The “game” Russian Roulette is a striking element in dramas and tense movies. Because of this, I preferred not to put any scene of these occasions… However, there is a rather interesting mathematics that I would like to explore in this “game”. For this we need to understand how this “game” is and how it works.
In the common context, there is a gun with a bullet drum, usually a revolver and a drum with space for 6 bullets. Add a bullet and turn the drum, so that no one knows in which drum the bullet is.
In case, with two players, the first player puts the gun on the head and fire! If the bullet is not in that drum, then the player survives and passes the gun to the other player, who repeats the procedure, until one of the two players dies.
A simple question we can ask in this context is which of the two players has a better chance of getting out alive in this “game”?
One way to solve this problem is to imagine that for each space in the drum there is the name of a player, so for 6 spaces and 2 players called player A and player B, we would have the following drum:
- Player A;
- Player B;
- Player A;
- Player B;
- Player A;
- Player B.
Then, of the 6 initial spaces, 3 are occupied by player A, and 3 are occupied by player B. Given that the bullet is placed in a random space, we have 50% chance of player A to live and 50% chance of player B to live. Since their life and death depend solely on the position in which the bullet is placed.
Now suppose for example that the drum has 5 spaces for bullet, in the case we would have:
- Player A;
- Player B;
- Player A;
- Player B;
- Player A.
Then of the 5 starting spaces, 3 are occupied by player A and 2 are occupied by player B. Given that the bullet is placed in a random space, we have 40% chance of player A to live and 60% chance of player B to live .
Interestingly, we can think of a situation with the same number of players as spaces. Thus, with 5 spaces, players A, B, C, D, E compete. For each one has a designated space.
- Player A;
- Player B;
- Player C;
- Player D;
- Player E.
Winning depends only on what position the bullet will initially be placed. Thus, each player has exactly 80% chance of winning. Regardless of whether you are the first to play or the last. The false belief that the latter will already be doomed when the gun reaches him is absurd because we need to consider that the gun’s chance of reaching the latter is:
(Player A probability to shoot and not die)*(Player B probability to shoot and not die)*(Player C probability to shoot and not die)*(Player D probability to shoot and not die).
Turning this into numbers, we have:
(4/5)*(3/4)*(2/3)*(1/2)=1/5.
That is, regardless of where we start in the “game”, in these conditions of victory are the same, what changes is at most our “impression of victory” when considering that the gun is closer or further away from us.
Another way to analyze the problem is to imagine a drum with 5 spaces and two bullets for 5 players. Now it is not something so trivial anymore. We have the same starting list
- Player A;
- Player B;
- Player C;
- Player D;
- Player E.
The simplest case is to analyze player A, his chance of losing initially is 2/5, so his chance to win is 3/5.
For player B, we have 4 situations.
- Pick up the gun with two bullets and win: (3/5) * (2/4) = 30%.
- Pick up the gun with a bullet and win: (2/5) * (3/4) = 30%.
- Pick up the gun with two bullets and lose: (3/5) * (2/4) = 30%.
- Pick up the gun with a bullet and lose: (2/5) * (1/4) = 10%.
Totalizing 60% of victory against 40% of defeat.
For player C, we have 7 situations:
- Pick up the gun with two bullets and win: (3/5) * (2/4) * (1/3) = 10%.
- Pick up the gun with a bullet and win, case 1: (2/5) * (3/4) * (2/3) = 20%.
- Pick up the gun with a bullet and win, case 2: (3/5) * (2/4) * (2/3) = 20%.
- Pick up the gun with two bullets and lose: (3/5) * (2/4) * (2/3) = 20%
- Pick up the gun with a bullet and lose, case 1: (2/5) * (3/4) * (1/3) = 10%.
- Pick up the gun with a bullet and lose, case 2: (3/5) * (2/4) * (1/3) = 10%.
- Win without picking up the weapon: (2/5) * (1/4) = 10%.
Totalizing 60% of victory against 40% of defeat.
For player D, we have 10 situations:
- Pick up the gun with a bullet and win, case 1: (2/5) * (3/4) * (2/3) * (1/2) = 10%.
- Pick up the gun with a bullet and win, case 2: (3/5) * (2/4) * (2/3) * (1/2) = 10%.
- Pick up the weapon with a bullet and win, case 3: (3/5) * (2/4) * (2/3) * (1/2) = 10%.
- Pick up the gun with a bullet and lose, case 1: (2/5) * (3/4) * (2/3) * (1/2) = 10%.
- Pick up the gun with a bullet and lose, case 2: (3/5) * (2/4) * (2/3) * (1/2) = 10%.
- Pick up the gun with a bullet and lose, case 3: (3/5) * (2/4) * (2/3) * (1/2) = 10%.
- Pick up the gun with two bullets and lose: (3/5) * (2/4) * (1/3) = 10%
- Win without picking up the weapon, case 1: (2/5) * (1/4) = 10%.
- Win without picking up the weapon, case 2: (2/5) * (3/4) * (1/3) = 10%.
- Win without picking up the weapon, case 3: (3/5) * (2/4) * (1/3) = 10%.
Totalizing 60% of victory against 40% of defeat.
For player E, we have 10 situations:
- Win without picking up the weapon, case 1: (2/5) * (1/4) = 10%.
- Win without picking up the weapon, case 2: (2/5) * (3/4) * (1/3) = 10%.
- Win without picking up the weapon, case 3: (3/5) * (2/4) * (1/3) = 10%.
- Win without picking up the weapon, case 4: (2/5) * (3/4) * (2/3) * (1/2) = 10%.
- Win without picking up the weapon, case 5: (3/5) * (2/4) * (2/3) * (1/2) = 10%.
- Win without picking up the weapon, case 6: (3/5) * (2/4) * (2/3) * (1/2) = 10%.
- Pick up the gun with a bullet and lose, case 1: (2/5) * (3/4) * (2/3) * (1/2) = 10%.
- Pick up the gun with a bullet and lose, case 2: (3/5) * (2/4) * (2/3) * (1/2) = 10%.
- Pick up the gun with a bullet and lose, case 3: (3/5) * (2/4) * (2/3) * (1/2) = 10%.
- Pick up the gun with a bullet and lose, case 4: (3/5) * (2/4) * (1/3) = 10%.
Totalizing 60% of victory against 40% of defeat.
Did you think the chances of winning and losing would be different between the players now that we had 2 bullets? Probability and its secrets!
Xoxelho 