on 1-by-1 square on $\mathbb{R}^2$ and make them intersect in uncountably many points, including a point such as $(p,q) = (1,1)$. I have been trying a lot to draw and come up with a corresponding game but I'm lost. I'm starting to think that it's not possible for a 2x2 game (these are the ones I've been trying with).

Any ideas? Thanks in advance!

Answer 1

It is possible for a 2 by 2 game. You are right that you need degeneracy, which in a 2 by 2 games means weak dominance. Here's an example, with solution, as given by http://banach.lse.ac.uk:

2 x 2 Payoff matrix A:

  1  0
  0  1



2 x 2 Payoff matrix B:

  1  1
  1  0

EE = Extreme Equilibrium, EP = Expected Payoff

Decimal Output

  EE  1  P1:  (1)  1.0  0.0  EP=  1.0  P2:  (1)  1.0  0.0  EP=  1.0
  EE  2  P1:  (1)  1.0  0.0  EP=  0.5  P2:  (2)  0.5  0.5  EP=  1.0

Rational Output

  EE  1  P1:  (1)  1  0  EP=    1  P2:  (1)    1    0  EP=  1
  EE  2  P1:  (1)  1  0  EP=  1/2  P2:  (2)  1/2  1/2  EP=  1

Connected component 1:
{1}  x  {1, 2}

We have uncountably many equilibria, namely all convex combinations of the 2 extreme equilibria. This is indicated by the part of the following part of the output at the end, which shows that player 1 can play his strategy $(1,0)$ (that is, row 1), against any convex combination of player 2's mixed strategies $(1,0)$ and $(1/2, 1/2)$:

Connected component 1:
{1}  x  {1, 2}

In other words, we can describe the full set of equilibria as:

• Player 1 plays row 1, and
• Player 2 players columns 1 and 2 with probabilities $(1-q)$ and $q$ respectively for any $q \in [0, 1/2]$.

Intuitively, as long as player 2 places no more that half her probability on column 2, it is a best response for player 1 to play row 1. And, when player 1 plays row 1, player 2 is indifferent and can herself mix. There are no other equilibria since, as soon as there is any probability on row 2, the first column is a unique best response, but, against the first column, the first row is the unique best response.