Is there an optimal game theory strategy for choosing the second lowest number? Say there is a group of n people, and they all must select an integer, x, between 1 and 100 with the goal being that in the set, y, of all numbers selected only one is less than the number you selected. So the person(s) who selected the second lowest number is the winner. 
If you tie you split the prize, but if everyone says the same number no one wins because they are all the lowest number. 
How does the solution change if the number is not limited to integers? 
 A: First of all, it's clearly a Nash equilibrium for everyone to choose $100$. Nobody wins, and nobody can make themselves win. Let's call that the trivial equilibrium.
If $n=2$, then there are also nontrivial equilibria where player 1 always chooses $100$ and player 2 does... anything at all. Player 2 never wins, so it doesn't matter what they choose; and player 1 has no incentive to choose anything less than $100$.
If $n>2$, there there is still a nontrivial equilibrium in which player 1 chooses between $99$ and $100$, possibly randomly, and every other player always chooses $100$. Player 1 can't win, so they are indifferent between all choices. The remaining players might win with $100$, or they might not, but they have no incentive to switch, because they definitely wouldn't win with anything else.
I claim that there are no other equilibria, pure or mixed, than those described above. This should be provable by inductively ruling out $0$, then $1$, then every other choice less than $99$. Since it's Friday, this proof is left as an exercise for the reader.
