Ice cream vendor problem

Question

PROBLEM

This is a question considering game theory.

Assume there is a beach with $n$ ice cream vendors on it who position themselves along the beach. For an arbitrary $n$, find a candidate Nash Equilibrium and prove its validity. (hint: distinguish the case where $n$ is even and $n$ is uneven)

You may assume that the customers are uniformly distributed over the beach.

My feeling says that the ice cream vendors will eventually all end up at the middle of the beach, as this happens with the case with two ice cream vendors, but I'm not sure about this, can anybody help me?

Is the beach continuous with a uniform distribution of customers? If not, what is the situation? — universalset, Nov 27 '13 at 13:48
@universalset yes, I think you may assume that the customers are uniformly distributed over the beach. — Nedellyzer, Nov 27 '13 at 13:50
@SjoerdSmaal You should include further assumptions in your original post. — BIS HD, Nov 27 '13 at 13:51
There is a hint here http://www.sites.carloalberto.org/gerardi/GameTheory/Problem-Set-4.pdf — oks, Nov 27 '13 at 14:07
The hint also gives the result for n=3 (though it doesn't derive it) — oks, Nov 27 '13 at 14:17

universalset · Answer 1 · 2013-11-27T16:33:38.353

Assuming the beach is continuous with a continuous distribution of customers:

If there are $3$ vendors, there is no equilibrium. If all three are in the center, one of them can move slightly to the left and capture nearly $\frac{1}{2}$ of the customers, which is more than the $\frac{1}{3}$ he had before. Otherwise there is some vendor which is not in the same place as either of the other two and is the closest to one edge. He can move slightly toward the center, capturing more customers.

If there are $4$ vendors, then there is an equilibrium when $2$ of them are $\frac{1}{4}$ of the way from the left end of the beach and the other two are $\frac{1}{4}$ of the way from the right end. I'm pretty sure this generalizes to even $n$, where pairs are at the same spot and the beach is divided into segments in the ratio $1:2:2:\cdots:2:1$.

This leaves the case for odd $n>3$. I haven't worked out the details, but you should probably be able to show that intervals can be arranged with $\lfloor \frac{n}{2} \rfloor$ pairs of vendors and a single vendor in the middle somewhere (doesn't need to be the exact center, just in between two pairs of doubled-up vendors), with segments in the ratio $1:2:2:\cdots:2:1$.

Could you please show how to derive the expression for the case for odd $n > 3$ — Nedellyzer, Nov 27 '13 at 14:54
The ratio of segment lengths should follow the same pattern of $1:2:2:\cdots:2:1$ — universalset, Nov 27 '13 at 16:28
3 vendors can be in equilibrium, but the beach would have to go around the whole island. :) — Řídící, Nov 27 '13 at 20:20
For n odd, it seems to have no pure strategies NR, see http://www.ww.uni-magdeburg.de/bizecon/files/cdmo/cdmo-st07_092-108.pdf — Bob, Nov 07 '16 at 06:13

Ice cream vendor problem

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