I'm assuming a game of Russian Roulette where you have a gun with 1 bullet and six chambers, and two people playing.

The rules are:

  1. You must shoot at least once on your turn.

  2. After shooting once, you can continue shooting as many times as you want until you decide to pass it to the other player.

  3. Whoever gets shot loses (obviously?)

  4. The chamber is spun once at the start of the game and the shots are taken in order from there without spinning. If the bullet is chamber 4 whoever shoots 4th will lose.

What are the optimal strategies for player 1 and player 2? Is it possible to figure this out or are there too many variables?

How about just for the first move? How many times should Player 1 shoot before passing it for an optimal chance at winning?

  • $\begingroup$ If the chamber chosen is random every shot then the optimal strategy should be obvious. Is there more to this game? Do you use each chamber in order until the bullet is fired? The optimal strategy will be the same but this detail matters for a proof given. $\endgroup$
    – Hugh
    Nov 12, 2016 at 21:05
  • $\begingroup$ @Hugh I was thinking of an in-order type game. As in, you put the bullet in the gun, spin the chamber and then never spin the chamber again. So if it's in the 4th chamber, whoever shoots the fourth shot is guaranteed to die. $\endgroup$ Nov 12, 2016 at 21:39
  • $\begingroup$ You might have seen it but there is also a paragraph on probability on the Wikipedia page en.wikipedia.org/wiki/Russian_roulette What I found interesting is the fact that due to gravity the round tends to end up in the bottom position after spinning the barrel. That tells you that if you are player 1 you could/should take more than one shot. $\endgroup$
    – Stefan
    Jan 18, 2018 at 5:02

2 Answers 2


What are the optimal strategies for player 1 and player 2? How many times should Player 1 shoot before passing it for an optimal chance at winning?

Once the barrel is spun, the position of the bullet is fixed, which means that there is a $\frac{1}{6}$ probability to be shot at turn number $i$, $i \in [|1,6|]$. As a result, the optimal strategy for player 1 and player 2 is to shoot only once on their turn. Following this strategy, the probability that player 1 wins is the same as the probability that player 2 wins, i.e. $0.5$, regardless of who starts.

  • $\begingroup$ If you are correct (and I bet you are) and I'm understanding you correctly (I'm not clear on the notation with vertical bars), I find it very counterintuitive that the probability would not increase with every passing round. $\endgroup$ Nov 13, 2016 at 20:33
  • $\begingroup$ @AntoniParellada regarding the notation with vertical bars: math.stackexchange.com/a/875294/24265 $\endgroup$ Nov 13, 2016 at 20:58
  • $\begingroup$ If the barrel of the gun (I don't know about guns, but...) is spun just once, the first time around the $\Pr(\text{firing})=1/6$. However, if the first person is lucky and the gun doesn't go off, the barrel advances automatically to the next chamber, leaving behind an empty slot, doesn't it? My guess would be, then, that the guy (or girl) going in second place has now a $\Pr(\text{firing})=1/5$. After five rounds without firing the probability of dying for the unfortunate player pulling the trigger for the sixth time would be $1$. But this is not what you are saying, is it? $\endgroup$ Nov 13, 2016 at 21:10
  • 1
    $\begingroup$ Kindly, I am able to "convince" myself more easily (I know the answer is obvious) following the reasoning in this answer, which can be tweaked ever so slightly for this question. I sense there is a subtlety in your answer that I am missing. $\endgroup$ Nov 13, 2016 at 22:12
  • 1
    $\begingroup$ What Franck is saying is there is a $\frac{1}{6}$ probability to be shot on turn $i$ as the bullet goes into one of the positions in the barrel with equal probability and the position of the bullet in the barrel completely determines on which turn someone will be shot. I believe you are thinking in terms of conditional probabilities i.e. the probability of being shot on turn 2 given you were not shot on turn 1. $\endgroup$
    – rwolst
    Nov 18, 2016 at 14:43

Here is an equivalent game:

There is a stack of 6 cards. The top card has the number 1, the second has the number 2, and so on through the bottom card, which has the number 6.

The rules are:

  1. You must take at least one card on your turn.
  2. After taking one card, you can continue taking as many cards as you want until you decide to pass your turn to the other player.
  3. After all cards are taken, a 6-sided die is rolled. Whoever holds the corresponding card loses.

There is only one difference between this game and the Russian roulette game. In this game, the losing card is not revealed until after all cards have been chosen; in the Russian roulette game, the losing chamber is revealed immediately when that chamber is chosen. But this doesn't matter for choosing a strategy, because none of the choices matter any more once the losing card/chamber has been taken.

In the card game, the probability of losing is clearly proportional to the number of cards taken, so the winning strategy is simply to take as few cards as possible.

This means that the winning strategy for both games is:

  • If there are an even number of cards/chambers remaining, take only one card/shot.
  • If there are an odd number of cards/chambers remaining, take either one or two cards/shots; it makes no difference.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.