For a 2 Player Game, it's obvious that player one will play, and 1/6 chance of losing. Player 2, has a 1/6 chance of winning on turn one, so there is a 5/6 chance he will have to take his turn. (I've intentionally left fractions without reducing them as it's clearer where the numbers came from)
Player 1 - 6/6 (Chance Turn 1 happening) * 1/6 (chance of dying) = 1/6
Player 2 - 5/6 (Chance Turn 2 happening) * 1/5 (chance of dying) = 1/6
Player 1 - 4/6 (Chance Turn 3 happening) * 1/4 (chance of dying) = 1/6
Player 2 - 3/6 (Chance Turn 4 happening) * 1/3 (chance of dying) = 1/6
Player 1 - 2/6 (Chance Turn 5 happening) * 1/2 (chance of dying) = 1/6
Player 2 - 1/6 (Chance Turn 6 happening) * 1/1 (chance of dying) = 1/6
So the two player game is fair without shuffling.
Similarly, the 3 and 6 player versions are fair.
It's the 4 and 5 player versions where you want to go last, in hopes that the bullets will run out before your second turn.
For a for 4 player game, it's
Now, the idea in a 2 player game is that it is best to be player 2, because in the event you end up on turn six, you KNOW you have a chambered round, and can use it to shoot player 1 (or your captor), thus winning, changing your total odds of losing to P1 - 3/6, P2 - 2/6, Captor - 1/6