# Nash Equilibrium for n-shops Location Game

So if two ice cream shops were to be placed in the location $$[0,1]$$, inorder to maximize their own pay offs, they both would finally come to the location $$[\frac{1}{2}, \frac{1}{2}]$$. This is also the Nash Equilibrium of the problem and does not require a lot of mathematics to understand.

Now my problem was when we are playing this game for 3 different shops, like there is no intuitive answer for the problem. Every possible answer I thought of ($$[\frac{1}{2}, \frac{1}{2}, \frac{1}{2}]$$, $$[\frac{1}{4}, \frac{1}{2}, \frac{3}{4}]$$ etc.) is not a Nash Equilibrium.

So like is there a mathematical way that I can use to find the Nash equilibrium location for this game, or prove that it doesn't exist.

Also can this be generalized to $$4,5 ...n$$ games?

The following two claims hold in the general $$n$$-shop case.

Claim 1. In equilibrium a shop closest to an edge (0 or 1) cannot be alone.

Proof. Such a shop could gain customers by moving slightly inward.

Claim 2. In equilibrium at most two shops can be in any location.

Proof. Assume there is an equilibrium where there are three or more shops in a location. Denote the number of customers coming to this location from the left and right by $$c_l$$ and $$c_r$$ respectively. (In the 3-shop case $$c_l + c_r = 1$$.) All shops in the location have a payoff of $$(c_l + c_r)/n$$. By moving slightly to the left, a shop can attain a payoff that is arbitrarily close to $$c_l$$, and by moving slightly to the right a shop can attain a payoff that is arbitrarily close to $$c_r$$, therefore it can attain (almost) $$\max(c_l;c_r)$$. It is easy to show that if $$n >2$$ then $$\frac{c_l + c_r}{n} < \max(c_l;c_r).$$

Using these two properties it is very easy to show that in the 3-shop case there is no pure Nash-equilibrium, so this is left as an exercise.

I happen to have a Desmos for this, if you'd like you can use it to validate the claims and reason about the proof.

Note: a mixed equilibrium does exist, see Shaked, A. (1982): Existence and Computation of Mixed Strategy Nash Equilibrium for 3-Firms Location Problem.

Interestingly, $$n = 3$$ is the only case without a pure equilibrium, $$n \in \left\{2,4,5\right\}$$ all have one, and for other $$n$$ values there are infinitely many equilibria. For a formal characterization one needs some additional properties, but no advanced mathematics. For a detailed discussion see Eaton, B.C., and R.G. Lipsey (1975): The Principle of Minimum Differentiation Reconsidered: Some New Developments in the Theory of Spatial Competition.