Suppose the game consists of only $2$ players, player $1$ and player $2$, and each of them has only $2$ strategies to choose between. This gives a $2$ by $2$ payoff matrix. Player $2$ has no preference when choosing one of his strategies, while player $1$ chooses the strictly dominant strategy for himself.

Questions are:

  1. In this case, is player $2$ going to mix her strategies? Given that whatever proportion player $2$ mixes her strategies, player $1$ would definitely gain positive earning.

  2. In this case, is there any mixed Nash equilibrium?

Thanks for help!


1 Answer 1


Let's call player's $1$ strategies A and B, and player's $2$ strategies C and D. Let C be the dominant one. So, the payoff matrix is: $$\small \begin{matrix} & & \textrm{player } 1&\\ & & \textrm{A} & \textrm{B} \\ \textrm{player } 2 &\textrm{C} & x_1, x_2 & y_1, y_2\\ & \textrm{D} & u_1, u_2 & z_1, z_2 \\ \end{matrix}$$

Since A is dominant, $x_2>y_2$ and $u_2 > z_2$. Since player $1$ always plays A we can ommit strategy B and the game's matrix actually looks like this:

$$\small \begin{matrix} & & \textrm{player } 1&\\ & & \textrm{A} \\ \textrm{player } 2 &\textrm{C} & x_1, x_2\\ & \textrm{D} & u_1, u_2 \\ \end{matrix}$$

Assuming that player $2$ goes only for maximising her own profit (i.e. she doesn't get any happier when player $1$ gets less), player $2$ would play the strategy that gives her most profit, that is: if $x_1>u_1$ player $2$ plays A, if $x_1<u_1$ player $2$ plays B.

If $x_1 = u_1$ then there are continuum different mixed stategy NEs since any pair $\{(1,0), (t, (1-t))\}$for $t\in[0,1]$ gives a NE.

  • $\begingroup$ thanks golob! say p and 1-p are the probability distribution for C and D respectively, there are infinite mixed stategy for p belongs to 0,1, since no matter how player 2 mix her strategy, player one would have positive earnings and player 2 would be indifferent between choosing which of these 2 strategies? $\endgroup$
    – Steve
    Commented Nov 16, 2012 at 10:55
  • $\begingroup$ something i am confusing about mixed strategy nash equlibrium. in this case, player 1 will definitely choose A because it is a strictly dominant strategy. Given player 1 will play A, player 2 would be indifferent in choosing C and D, so thats why she mix? what actually means by mixed nash equilibrium in this case? $\endgroup$
    – Steve
    Commented Nov 16, 2012 at 10:59
  • $\begingroup$ and sorry that i don't really understand this math- {(1,0),(t,(1−t))} for t∈[0,1] $\endgroup$
    – Steve
    Commented Nov 16, 2012 at 11:02
  • $\begingroup$ player 1 will definitely choose A NOT because it is a strictly dominant strategy, but becasue it is assumed so (player 1 chooses the strictly dominant strategy for himself). Mixed means that some players use a mixed strategy, that is they choose at random (with some given distribution) on of their strategies. $\{(1,0),(t,(1−t))\} for t\in[0,1]$ is a notation for a global strategy. First parenthesis means player 1 plays A with prob. 1 and B with prob. 0, while second parenthesis means player 2 plays C with prob. $t$ and B with prob. $1-t$. So for strategy to be mixed, must $t \in (0,1)$. $\endgroup$
    – Golob
    Commented Nov 16, 2012 at 11:21

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