N-Player Iterated "And Game"

This started as a write-up of what Vlad explained to me on Monday. I will add details as we further explore.

You have a n-player game where players can choose to be on or off. The mechanism can only detect whether all players have chosen on or not. Players can deposit some amount of value and expect to get some return on their deposit based on how the mechanism decides to payout. The game is played repeatedly on an open market.

Let R_on_i_p and R_off_i_p be the payout for round i for player p if everyone chooses to be on and off respectively. R_on_i_p and R_off_i_p are functions of previous round outcomes, the player p's deposit $D_p$, and the other player's deposits.

If we want to incentivize everyone to choose on, for the current iteration, we only need to look at mechanisms where R_on_i_p >= R_off_i_p. This is because if we have R_off_i_p > R_on_i_p and all players decide to choose to be on, we can increase the R_on_i_p so that R_off_i_p =< R_on_i_p and this will only further incentivize all players to choose to be on.

Similarly, we only need to look at mechanisms where R_on_i_p > 0.

We also want R_on_i_p to increase at least linearly with $D_p$ as otherwise players will just split up $D_p$ to across multiple players.

## Exploration of specific reward functions

Let R_on_i_p = $\frac{D_p}{\sum_k{D_k}} X$ where X is some fixed value and R_off_i_p <= 0. Assume player p has their own function S_p that takes previous round outcomes and reduces them to a single probability that represents the chance that all players will choose to be on in the next round. Thus player p's expected rate of return is $\frac{S_pD_pX}{\sum_k{D_k}}$. If the external world's rate of return is $Out$, then a player p will deposit up to $\frac{S_pX}{Out} - \sum_{k \neq p}D_k$. For example, if we are the only player in the game (and thus $\sum_{k \neq p}D_k = 0$ and $S_p = 1$) and $Out$ is 1 (aka 100% rate of return per round), then we want to deposit up to X so we will get a better return than 100%.