Guessing a number among K

Question

Consider two players $a$ and $b$. Player $a$ moves first and picks a number $n\in\{0,1,2,...,K\}$. Then moves player $b$ who guesses at the number picked by $a$. If the guess is correct, $b$ wins a prize equal to $\pi=K−n$ and $a$ earns nothing. Otherwise player $a$ earns $\pi=K−n$ and $b$ earns nothing. What number should $a$ optimally pick to maximize her expected reward? what should do $b$?

Answer 1

Suppose $K=1$. Then $\sigma_a=0$ and $\sigma_b=0$ is a Nash equilibrium ($\sigma_i$ denotes the strategy of player $i$). In this equilibrium, $b$ earns 1 while $a$ doesn't.

For $K=2$, this strategy profile is not an equilibrium any more: If $a$ thinks $b$ thinks $a$ is going to play $\sigma_a=0$, and will best respond by $\sigma_b=0$, then $a$ is better off by playing $\sigma_a=1$, giving a payoff of 1.

Indeed, for $K>1$, there is no pure strategy equilibrium. That is because for any $\sigma_b\in\{0,\ldots ,K\}$, there is a $\sigma_b\in\{0,\ldots, K-1\}$ which gives a positive payoff to $a$.

Consequently, for $K>1$ there is no number which $a$ should choose with probability 1, because then he would be predictable. Only mixed strategy equilibria exist, so there is no single optimal number to pick. For $K=2$, for example, the only mixed strategy equilibrium is $$Pr(\sigma_a=0)=1/3, Pr(\sigma_a=1)=2/3, Pr(\sigma_b=0)=2/3, Pr(\sigma_b=1)=1/3.$$ Interestingly, $a$ does not choose 0 more often than 1, even though it would give a higher payoff if $b$ guessed wrong. That's because $a$ anticipates that $b$ sees 0 would be more profitable, and more often guesses 0 than 1.