Assume that we are playing a game of Russian roulette (6 chambers) and that there is no shuffling after the shot is fired.

I was wondering if you have an advantage in going first?

If so, how big of an advantage?

I was just debating this with friends, and I wouldn't know what probability to use to prove it. I'm thinking binomial distribution or something like that.

If $n=2$, then there's no advantage. Just $50/50$ if the person survives or dies.

If $n=3$, then maybe the other guy has an advantage. The person who goes second should have an advantage.

Or maybe I'm wrong.

Christian Chapman
  • 4,427
  • 2
  • 22
  • 43
  • 2,669
  • 2
  • 13
  • 5
  • 3
    @lhf impossible to get a real gun in the UK. Was just having an argument about how fear makes you do illogical things. Surely it's illogical to go second, yet people would fear going first. – nikkita Jan 04 '12 at 12:43
  • Is n the number of participants? – LarsH Jan 04 '12 at 16:44
  • 10
    I think it would be appropriate to specify the rules of whatever variant of Russian Roulette you have in mind and what *n* represents (I assume you mean number of players, but it could also be number of bullets). – slim Jan 04 '12 at 16:53
  • 17
    I would think there would be a *lot* of shuffling after the shot went off: shuffling of brain matter into the air, shuffling of feet as people got out of there before the cops show up, etc. – Justin ᚅᚔᚈᚄᚒᚔ Jan 04 '12 at 16:56
  • 54
    There's an implicit assumption here that it's an advantage to survive. [Apparently](http://www.brightreview.co.uk/ARTICLE-The-Man-Who-Invented-Russian-Roulette.html) what's now called "Russian roulette" originated in circumstances where the players were sometimes weary of life and didn't mind "losing". – joriki Jan 04 '12 at 18:03
  • 22
    If after the first guy fires a blank, the host offers you to swap turns, should you? – Larry OBrien Jan 04 '12 at 23:35
  • 15
    Not a mathematician but if I may put a non-math spin on this one... In real life, the added mass of the bullet will bias the spin and probably keep the bullet in one of the lower chambers. – Everyone Jan 05 '12 at 12:23
  • 1
    That depends. Are you using an automatic or revolver? – jacknad Jan 05 '12 at 17:25
  • @Everyone that's an excellent question for the physics SE – AncientSwordRage Jan 06 '12 at 00:49
  • I think is better not go. – leo Jan 06 '12 at 02:09
  • @JackN: The magazine in an automatic wouldn't leave much room for probability/chance. The person who loaded the magazine would be aware of the initial state. – Everyone Jan 06 '12 at 04:29
  • 2
    I think it's just best not to go last. – daniel Jan 06 '12 at 16:21
  • http://www.smbc-comics.com/?db=comics&id=7#comic Does Zach Weiner read math.SE? – fosco Jan 07 '12 at 12:09
  • I remember reading in the literature, quite some years ago, the remark (actually a complaint by a Russian) that this pastime is actually misnamed, as it actually originated in the American Civil War. Can anyone verify or refute this claim? – Hexagon Tiling Jan 09 '12 at 12:56
  • I am pretty sure that there is a Russian Roulette scene (although it's not called that) in Lermontov's "A Hero of Our Time", which was written in 1839. – Flounderer Jan 26 '12 at 23:39
  • For those interested in probability theory, note that this problem is equivalent to sampling without replacement from the set of 6 chambers. It is a general property of sampling w/o replacement that the probability for a certain sequence of draws is only dependent on how many draws there were of each "type", and not on their order ("exchangeability") – fiftyeight Jul 21 '17 at 16:40
  • 1
    @ChristianChapman What is the point in bumping this old question just to add absolutely unnecessary white space in two LaTeX equations? This is not an appropriate use of editing. – TheSimpliFire Aug 03 '21 at 08:44

5 Answers5


For a $2$ Player Game, it's obvious that player one will play, and $\frac16$ chance of losing. Player $2$, has a $\frac16$ chance of winning on turn one, so there is a $\frac56$ chance he will have to take his turn. (I've intentionally left fractions without reducing them as it's clearer where the numbers came from)

Player 1 - $\frac66$ (Chance Turn $1$ happening) $\times \ \frac16$ (chance of dying) = $\frac16$

Player 2 - $\frac56$ (Chance Turn $2$ happening) $\times \ \frac15$ (chance of dying) = $\frac16$

Player 1 - $\frac46$ (Chance Turn $3$ happening) $\times \ \frac14$ (chance of dying) = $\frac16$

Player 2 - $\frac36$ (Chance Turn $4$ happening) $\times \ \frac13$ (chance of dying) = $\frac16$

Player 1 - $\frac26$ (Chance Turn $5$ happening) $\times \ \frac12$ (chance of dying) = $\frac16$

Player 2 - $\frac16$ (Chance Turn $6$ happening) $\times \ \frac11$ (chance of dying) = $\frac16$

So the two player game is fair without shuffling. Similarly, the $3$ and $6$ player versions are fair.

It's the $4$ and $5$ player versions where you want to go last, in hopes that the bullets will run out before your second turn.

For a for $4$ player game, it's:
P1 - $\frac26$,
P2 - $\frac26$,
P3 - $\frac16$,
P4 - $\frac16$

Now, the idea in a $2$ player game is that it is best to be player $2$, because in the event you end up on turn six, you KNOW you have a chambered round, and can use it to shoot player $1$ (or your captor), thus winning, changing your total odds of losing to P1 - $\frac36$, P2 - $\frac26$, Captor - $\frac16$

Plus Twenty
  • 838
  • 8
  • 24
Chris Cudmore
  • 3,798
  • 1
  • 12
  • 17
  • With shuffling, the first column is all 5/6 (except turn 1), and the right column is always 1/6, giving odds/turn 5/36 (1 / 7.2), which gives player 2 a slight advantage. – Chris Cudmore Jan 04 '12 at 18:04
  • 164
    By that logic, if player one is holding the gun at chamber five, he either shoots himself or waits to get shot by you. Thus he should just shoot you at that point. Which, of course, means you need to shoot him at chamber four, but then... – Joshua Shane Liberman Jan 04 '12 at 21:13
  • 17
    @Joshua - but if he tries to shoot you at five and five is empty, you have a tiny instant to react and duck/go ballistic on him. The same goes for the others, each has a decreasing chance that the chamber can be used to shoot the other person in time. – Daniel G. Wilson Jan 04 '12 at 21:29
  • 21
    This is easily solved: Instead of shooting yourself, you just shoot the other player each turn. – Vinko Vrsalovic Jan 05 '12 at 13:41
  • 9
    @XenElement, I think he meant, after you shoot chamber 5 at yourself (and its empty), before you hand the gun over to player 2, you will KNOW chamber 6 is loaded and have your hands on the gun first. So why hand it over? Just shoot player 2. – JD Isaacks Jan 05 '12 at 15:04
  • @ChrisCudmore The deer hunter ^_^ – Gabriel Jan 05 '12 at 17:25
  • 1
    There's also a slight advantage to go as late as possible due to the small but real possibility that the gun will jam at at any given round. – Zarrax Jan 06 '12 at 02:57
  • @DilipSarwate: I'm reminded of COD Black Ops – qwertymk Jan 06 '12 at 03:40
  • @Zarrax, agreed. Always procrastinate. Who knows what win-win deal you (or your captor, or your opponent) may come up with to replace the decidedly zero-sum game of RR. And you'd delay if you thought the "ref" may have "loaded the deck" in favor of death (filled all chambers). And the moment of intense focus everyone has on the gun during the first "play" may give you enough time to catch someone off guard and get out of it altogether. – hobs Jan 06 '12 at 11:35
  • 1
    So you should not bring a knife to a gunfight, but in some cases bringing an extra gun would seem beneficial. – Kjensen Jan 08 '12 at 14:50
  • 8
    Math saving lives and killing captors. Now that is practial mathematics. – Nils Munch Jan 09 '12 at 14:29

Your best bet is actually to go last, because this will either make no difference or decrease the probability of shooting yourself. Why? If there are $n$ people and $n$ divides $6$, then the probability of any one person taking the bullet is equal, since the probability distribution is assumed to be uniform over the number of times the trigger is pulled. If $n \le 6$ does not divide $6$, i.e. if $n=4$ or $5$, then the turns wrap around and begin with the first person again, so the first one or two people have to shoot again; naturally, this doubles their probability of being shot. And finally, if $n>6$ then the bullets will have run out before they come to you if you go last.

Clive Newstead
  • 62,292
  • 4
  • 91
  • 158
  • 9
    Reminds me of the baseball manager's strategy of putting the best hitters first in the lineup – JoelFan Jan 04 '12 at 15:46
  • 10
    Of course, once the first shot has fired, the probability changes given those results. Another advantage of going last is you can run away after there have been 4/5 shots fired and the game has gotten too risky for your enjoyment. – user606723 Jan 04 '12 at 17:08
  • 103
    If you're playing a variant in which the game ends when someone dies, then you certainly *don't* want to go last... – slim Jan 04 '12 at 17:20
  • 14
    @slim LOL! That's like the joke about tennis: Winning the last ball is a guarantee to win the game. – Tormod Jan 04 '12 at 17:27

This reminds me of a similar interview question; you see your interviewer load two bullets into consecutive chambers of a gun, point at his head and pulls the trigger, but survives as it was an empty chamber. He then gives you the gun to fire at your head, giving you the option to either spin the barrel before shooting or to just take the next shot.

You then decide not to spin the barrel; as with spinning you have a 2/6 ~ 33$\frac{1}{3}$% chance of killing yourself. Without spinning and knowing that the last shot was nonlethal you leave only 5 chambers; two of which have bullets; however knowing the bullets are in consecutive chambers, you are only afraid of running into the group of bullets which appears in only 1 of 4 spots (of consecutive bullets) so there's only a 1/4 ~ 25% chance of dying.

dr jimbob
  • 963
  • 5
  • 14
  • 40
    Inspired by @Chris Cudmore's problem solving techniques, I'd empty the whole gun at the interviewer and run like hell. – Gunnar Þór Magnússon Jan 04 '12 at 22:26
  • 8
    Depending on the gun type, if it's well maintained and oiled, spinning it (and letting it roll out) will ensure that the bullets will end up far from the barrel because of their weight. – vsz Jan 05 '12 at 07:35
  • 8
    And all this goes wrong when the 2/6 chance hits the interviewer first, in which case you also still need a job… – poke Jan 05 '12 at 09:54
  • 1
    The postscript to the question was that the gun was a fake (couldn't fire anything) and that it was additionally testing (a) your attention to detail (noticing bullets were in consecutive chambers), (b) your desire to have the job (you supposedly would have the job if you pass this one last test), and (c) the level of hostile working conditions you'd be fine with (anyone who would point a gun at their head will have no problem with 80+ hour work weeks). – dr jimbob Jan 05 '12 at 14:40
  • This might be a non-intuitive question, but I don't follow the reasoning here: the interviewer has just fired a blank (at a 2 in 6 chance), so now there are 5 chambers left with 2 guaranteed to be loaded: a 2 in 5 chance. If you spin the barrel, you improve your chances back to 2 in 6, since you add the blank chamber back into the draw. – Cygon Jan 05 '12 at 15:07
  • 4
    @Cygon: the point is that the gun fires chambers consecutively, so it is not just 2 in 5: since the interviewer is still alive, he fired one of the 4 empty chambers. Only one of those 4 have a bullet in the "next" chamber (since the bullets are placed in consecutive chambers initially). So the chance you hit the jackpot is actually 1 in 4, not 2 in 5, which is certainly better than 1 in 3 (or 2 in 6) if you spin the barrel. – Willie Wong Jan 05 '12 at 15:19
  • 14
    @Cygon: maybe the following will help: since the interviewer is still alive and the bullets were loaded in consecutive chambers, it is actually *impossible* for you to die from shooting the "second" of the two bullets. So you only have to worry about the "first" bullet, and not both. – Willie Wong Jan 05 '12 at 15:21
  • 1
    Well, this is a fun post, but not an answer to the question, rather an elaborate comment... – Konerak Jan 06 '12 at 10:35
  • @WillieWong You just summed it up perfectly. – Dunhamzzz Jan 06 '12 at 15:34
  • 3
    @Konerak: true. But it may be better for this post to stand alone rather than to have the entire thread merged into the comment section of the main question, which already has enough comments as it is. – Willie Wong Jan 06 '12 at 15:47
  • @Konerak - I also agree. I originally was writing a comment but couldn't fit it in the length limitations and rather than not post it; left it as answer. I'm quite surprised at the many +1s (though that seems that people were interested in seeing this; even if it was more of an aside than an answer to the asked question). – dr jimbob Jan 06 '12 at 16:09

They are all the same, because you only shuffle once.

The order of blank chambers and bullet are determined in the very beginning of the game.

Prob(Player A got shot) is just the probability of that bullet lies on the chamber player A will shot, for every one it is (number of bullets)$/$(number of chambers)

  • 241
  • 2
  • 7

For 2 players and n chambers player 1 loses if the round is in the odd-numbered chamber. Hence if $n$ is even, chances are $50/50$ and if n is odd, chances of losing are $(\frac{n+1}{2n},\frac{n-1}{2n}$) - it's a disadvantage to go first.

  • 3,955
  • 1
  • 9
  • 22