Two players bargain over how to split \$10. Each player $i \in \{1, 2\}$ chooses a number $s_i \in [0, 10]$ (which does not need to be an integer). Each player's payoff is the money he receives. We consider two allocation rules. In each case, if $s_1 + s_2 \leq 10$, each player gets his chosen amount s; and the rest is destroyed.
- In the first case, if $s_1 + s_2 > 10$, both players get zero. What are the (pure strategy) Nash equilibria? Answer: All couples such that $s_1 + s_2 = 10$ and $(s_1= 10, s_2= 10)$. My question; why is $(10,10)$ Nash equilibrium? Neither one will get money in that case?
Reminder: A Nash Equilibrium is a strategy profile such that no player has a profitable deviation. It has nothing to do with global optimality or global maximization of payoff. Instead, for each player, it is the strategy that maximizes his payoff given the rest are following the equilibrium strategies.
In our case: Suppose we both say $s_i=10$. Then as you mentioned, we get $0$.
Do any of us have a profitable deviation?
No. Regardless of what you say, you will still get $0$, so you don't have a profitable deviation, and $(10,10)$ is an equilibrium.
To clarify, $(10,5)$ is not an equilibrium, since in this case, I get $0$, but if I offer $5$ instead of $10$ I'll get more strictly, so it is a profitable deviation. This is why only $(10,10)$ is an equilibrium because no one can unilaterally reduce the sum below $10$ and actually get a nonzero payoff.