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We have a game with two participants, $A$ and $B$. There are two options in the game: $\$60$ or $\$40$.

  • If both choose $\$60$, then each of them gets $\$30$.
  • If both choose $\$40$, then each of them gets $\$20$.
  • If one chooses $\$40$, another chooses $\$60$, then each of them gets $\$50$.

What is the optimal strategy for $A$? Any thoughts? Thanks!

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    Hint: $B$ will, or ought to, follow the exact same strategy. What's the best deterministic strategy? Can they do better if they follow a probabilistic one? – 2017-01-09

2 Answers 2

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the best option for A would be to always choose $60.

to make this a lot harder, you would have to make that if both of them chooses 60, they would both lose 30.

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Let

  • $A$ play $\$40$ with probability $p \in [0,1]$ and $\$60$ with probability $1-p$.
  • $B$ play $\$40$ with probability $q \in [0,1]$ and $\$60$ with probability $1-q$.

The expected payoff for $A$ (and also for $B$) is

$$20 pq + 50 p (1-q) + 50 (1-p) q + 30 (1-p) (1-q) = \color{blue}{-50 pq + 20 p + 20 q + 30}$$

Taking the partial derivatives and finding where they vanish, we obtain the critical probabilities

$$p^*, q^* := \frac{2}{5}$$

We have a candidate mixed Nash equilibrium with expected payoff

$$-50 \left(\frac{2}{5}\right)^2 + 20 \left(\frac{2}{5}\right) + 20 \left(\frac{2}{5}\right) + 30 = - 8 + 8 + 8 + 30 = \color{blue}{\$38}$$

Let $q^* = \frac 25$ be fixed and $p$ be free. The expected payoff for $A$ is

$$-50 \left(\frac{2}{5}\right) p + 20 p + 20 \left(\frac{2}{5}\right) + 30 = - 20 p + 20 p + 8 + 30 = \color{blue}{\$38}$$

We conclude that $A$ has no incentive to unilaterally deviate from $p^* = \frac 25$ when $B$ chooses $q^* = \frac 25$, as his expected payoff remains $\$38$. Likewise, $B$ has no incentive to unilaterally deviate from $q^* = \frac 25$ when $A$ chooses $p^* = \frac 25$. Hence, $(p^*, q^*)$ is indeed a mixed Nash equilibrium.

There are also two pure Nash equilibria, each with expected payoff $\color{blue}{\$50}$.