In a finite game, suppose player 1 has strategies $\{\alpha_1,\alpha_2\}$ and player 2 $\{\beta_1,\beta_2\}$ with payoffs as below.

\begin{array}{c|c|c} &\beta_1&\beta_2\\ \hline\\ \alpha_1&1,1&0,0\\ \hline \alpha_2&0,0&0,0 \end{array}

A friend of mine told me that the only Nash equilibria are $(\alpha_1,\beta_1)$ and $(\alpha_2,\beta_2)$. But I doubt that.

Considering mixed strategies, if $\sigma_1=(x,1-x),~x\in[0,1],$ is the mixed strategy of player 1 and $\sigma_2=(y,1-y),~y\in[0,1],$ is the mixed strategy of player 2, then player 1's payoff function is $U_1=xy$ and player 2's is $U_2=xy$. Therefore one of Nash's equilibria is $(\sigma_1,\sigma_2)=(1,0,1,0)$, corresponding to pure strategy equilibrium $(\alpha_1,\beta_1)$. But I found that other equilibria are $(\sigma_1,\sigma_2)=(x,1-x,0,1),~x\in[0,1]$.

Then there are equilibria other than $(\alpha_1,\beta_1),~(\alpha_2,\beta_2)$. Is there something wrong with my thought?


Your friend is correct. Consider this intuitively.

If the first player has any chance of playing $\alpha_1$, then the second player's dominant strategy is to play $\beta_1$ all of the time, since he gains a positive expected utility compared to zero utility from $\beta_2$. Similarly, if the second player plays $\beta_1$ with positive probability, the first player will always want to play $\alpha_1$. Only when the other plays his second strategy with $100\%$ probability does a player not gain incentive from deviating to his first strategy.

For the actual calculation: Let $p$ be player 1's probability of choosing $\alpha_1$, and let $q$ be player 2's probability of choosing $\beta_1$. For player 1 to mix, he must be indifferent between the two options. So $$\begin{cases} \pi_1(\alpha_1, q) = (1)(q) + (0)(1 - q) = q \\ \pi_1(\alpha_2, q) = (0)(q) + (0)(1 - q) = 0. \end{cases}$$ Setting $\pi_1(\alpha_1, q) = \pi_1(\alpha_2, q)$ gives us $q = 0$. A similar derivation gets us $p = 0$. So $(0, 1, 0, 1)$ is the unique MNE. Basically, in this case a player will not be willing to mix unless he has no chance of getting positive utility.

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  • $\begingroup$ Tony, I understand what you said. But there's a gap between intuition and calculation. With payoff functions given above, when y=0, both of players have payoffs 0 no matter how player 1 chooses his mixed strategy. Allow me to put it this way, how do I compute all Nash equilibria merely using payoff functions and the definition of Nash equilibrium? $\endgroup$ – Rita Tseng Mar 18 '14 at 16:23
  • $\begingroup$ Edited to add calculation. $\endgroup$ – Tony Mar 18 '14 at 17:54
  • $\begingroup$ Tony, I think I understand what you mean. Thanks for your detailed explanation. But (0,0,0,0) is ? $\endgroup$ – Rita Tseng Mar 19 '14 at 5:37
  • $\begingroup$ Oops. Slip-up on my part. $\endgroup$ – Tony Mar 19 '14 at 5:50
  • $\begingroup$ Thanks again for patiently answering my question. :) $\endgroup$ – Rita Tseng Mar 19 '14 at 6:51

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