Setup: players must chose a number between $0$ and $100$. The winner of the game is the player whose chosen number is closest to the average of all chosen numbers multiplied by $p$. Assume that in a tie the reward is split equally among winners. I know that if $p<1$ then the Nash equilibrium is 0. If $p>1$ the Nash equilibrium is $100$.
What about if $p=1$?
Is the Nash equilibrium a mixed strategy that assigns equal probability to all numbers between $0$ and $100$?
OR
Is the Nash equilibrium simply choosing $50$ (the mean of all numbers between 0 and 100)?
I am confused and so is my professor, please help :)
There are a lot of Nash equilibria, it makes no sense to speak of the Nash equilibrium.
For example, everyone choosing the same number is an equilibrium, as is everyone randomizing between the same two consecutive numbers with equal probability. With only two players, every pair of mixed strategies is an equilibrium.