$N$ players are playing paintball tag. They stand in a circle and take turns clockwise to act, starting from player 1, as illustrated below. The last untagged participant wins
The rules are simple. In his turn, a player can either randomly tag one of the other players, or shoot in the air. But if nobody is tagged in a whole round (i.e. all players have chosen to shoot in the air their last turn), the referee will randomly pick one player as the winner. Players act to maximize their own winning probabilities, but prefer shooting in the air to tagging if both are equally good.
For sufficiently large $N$, will player 1 always choose to tag at the start of the game?
Notice that if it's optimal for player 1 to shoot in the air at the start of the $N-1$ game, then it'll be optimal for them to tag at the start of the $N$ game. So for large $N$, tagging will happen at least as frequently as shooting in the air at the start of the game.
Best Answer
A strange game.
For $n=2$, obviously tag and win. For $n=3$, tag and the third player tags you back, the classic standoff. Referee-selection happens, everyone wins with probability $1/3$.
For $n>3$, suppose not tagging is correct. Then it is equally correct for everyone, and referee-selection happens - everyone wins with probability $1/n$.
However, suppose you do tag, and further suppose tagging is correct until it reaches $3\leq m$ standoffs. The standoffs each win with probability $1/m$, so what's the probability of being one of them?
Since you can't tag yourself, after the first turn there are $n-1$ players, you guaranteed to be one of them. Everyone after that point will be tagging until $m$ standoffs, and since $n-m<n$, you won't get a chance to tag again (all the player-tags will have happened).
Whoever's next fires, with $n-2$ targets, so you dodge that paint with probability $(n-3)/(n-2)$. The person after that aims at $n-3$ targets and you dodge that with probability $(n-4)/(n-3)$. The last tag occurs with $m$ targets (from $m+1$, to leave $m$ standing). You make it through to the last $m$ with probability $$\frac{n-3}{n-2}\frac{n-4}{n-3}\dotsm\frac{m}{m+1}\frac{m-1}{m}=\frac{m-1}{n-2}$$ and counting final jeopardy, that gives a winrate when tagging of $$\frac{m-1}{m(n-2)}\text{.}$$
To not tag, you want $$\frac{m-1}{m(n-2)}\leq\frac{1}{n}\quad\Rightarrow\quad n\geq2m$$ where $m$ was the last standoff population size, which means you refuse to tag at $n=3\cdot2^k$. Therefore, there's no berzerker population limit.