n players numbered 1~n play a shooting game. Their accuracy rates are 0<${p}_{1}$<${p}_{2}$<...<${p}_{n}$<1, respectively. This is common knowledge.

Before the game starts, the referee arranges the n players in some order. When game starts, players take turns to fire at one another according to that order. (For example, if n=4, and the referee arranges the players in order (3,4,1,2), when game starts, 3 fires first, then 4 fires, then 1, then 2, then 3 again, so on and so forth, as long as they are alive) The last person left is the winner of the game.

To be more specific, define $S_{i}={1,2,3,...,i-1,i+1,i+2,...,n}$. Let $S_{i}^{k}$ be any subset of $S_{i}$ that contains k elements (0<k<n). Then the strategy of player i is a function that maps $S_{i}^{k}$ to one of its elment, for any $S_{i}^{k}$. What this definition means is that, given any k players (excluding i himself) left, player i's strategy tells him whom to shoot first. Notice that we rule out the possibility that any player can hold fire in his turn: he must choose someone to shoot.

After the referee arranges the firing order, all players must announce their strategies simultaneously. A player's payoff is his winning probability.

Question 1: Is there always an Nash equilibrium in this game, for any firing order?

Question 2: Suppose firing order is (1,2,3,...,n). Which players have dominant strategies for $(0,0,...,0)$<(${p}_{1}$,${p}_{2}$,...,${p}_{n}$)<$(1,1,...,1)$, 0<${p}_{1}$<${p}_{2}$<...<${p}_{n}$<1 ? (When n=3, all have dominant strategies; when n=4, player 1,2 and 4 have dominant strategies. I guess there could be regularities as n gets larger? At least can we say 1 always has dominant strategy? )