There are three players and each stakes 10 dollar. The players simultaneously choose a number from the set {1, 2, 3, 4}. If three numbers are different, the player choosing the middle number wins the pot and the payoff is 20 dollar; while the players with the largest and smallest numbers each have a payoff of -10 dollars. If the numbers are not all different the house keeps all stakes, so all players get a payoff of -10 dollars.
I would like to find the pure strategy Nash equilibrium.
Here, "1" and "4" are strictly dominated by "2"and "3". I see this. However, I could not find the pure strategy NE. How can I think to solve this?
There is no single pure strategy equilibria, there are many.
For example, each strategy profile where they bid $1-2-3$ or $2-3-4$ should be a Nash Equilibria. The ones that bid the extremes have no profitable deviation (they get $-10$ no matter what) and the one that bids the central has no profitable deviation (he gets $20$ now, and $-10$ if he deviates).
(Bids like $1-2-4$ and $1-3-4$ are not equilibria, because there is "room" for an extreme bidder to become the central and win).
In addition, there are many other nash equilibria where they all lose. For example, when they all choose the same number or when two chooses one number and the other one choose the next or the previous (2-2-3 for example).