Nash equilibria problem

Question

I have a problem to solve the next exercise.

Two players call a number from 1 to 100. The winnings are distributed as follows: players always receive no more than 100 rubles in total, the most greedy gets what he asks for. More formally: if the players have chosen numbers a and b, then the payoff will be $(a,b)$ if $a+b≤100$. If $a+b>100$, then the payoffs are $(a,100-a)$, if $b>a$,the payoffs $(100-b,b)$ , if $a>b$ and (50,50) if $a=b$.

Find all Nash equilibria in this game.

Answer 1

Let $(a, b)$ denote the selections of player $A$ and $B$.

I found the pure (deterministic) strategy combinations $(50, 50)$, $(51, 50)$, $(50, 51)$, $(51, 51)$ and all mixed strategy combinations where both players pick either $50$ or $51$ with some probability given to each to be the only Nash equilibria. All of these strategies will reward both players with $50$ payoff.

Any strategy with $a < 50$ or $b < 50$ is not a part of a Nash equilibrium due to the fact that the player selecting $< 50$ can always select $50$ or $51$ to get higher payoff. If $a = 50$, $A$ will always get $50$ (and the same obviously holds for $B$). Either due to $a + b \leq 100$ or due to the fact that if $a + b > 100$, then $50$ will be the lowest selection.

Any strategy with $a > 51$ or $b > 51$ is not a part of a Nash equilibrium due to the fact that the one player can improve by selecting one less than the other. For example, if $b \geq 52$, $A$ can improve by selecting $a = b - 1$.

If any player selects $50$ or $51$, the other can't get more than $50$ payoff. If $a = 50$, player $B$ can select any number $\geq 50$ to get $50$.

Thus, the only pure strategy combinations that are Nash equilibria are $(50, 50)$, $(51, 50)$, $(50, 51)$, and $(51, 51)$.

This might be made more clear by showing some cases:

• $(a, b) \rightarrow$ (Payoff for $A$, Payoff for $B$)
• $(50, 50) \rightarrow (50, 50)$ No player can improve.
• $(51, 50) \rightarrow (50, 50)$ No player can improve.
• $(50, 51) \rightarrow (50, 50)$ No player can improve.
• $(51, 51) \rightarrow (50, 50)$ No player can improve.
• $(49, 50) \rightarrow (49, 50)$ $A$ can improve by selecting $a = 50$ or $a = 51$. $B$ can improve by selecting $b \geq 51$.
• $(52, 50) \rightarrow (50, 50)$ $B$ can improve by selecting $b = 51$.
• $(52, 51) \rightarrow (49, 50)$ $A$ can improve by selecting $a = 50$ or $a = 51$.
• $(52, 52) \rightarrow (50, 50)$ Both players can improve by selecting $51$.