Prove that there are finite Nash Equilibria in the following game Three players are competing on a game show. Each wagers an integer dollar amount between 1 and 100.
If all three players wager different amounts, the player who wagered the median wins the pot. If two players wager the same amount and the third wagers a different value, the third player wins the pot. If all the players wager the same value, nobody wins the pot.
I've so far found two NEs, both involving all the players playing either 1 or 2, and think I might have one that's a mixed strategy playing 1, 2, and 3 in a symmetric NE, but I suspect these might be all. Unfortunately, I don't know how to prove this intuition.
 A: You have not really provided a payoff or utility function. I assume that the winner has a utility equal to the pot minus his own contribution.
Then the following strategy profile is a Nash equilibrium: $(1,1,2)$, where the first player bets $1$, the second also bets $1$, and the last bets $2$.
Can those playing $1$ unilaterally deviate and do better? Switching to $2$ still means he loses, since the other one playing $1$ now wins. Switching to $>2$ means the one playing $2$ wins, so that's no better. So he can't do better.
Can the one playing $2$ unilaterally deviate and do better? He wins in this equilibrium. Switching to $1$ means nobody wins, that's no better. Switching to $>2$ does not change anything, he still wins the bet of the two others. So he can't do better either.
Hence, $(1,1,2)$ is a Nash equilibrium. Obviously, $(2,1,1)$ and $(1,2,1)$ are as well. There might be more.
Clearly, any symmetric pure strategy profile $(x,x,x)$ cannot be an equilibrium for any $x\in[1,100]$, as any deviation would be profitable.
