# A variant of Russian Roulette

Rules: When it is your turn you can either shoot yourself or shoot your opponent and then shoot yourself. When your turn ends you spin the cylinder and hand the gun to the other player. The cylinder holds 6 rounds.

How should you play the game so that you have the greatest chance of survival?

This is my attempt at the problem:

I’m going to cover only some cases.

Case 1: You start and only shoot yourself but your opponent always shoots you then himself when it’s his turn.

Then at the end of each round you have a 1 - (5/6)^2 = .306 probability of death and your opponent has a 1/5 = .2 probability of death (you shoot yourself (1/6) then opponent spins cylinder and shoots you (1/6) then opponent shoots himself (1/5)).

In this case your opponent has a greater chance of “beating you” at the game.

Case 2: Your opponent starts and you only shoot yourself but your opponent always shoots you then himself when it’s his turn.

The round-end probabilities here are the same as in Case 1. Your opponent has an advantage still, but less of an advantage than in Case 1 because he could shoot himself at the start of the game on the first shot.

Case 3: You start and always shoot opponent then yourself. Your opponent always shoots you and then himself.

Then at the end of each round the probability of death is the same for you as it is for your opponent, so you have an advantage since you could kill your opponent in round one when you start the game by shooting him first.

Case 4: Opponent starts and always shoots you then himself when it’s his turn. You always shoot opponent then yourself when it’s your turn.

Round-end probabilities are the same and you have a disadvantage since your opponent could shoot you on the first shot in the first round.

So it seems that it is always in your best interest to shoot your opponent and then yourself when it is your turn. Using that method, you could only be more likely to lose than your opponent IF your opponent starts off the game. This would be a great question for mathstackexchange. I might ask it there and get back to you because now I’m curious.

Let $$P_n$$ be the probability of the first person surviving, when there are $$n$$ bullets, if both participants always shoot at the opponent first. Then $$P_0=P_1=1$$.

For $$n\ge2$$,

$$P_n=\frac{1}{n}+\frac{n-2}{n}(1-P_{n-2})$$

and so $$nP_n+(n-2)P_{n-2}=n-1.$$

From this it is easy to see that $$nP_n$$ is the non-decreasing sequence $$0,1,1,1,2,3,3.$$

If the first deviation from this policy occurs with $$n$$ bullets remaining then the person making that choice changes their probability of survival from

$$\frac{1}{n}+\frac{n-2}{n}(1-P_{n-2})$$ to $$\frac{n-1}{n}(1-P_{n-1}).$$

This is a change of $$\frac{1}{n}((n-2) P_{n-2}-(n-1) P_{n-1} ).$$

Since this change is never positive, there is no (mathematical) reason to not shoot at the opponent first.

The general case

This analysis applies irrespective of how many bullets are initially used. In particular, the pattern $$0,1,1,1,2,3,3,3,4,5,5,5,6,7,...$$ means that the probability of survival for the first person is precisely $$\frac{1}{2}$$ when $$n$$ is even. For an odd number of bullets the probability alternates between being above and below $$\frac{1}{2}$$ and tends to $$\frac{1}{2}$$ as $$n$$ tends to infinity.