Game theory problem, 3x3 matrix: pure and mixed strategies

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From the following matrix: $$ \left( \begin{array}{c|ccc} & L & C & R\\ \hline T &3,0& 1,1 &4,2\\ M &3,4& 1,2 &2,3\\ B &1,3& 0,2 &3,0 \end{array} \right) $$

I have derived the following Nash Equilibrium: ($M$,$L$) and ($T$,$R$). But I cannot, it seems, solve it w.r.t. mixed strategies. I have used an embarrassing amount of time trying to do so. Could somebody please help me solve the matrix for mixed strategies?

Thanks in advance.

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Here's one sensible sequence of steps:

Step 1: Notice that T strictly dominates $B$, since $(3,1,4)$ is componentwise strictly greater than $(1,0,3)$. Remove $B$ and we are left with a $2 \times 3$ game.

Step 2: In this new game, with $B$ removed, $R$ dominates $C$, since $(2,3)$ is componentwise strictly greater than $(1,2)$. After removing $C$ we are left with a $2 \times 2$ game:

$$ \left( \begin{array}{c|cc} & L & R\\ \hline T &3,0& 4,2\\ M &3,4& 2,3\\ \end{array} \right) $$

Step 3: Having found two pure equilibria already, look for non-pure equilibria. Player 2 can be made indifferent between $L$ and $R$ as we see below. But, player 1 cannot be made indifferent between $T$ and $M$ because $T$ weakly dominates $M$: as soon as there is any positive probability on $R$, player 1 strictly prefers $T$. Thus player 2 cannot mix in equilibrium, and actually the pure equilibrium $(M, L)$ is actually only the endpoint of a range of equilibria:

$$ ((1-p, p), L)\ \text{where } p \in [2/3, 1] $$

The threshold of $p=2/3$ is the point at which player II is indifferent between $L$ and $R$ against $(1-p,p)$. When $p=2/3$ both L and R give expected payoff $1/3 \cdot 0 + 2/3 \cdot 4 = 1/3 \cdot 2 + 2/3 \cdot 3 = 8/3$.

A range of equilibria like this is only possible is a degenerate game. A $2 \times 2$ game is necessarily degenerate. More generally, a game is degenerate if there exists a mixed strategy $x$ with support size $k$ (i.e. , $|\{x_i \ | \ x_i >0\}| = k$) and more than $k$ best responses against $x$. In this example, the strategy $x$ is the pure strategy $L$, which has support size 1 but two best responses against it, $TM$ and $M$.

You can check the equilibria at:

The output from the former for this game is:

2 x 2 Payoff matrix A:

  3  4
  3  2



2 x 2 Payoff matrix B:

  0  2
  4  3

EE = Extreme Equilibrium, EP = Expected Payoff

Decimal Output

  EE  1  P1:  (1)  0.333333333333  0.666666666667  EP=  3.0  P2:  (1)  1.0  0.0  EP=  2.66666666667
  EE  2  P1:  (2)             1.0             0.0  EP=  4.0  P2:  (2)  0.0  1.0  EP=            2.0
  EE  3  P1:  (3)             0.0             1.0  EP=  3.0  P2:  (1)  1.0  0.0  EP=            4.0

Rational Output

  EE  1  P1:  (1)  1/3  2/3  EP=  3  P2:  (1)  1  0  EP=  8/3
  EE  2  P1:  (2)    1    0  EP=  4  P2:  (2)  0  1  EP=    2
  EE  3  P1:  (3)    0    1  EP=  3  P2:  (1)  1  0  EP=    4

Connected component 1:
{1, 3}  x  {1}

Connected component 2:
{2}  x  {2}

Here one of the two equilibrium components (the range of equilibria described above in terms of $p$) is denoted by

Connected component 1:
{1, 3}  x  {1}

The other equilibrium component is the pure strategy equilibrium $(T,R)$.