In Russian roulette, is it best to go first? Assume that we are playing a game of Russian roulette (6 chambers) and that there is no shuffling after the shot is fired.

I was wondering if you have an advantage in going first?

If so, how big of an advantage?

I was just debating this with friends, and I wouldn't know what probability to use to prove it. I'm thinking binomial distribution or something like that.

If $n=2$, then there's no advantage. Just $50/50$ if the person survives or dies. 

If $n=3$, then maybe the other guy has an advantage. The person who goes second should have an advantage. 

Or maybe I'm wrong.
For a $2$ Player Game, it's obvious that player one will play, and $\frac16$ chance of losing.  Player $2$, has a $\frac16$ chance of winning on turn one, so there is a $\frac56$ 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 - $\frac66$ (Chance Turn $1$ happening) $\times \ \frac16$ (chance of dying) = $\frac16$

Player 2 - $\frac56$ (Chance Turn $2$ happening) $\times \ \frac15$ (chance of dying) = $\frac16$

Player 1 - $\frac46$ (Chance Turn $3$ happening) $\times \ \frac14$ (chance of dying) = $\frac16$

Player 2 - $\frac36$ (Chance Turn $4$ happening) $\times \ \frac13$ (chance of dying) = $\frac16$

Player 1 - $\frac26$ (Chance Turn $5$ happening) $\times \ \frac12$ (chance of dying) = $\frac16$

Player 2 - $\frac16$ (Chance Turn $6$ happening) $\times \ \frac11$ (chance of dying) = $\frac16$

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:
P1 - $\frac26$,
P2 - $\frac26$,
P3 - $\frac16$,
P4 - $\frac16$

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 - $\frac36$, P2 - $\frac26$, Captor -  $\frac16$
For 2 players and n chambers player 1 loses if the round is in the odd-numbered chamber. Hence if $n$ is even, chances are $50/50$ and if n is odd, chances of losing are $(\frac{n+1}{2n},\frac{n-1}{2n}$) - it's a disadvantage to go first.