Russian Roulette: is the practice of placing a single round in a revolver, spinning the cylinder and closing it into the firearm without looking, aiming the revolver at one's own head in a suicidal fashion, and pulling the trigger.
Once the cylinder is spun, the weight of the bullet tends to make the cylinder rest with the bullet toward the bottom, thus increasing the odds that the shot taken will be a blank. Therefore, the argument that each hole is equally likely to be under the hammer is contestable. One way that this 'bullet bias' could be eliminated is to spin the cylinder with the barrel pointed down, so the cylinder spins on a vertical axis instead of a horizontal one.
For the purpose of this section, the effect bullet bias would have on the odds is ignored for simplicity reasons.
The terminology for this section:
Player: One participant in the game
P1, P2 ... Pn: Player 1 to Player n respectively
T: The total number of players in the game.
B: The number of bullets in the gun
C: The number of chambers in the gun
Round: A round occurs when a player takes one shot at his head with the gun. For example, the normal game with B = 1 and C = 6 and the cylinder isn't being spun would have a maximum of 6 rounds. It is assumed that P1 goes first, then P2 and so on.
R1, R2 ... Rn: Rounds 1 to n respectively
Losing a round: the gun gets fired.
Winning a round: the gun wasn't fired.
The game stops on the first losing round.
A player Pn dies if Rx results in a death and {(x - n)} \mod {T}. For example, if there are 2 players (T = 2) then player 1 dies if round 13 is a death : {(13 - 1)} \mod {2}. Put another way, Pn loses if any of the rounds n, n + T, n + 2T... results in a loss (these can be represented by the formula n + xT where x is a positive integer or 0).
The most common russian roulette game has T = 2; B = 1; C = 6; P1 loses on rounds 1, 3, 5 and P2 loses on rounds 2, 4, 6.
If the cylinder is spun after every shot, the odds of losing a round is BC. Alternatively, the odds of winning a round is 1 - \frac{B}{C}. However, the odds of making it to round n drop as n gets larger. This is because to make it to round n, rounds n-1, n-2... must have been won. So the odds for the game to stop on round n is (1 - \frac{B}{C})^{(n-1)} * \frac{B}{C}. Then, the odds of Px to lose is \sum_{k=0}^n (1 - \frac{B}{C}) ^{(x + kT - 1)} * \frac{B}{C} as n approaches infinity. This can be simplified to A^{x-1} . \frac{B}{C} . \frac{1}{(1-A^T)} where A = 1 - \frac{B}{C}.
For a standard game, P1 has a 6/11 chance of losing, while P2 has a 5/11 chance. Hence it is better to go last. Also, note the Ax - 1 part of the equation. A is always less than 1, so as (x-1) increases, the chance to lose decreases. Hence it is always better to go last independent of number of players, and other parameters.
If the cylinder is not spun after each shot, the probability of losing a game can be determined by looking at each possibility of the bullet configuration in the gun. For example, in a standard game, if the bullet was in position 3, player 1 would lose. There are six possible positions for the bullet to be in a standard game: 1,2,3,4,5 or 6. Player 1 would lose if it is in position 1,3,5 (a 3/6 chance) and player 2 would lose if it is in position 2,4 or 6 (a 3/6 chance). Therefore both have an equal probability of losing (1/2).
Another example is with 6 players and 9 chambers with 1 bullet. There are seven possible positions for this game: 1,2,3,4,5,6,7,8 or 9. Player 1 would lose if it is in position 1,4,9 (a 3/9 chance), player 2: 2,5 (2/9), player 3: 3,6 (2/9), player 4: 4,7 (2/9), player 5: 5,8 (2/9), player 6: 6,9 (2/9) and player 7: 7 (1/9). In this case, it is much better to go last as compared to going first.