# Nash Equilibria of P-Beauty

I'm a little confused with the work I am currently doing in Game Theory.

Here is the questions I'm working on:

1) Suppose we modify the $p$-Beauty contest by requiring every guess to be an integer. What are all of the Nash equilibria? Why?

2) The $1$-Beauty contest is the $p$-Beauty contest with $p = 1$, i.e., a winner is anyone who chooses a number that’s closest to the average of all the choices. What are all of the Nash equilibria? Why?

I don't understand how these change anything. The Nash equilibrium of the $p$-Beauty contest is $0$. Requiring every guess to be an integer shouldn't change anything correct? The nash equlibria would just be $\{ 0 \}$?

How does changing the $P$ to equal $1$ change this? $1\times 0$ is still $0$. So would the nash equilibria stay as $0$? I feel like it should change, but I don't understand how. I feel like it could be any number

In the $p$-beauty contest game (Moulin $1986$), all participants are asked to simultaneously pick a number between $0$ and $100$. The winner of the contest is the person(s) whose number is closest to $p$ times the average of all numbers submitted, where $p$ is some fraction, typically $\frac23$ or $\frac12$.

• What is the $p$-beauty contest? – Théophile Mar 3 '16 at 20:47
• @Théophile Sorry I added a description to the bottom of OP – Joseph Mar 3 '16 at 20:49
• – joriki Mar 3 '16 at 20:50

1. Can you first prove that all players must guess the same number in a Nash equilibrium? Now, for a given $p < 1$, the players will be satisfied if they all guess $g \in \mathbb Z$ where the closest integer to $pg$ is $g$. You're right that $g=0$ will always be a solution. But for $p=\frac23$ (for example), is there any other $g \in \mathbb Z$ for which $\frac23 g$ is close to $g$?