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.
 A: This reminds me of a similar interview question; you see your interviewer load two bullets into consecutive chambers of a gun, point at his head and pulls the trigger, but survives as it was an empty chamber.  He then gives you the gun to fire at your head, giving you the option to either spin the barrel before shooting or to just take the next shot.

You then decide not to spin the barrel; as with spinning you have a 2/6 ~ 33$\frac{1}{3}$% chance of killing yourself.  Without spinning and knowing that the last shot was nonlethal you leave only 5 chambers; two of which have bullets; however knowing the bullets are in consecutive chambers, you are only afraid of running into the group of bullets which appears in only 1 of 4 spots (of consecutive bullets) so there's only a 1/4 ~ 25% chance of dying.
A: They are all the same, because you only shuffle once.
The order of blank chambers and bullet are determined in the very beginning of the game.
Prob(Player A got shot) is just the probability of that bullet lies on the chamber player A will shot, for every one it is (number of bullets)$/$(number of chambers)
A: 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 -  $\frac16$
A: Your best bet is actually to go last, because this will either make no difference or decrease the probability of shooting yourself. Why? If there are $n$ people and $n$ divides $6$, then the probability of any one person taking the bullet is equal, since the probability distribution is assumed to be uniform over the number of times the trigger is pulled. If $n \le 6$ does not divide $6$, i.e. if $n=4$ or $5$, then the turns wrap around and begin with the first person again, so the first one or two people have to shoot again; naturally, this doubles their probability of being shot. And finally, if $n>6$ then the bullets will have run out before they come to you if you go last.
A: 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.
