I've seen a discussion on a two level tug of war. There are N*N players. Each player has resources associated with them. Each player splits resources between the "team game" and the "individual game", so that it sums to the amount of resources the player controls. In the first stage N groups of N players divide the prize. The prize is divided among groups proportional to the sum of "team game" contribution of the members of a group. Then the group share is divided among players proportional to "individual game" contribution of each player.
Each player decides how to split the resource between the "team game" and "individual game". If everyone on the team puts all the resource to "individual game" then their team will get no prize. At the same time, if the player over invests in the "team game" he will be scooped by tricker players in the second stage of the game.
What is the optimal strategy for a player in this game?
Edited: The answer depends on the function F (investment1,investmnet2,..investmentN) -> payoffs(p1,p2,..pN) and in some cases th4e strategy might be trivial. Which functions are typically considered for this game?
There is no globally optimal strategy, as the optimal strategy depends on the strategies of the other players. Instead, in game theory solution concepts are used to predict how rational self-interested players will play. The most widely applied solution concept is that of a Nash equilibrium, in which each player’s strategy is a best response to the strategies of the other players.
So consider the strategy of player $P$. For simplicity, take the values of the resources available to each player to be $1$. Given the other players’ contributions to the “team game”, we can form the sum $S_P$ of the contributions of $P$’s teammates and the form $S_O$ of all other players. Then if $P$ invests a proportion $p$ in her team contribution, her team will get a proportion $(S_P+p)/(S_P+P+S_O)=1-S_O/(S_P+p+S_O)$ of the prize, and $P$ will get a share $(1-p)/(N-S_P-p)=1-(N-S_P-1)/(N-S_P-p)$ of the team’s spoils. Thus the payoff for $P$ is
$$ \left(1-\frac{S_O}{S_P+p+S_O}\right)\left(1-\frac{N-S_P-1}{N-S_P-p}\right)\;. $$
Assuming a symmetric equilibrium, we can set the derivative with respect to $p$ to $0$ and then substitute $S_P=(N-1)p$ and $S_O=N(N-1)p$, which makes the denominators $S_P+p+S_O=N^2p$ and $N-S_P-p=N(1-p)$ and makes both factors come out as $\frac1N$. Thus the stationarity condition is
$$ \frac{N(N-1)p}{(N^2p)^2}-\frac{(N-1)(1-p)}{(N(1-p))^2}=0\;. $$
Most of the factors cancel, and we’re left with
$$ \frac1p-\frac N{1-p}=0\;, $$
with solution $p=\frac1{N+1}$.