# Calculating probability and optimization - russian roulette

In a game of Russian Roulette, there are 6 chambers, one of which has a bullet in it.

You are playing with one other person.

You go first. Now, your chance of getting hit is 1/6. After firing (and not getting hit), you can either pull the trigger again (1/5 probability of getting hit), or you can pass the gun and let Player 2 try.

My question is this: if you are playing multiple rounds of this game (or a game like this), is there some optimal number of shots you should be taking, or is it always better to just pass the gun as soon as you've fired it (and survived)?

EDIT 1: So after each shot, the chambers are not re-spun. And the objective of the game is to live.

• What does your intuition tell you? – Bram28 Mar 24 '17 at 17:23
• Based on the probability moving from $1/6$ to $1/5$, I'm assuming you don't respin the chambers. – Brian Tung Mar 24 '17 at 17:23
• What is in the game? Is it only about life and death or does the choice affect any other wins? For example if I take two shots will I win much more of something than if I don't? Either way we must define an objective function to be able to do any optimization ! – mathreadler Mar 24 '17 at 17:31
• At the start of the game, you each have a 50-50 chance of winning, as you are each responsible for firing 3 times. If you play first and then shoot again before passing, now you are responsible for firing 4 times (shots 1, 2, 4, and 6). Definitely a bad bet, as it means you now have a $\frac23$ chance of shooting yourself. – paw88789 Mar 24 '17 at 17:48

## 2 Answers

Assuming no one ever respins the chambers (because otherwise the strategy is trivial—always pass the gun off):

Work backwards. Obviously, with only one chamber remaining, you would pass the gun off. Good luck captain!

With two chambers left, if you pass the gun, you have a $1/2$ chance of surviving, no matter what you do (pass the gun, or shoot again).

With three chambers left, if you pass the gun, your opponent has a $1/3$ chance of surviving that shot, after which your chances are both $1/2$. So your chances of survival by passing the gun off is $2/3$. So you pass the gun off.

With four chambers left, if you pass the gun, your opponent has a $1/4$ chance of surviving that shot, after which they will pass the gun off, and your chances of survival at that point are $1/3$. So the chances are $1/2$ for both, and it doesn't matter what you do.

With five chambers left, if you pass the gun, your opponent has a $1/5$ chance of surviving that shot, after which their chances of survival are $1/2$. So your chances of survival if you pass the gun off are $3/5$, so you pass the gun off.

You should be able to show by induction that passing the gun off is never a bad choice. With an even number of chambers left, it's always a $50$-$50$ proposition, and with an odd number of chambers left, it's strictly better to pass the gun off.

There might be an analysis that short-circuits the induction and makes the pattern obvious upfront, but I don't see it yet.

Given that you taking a second shot or not taking a second shot has no further consequences on the chances of you or your opponent dying (assuming you do survive the second shot), the choice is really between taking a second shot with a chance of $\frac{1}{5}$ of dying, or not taking shot, with a chance of $0$ of dying ( at least for that moment).