Talk:Russian roulette

Turkish roulette

The line "This version is highly fatal since the probability of death is 5 in 6." should probably refer to the probability of being shot, rather than the probability of death. A few people survive being shot in the head.

--Ilnyckyj (talk) 22:37, 30 January 2009 (UTC)

"World War I"?

The following appears deeply suspect to me. As I understand it, World War I was not called "World War I" until some time after the start of World War II in 1939. Maybe the date is wrong?

The earliest known use of the term is from "Russian Roulette," a short story by Georges Surdez in the January 30, 1937, issue of Collier's Magazine. A Russian sergeant in the French Foreign Legion asks the narrator,

“ "Feldheim… did you ever hear of Russian Roulette?" When I said I had not, he told me all about it. When he was with the Russian army in Romania, around 1917 during the Romanian Campaign of World War I.... ”

98.204.24.131 (talk) 09:07, 16 April 2008 (UTC)

I totally noticed the same thing and removed the paragraph in question from the article. It is presented here for posterity:

The earliest known use of the term is from "Russian Roulette," a short story by Georges Surdez in the January 30, 1937, issue of Collier's Magazine. A Russian sergeant in the French Foreign Legion asks the narrator,

“ "Feldheim… did you ever hear of Russian Roulette?" When I said I had not, he told me all about it. When he was with the Russian army in Romania, around 1917 during the Romanian Campaign of World War I, and things were cracking up, so that their officers felt that they were not only losing prestige, money, family, and country, but were being also dishonoured before their colleagues of the Allied armies, some officer would suddenly pull out his revolver, anywhere, at the table, in a cafe, at a gathering of friends, remove a cartridge from the cylinder, spin the cylinder, snap it back in place, put it to his head, and pull the trigger. There were five chances to one that the hammer would set off a live cartridge and blow his brains all over the place. Sometimes it happened, sometimes not. ”

199.181.134.212 (talk) 21:24, 28 April 2008 (UTC)

Re: the question about blank rounds -

where does the lethal remark come from?

In 1984, Jon-Erik Hexum, a television actor, shot himself in the head with a blank round and died. See 1984 in television.

---

I have heard the game this way also: American roulette is placing one round in a revolver, but Russian roulette means taking one round out of the revolver :). Joakim 19:57, 2 Jul 2004 (UTC)

Started by White Army soldiers (Russian civil war)

To escape boredom... that and to replace the feeling of losing ones nation.

-G

There needs to be a source. Seems unverifiable anyhow. --Edwin Herdman 08:33, 26 March 2007 (UTC)

Who the heck is Valerie Douglas?

Is this supposed to be a joke: "According to one website claiming to offer insight into the practice of Russian roulette, Valerie Douglas, whose father's cousin and father were in the Vietnam War states that Russian roulette occurred both for gambling and murder."

My father's third cousin's grade school teacher's brother's stepfater was over there and he never saw anything like that.69.19.14.33 (talk) 19:15, 27 January 2008 (UTC)Me

Yeah the link is now dead which is a real shame as I have heard it said that russian roulette never occured in vietnam (that we know of: comments made in relation to The Deer Hunter) so any evidence to the contrary would be most interesting but this evidence seems now gone: if anyone knows of a new link or an achive of this one it would be much appreciated I am sure!... 122.148.173.37 (talk) 14:14, 6 August 2008 (UTC)

Video Games

It is actually Boss who diverts Ocelot's shot and takes out Snake's eye, not Snake himself. I fixed this. --NLUT

Odds

What about the odds? I think each round has the same odds of 1/6; odds are not modified by successive actions of something. Just as each time you flip a coin there's a 1/2 chance it will end up heads (see Rosencrantz and Guildenstern Are Dead) --Xinoph 17:47, Sep 12, 2004 (UTC)

I think it depends what the odds are supposed to mean. In advance, there is a 1/6 chance of each chamber having the bullet. But once after the first shot, for instance, we know the bullet isn't in the first chamber (unless the gun goes off), so there is a 1/5 chance of a shot being fired on the 2nd shot, assuming the first shot isn't 'successful'. Alternatively, if the revolver is spun before each shot, it is always 1/6 anyway.

I think the odds section should be re-written anyway, as it isn't clear what information it is supposed to provide.

This sounds like a misinterpretation of the Monty Hall problem. As Xinoph wrote, 1/6 remains constant if no spinning of the cylinder is performed. Only 1 initial selection was made, rotations of the cylinder (due to firing) do not change the initial selection, and thus successive trigger-pulls do not increase the chance of the next chamber containing the bullet.

Errr? It all depends on whether the cylinder is spun after every round or not. I just rewrote the odds section to be clearer, but if the cylinder IS spun, the odds remain constant (whether they're 1/6 or slightly less due to bullet weight). If it's not spun, the odds of losing increase on each trigger pull.

What I didn't understand was the part about the odds being 1/2 for each side. Surely $1/6+1/4+1/2\neq 1/5+1/3+1$? I changed this to say that the early players are favored, until someone can explain it... -- nae'blis _(talk) 21:24, 24 February 2006 (UTC)

The 1/6th is just plain wrong. The bullet has a certain weight associated with it. It will have a (very) strong tendency to end up at the BOTTOM when the cylinder is spun. You can prove this easily from basic mechanics. This completely changes the odds for the cylinder being re-spun every time. The fact no one is mentioning this is slightly worrying. —Preceding unsigned comment added by 76.19.65.229 (talk) 04:40, 25 October 2008 (UTC)

Rewritten on April 30, 2006 for odds (1/6 every time except the 6th trigger pull with NO RESPIN of the cylinder). Please try the simulation [1] if you're still confused. This simulation plays a large number of games (you can set how many), and tells you which trigger pull generated which percent losses. I admit, this is not intuitive (heck, I had to write a program to convince myself!), but understanding Monty Hall and similar statistical mind-twisters helps to clarify this.

Actually, Monty hall dictates that the odds on each trigger pull without re-spin (assuming you were calculating the odds just before you pull the trigger, rather than at the beginning of the game) would be 1/6, 1/3, 1/2, 2/3, 5/6 and 1. --67.193.163.65 (talk) 04:32, 13 January 2009 (UTC)

i remember reading an article somewhere which indicated that the odds are not as simple as 1:6, rather the chances are less than that of actually getting shot, this is due to the weight of the bullet in the cylinder and the effect of gravity which makes it more likely that the bullet will end up at the bottom of the cylinder when it finishes spinning. Philbentley 02:58, 26 June 2006 (UTC)

I rewrote this section to be much more general, accurate and to have a mathimatical basis. However, I'm not very good at doing maths formulas in wiki yet so the section looks fairly ugly. --WikiWizard 12:24, 28 September 2006 (UTC)

Besides the "spin between turns" issue, there's also the question whether we're talking the odds of your survival on your turn of the game, versus odds of your survival over the game as a whole. If the latter, the trick here is that the game stops with the first loser... so the more players ahead of you in line increases your chances that they'll be a loser before you. JeramieHicks 12:25, 24 November 2006 (UTC)

The odds are 1/6 for each contestant no matter how you do it (if gravity is ignored). Once you spin the cylinder the chamber is chosen and so is the number of the killing shot. You can also look at it this way. Even if you are allowed to pick what chamber to use (without reusing a chamber) your odds are still 1/6. Your chance of survival is not just your chance with your shot, but also teh chance that someone else dies before you. That is person one has 1/6 while person 2 has 1/5*5/6 = 1/6. 1/5 for one chamber used and only five left and 1/6 for the chance you never get to shoot because person one dies. It will always be 1/6.

Subdivision

I've noticed that the Legends section includes things that aren't really legends. In particular, the Deer Hunter reference would probably be under Popular Culture or something like that. Similarly, the Derren Brown bit isn't a legend at all, as it actually happened, and the semi-automatic pistol reference is presumedly just true, and might fit in the intro.

I'm not sure how to divide things, as the things I've talked about aren't really in the same category, IMO, but if not then it would seem a bit odd to have just one item under a heading.

A question about how the game would work as a form of gambling

I saw that Toy Gun version, but that isn't how it would work like it did in the Deer Hunter. Obviously there would have to be a monetary incentive for the players to play, unless they were being forced. How does the "house" then make money? Do they take a percentage of all bets made? Or is there more of an elaborate odds structure, like race track betting? If anyone has some insight, it seems facinating, but I don't see how everything fits together. MicahMN | Talk 16:21, 7 Feb 2005 (UTC)

Some poker rooms rent tables per half hour. I'd imagine that toy gun rental in such a manner would work as well. --Damian Yerrick 00:59, 15 October 2005 (UTC)

Russian roulette with Soviet handguns

I heard a rumor once that Soviet handguns were designed so that the weight of the loaded chambers would cause the empty chamber to be the first one that was fired. Anyone know if that is true?

I don't know if that's true. I removed this rumor from the article, because it was stated as a fact. If someone can verify whether this is a fact or not, we should mention it in the article. Just going by common sense, I doubt that every Soviet revolver had this characteristic. Rhobite 02:03, Apr 22, 2005 (UTC)

As said, Nagant revolver isn't suited for Russian Roulette. However, I've heard that if the revolvers' design allows the cylinders to roll freely (true for most other revolvers), weight of the single round indeed tends to leave the loaded chamber in the bottom, IF the cylinder is allowed to roll unrestricted. No personal observations whether this is true, though. Btw, Russian Army did use other revolver designs before Nagant, which were more conventional.--Mikoyan21 01:56, 8 March 2006 (UTC)

I just made a correction. The Nagant's cylinder DOES INDEED roll freely until the hammer is engaged. I've owned several, and they all have this characteristic. It does not swing out, however. These characteristics would actually make it more likely to be suitable for roulette, since unlike a hand-ejector type DA revolver you cannot easily see where the cartridges are as the cylinder spins. The long brass and deeply sunk bullet of the uique cartridge makes it more difficult to see which chamber is hot. However, the Nagant is NOT a six-gun. This makes me think the Collier's reference was made up by an American writer.

(This next comment is by a different guy)

Before the Nagant was made standard issue, Russia used Schofield-type S&Ws in the now obsolete .44 Russian. Those were sixguns.

[This next one by an old Rhodesian. Remember the World forced us to hand over to Robert Mugabe?] —Preceding unsigned comment added by 196.207.33.197 (talk) 04:53, 29 May 2008 (UTC)

I never met an armed Russian, but in the seventies I owned a Taurus .38 Special -- a six shot, Brazilian revolver, the accuracy of which continually surprized me. I quickly found out that the cylinder was so finely balanced that a single live round in the cylinder would almost always cause the cylinder to stop with the round at the bottom. This allows at least two and usually three "safe" shots. —Preceding unsigned comment added by 196.207.33.197 (talk) 04:49, 29 May 2008 (UTC)

Revolver Ocelot

I added his version of the game. I have bad spelling so you may find it neccessary to clean it up a little. MegaloManiac 18:43, 1 May 2006 (UTC)

Another Film

Dear Lord, I am so upset that I forgot the name of this movie. But it's about this crazed landlord who pretty much admits only young women to his complex and spys on them. But I digress, at the start of each day he take a gun, spins the chamber, places the barrel in his mouth and pulls the trigger. After he survives, he says "So be it." It was decent and I saw it on IFC. Please someone must have the title...! Yanksox 15:21, 30 May 2006 (UTC)

No idea which is that movie. Although I agree to maintain short the list of cultural references on Russian Roulette, I would suggest Intact ([2], as odds and Russian roulette are integral to the movie. By the way, this is my first collaboration in Wikipedia ;-)

Rushing Roullette

Copied from Wikipedia:Reference desk/Mathematics for May 26, 2006. --Lambiam^Talk 18:43, 30 May 2006 (UTC)

What are the that someone would lose Rushing Roullette if you played 2 times in a row? What about 3,4,5 times in a row? More importantly, what equations would one use to solve this?199.201.168.100 15:53, 26 May 2006 (UTC)

Do you mean Russian roulette ? First, you would need to know how many chambers the revolver had and how many bullets were loaded. If there was 1 bullet and 6 chambers, then the odds of dying on the first shot would be 1/6 (assumming a 100% death rate if the bullet is in the chamber you try to fire). If the player survives and the barrel is spun after each trigger pull, then the odds would be the same (1/6) every time after, as well. Otherwise, if the player survives and barrel is NOT spun, the revolver advances to the next chamber, the odds would be 1/5, then 1/4, 1/3, 1/2, 1/1 (this means, if the first 5 players survived, the 6th would be guaranteed to die). Of course, the chances someone will die before it gets to be your turn go up the later your turn is. If you had the second turn, your chances of having to play (the chance the first player survived) would be 1 - 1/6 or 5/6. This, multiplied by the 1/5 chance of dying if you have to pull the trigger, gives you (5/6)(1/5) or 1/6 chance of dying, which is the same as the first player. In fact, the odds of dying are the same 1/6 for all six trigger pulls if the barrel is not spun between them. Thus, if you plan to fire a maximum of twice (it doesn't matter if it's in a row or not), your chance of dying is 2(1/6) or 2/6, three times is 3/6, four times is 4/6, five times is 5/6 and if you fire six times in a row your chance of dying is 6/6, or 100% guaranteed. I don't suggest that you try to verify this at home. StuRat 16:55, 26 May 2006 (UTC)

If this is a single player, spinning the cylinder (or whatever it is that revolves on a revolver) each time, the odds of survival in the N+1^th round given survival up to and including the N^th round is 5/6. Because each round is independent, you find then survival odds of (5/6)^N for N rounds, which means fatality odds of 1 – (5/6)^N. For the first few values of N this amounts to:

N = 0: p_fatal = 0.00000

N = 1: p_fatal = 0.16667

N = 2: p_fatal = 0.30556

N = 3: p_fatal = 0.42130

N = 4: p_fatal = 0.51775

N = 5: p_fatal = 0.59812

N = 6: p_fatal = 0.66510

N = 7: p_fatal = 0.72092

N = 8: p_fatal = 0.76743

N = 9: p_fatal = 0.80619

Personally I'd call any player of Russian roulette a loser even before they begin to play. --Lambiam^Talk 17:38, 26 May 2006 (UTC)

Postscriptum. I just read the article Russian roulette, and my impression is that the section Odds is at odds with probability theory, or else so unclear as to be misleading. It should be rewritten, which ought to be a cinch for any probabilist out there (or should I say "in here"?). --Lambiam^Talk 17:45, 26 May 2006 (UTC)

What is the problem? Just as StuRat shows, it is correct under the assumption that the barrel is not spun after every shot, and it displays the probabilty that you will die on round n, given that you have made it there. -- Meni Rosenfeld (talk) 18:06, 26 May 2006 (UTC)

I'm not a gun expert, but isn't it the cylinder that is spun (or not)? The article states: "If the cylinder is spun after every shot, the odds of losing remain the same". Still correct? --Lambiam^Talk 23:23, 26 May 2006 (UTC)

I have no idea what is spun. And yes, if the whatever is spun after every shot, the probability of losing on round n, given that you have made it there, is 1/6 - The same as in the first round ("remain the same" does not mean "the same as in the previous case", but rather "the same as in the first round" - As opposed to the first case, where the probability increases at each round). -- Meni Rosenfeld (talk) 12:57, 27 May 2006 (UTC)

Right. You know what it is that it is the same as, I know what it is that it is the same as, but does the reader know? It could very easily be taken to mean: "the same as in the previous case". I think it is also confusing that the probabilities are conditional: "given that you survived this far". The statistically uneducated reader could easily understand the given table to imply that the odds of surviving 2 rounds are 1-1/5 = 4/5 instead of 4/6. --Lambiam^Talk 14:06, 27 May 2006 (UTC)

I agree that it could have been clearer and that a rewrite would be useful, but still, I think it's clear enough and that a rewrite is not necessary. -- Meni Rosenfeld (talk) 15:51, 27 May 2006 (UTC)

That reminds me, I must rent a copy of The Deerhunter and reminisce about these scenes. JackofOz 14:26, 27 May 2006 (UTC)

I agree that our article wasn't clear on whether they were giving the odds assuming you get to the second (and subsequent) trigger pull or the odds including the probability that the game may not progress that far. StuRat 02:26, 29 May 2006 (UTC)

Real life incidents

I notice that the vast majority of the incidents listed here are American teenagers. Is this a bias in the article or an accurate reflection of who most of the victims are? Tellkel 14:13, 30 June 2006 (UTC)

I'm removing the Darwin Award for 'Rasheed from Houston' incident. It's almost certainly apocryphal, because nobody is that stupid. It was put under the 'reality' heading, it's not real, therefore it atleast doesn't belong under that heading, but I didn't see where else to put it. Besides, The Darwin Award has about as much integrity as The Onion. --Stevekl 20:13, 26 July 2006 (UTC)

Massachusetts v. Atencio, 189 N.E.2d 233 (1963), Pennsylvania v. Malone, 47 A.2d 445 (1946), Washington v. Brubaker, 385 P.2d 318 (1963), Lewis v. Alabama 474 So. 2d 766 (1985).

Would the accidental death of Chicago guiarist Terry Kath be considered Russian roulette?

I couldn't find a reference for a Samantha Goodson. The only thing I found was one for a Nadera Goodson, whose account otherwise matches (Jamaica area of Queens, abandoned house, boyfriend shot, etc). Even then, the only record I could find was this: http://query.nytimes.com/gst/fullpage.html?sec=health&res=9B02E0D6113CF933A2575BC0A9629C8B63 -- I'm not going to update the article cuz I'm not really familiar with the whole event. JeramieHicks 12:29, 24 November 2006 (UTC)

fictional accounts

The film, entertainment, and television sections need to be drastically trimmed to a few notable examples. There's no point in trying to include every time anyone anywhere has depicted a game of Russian Roulette, and it just makes the article unbalanced toward the fictional, and harder to maintain. Anything kept should have a source, or at least an article on Wikipedia backing up the incident. -- nae'blis 20:23, 28 July 2006 (UTC)

   I handled it.

13 Tzemati

Somebody add "13 Tzemati" to this page...it is a movie related to russain roulette

And someone add that eppisode of 24 in season 3 where Jack Bauer and a prison gaurd are forced to play Russian Roulette during a prison riot.

Oh and there was a cut scene where the game was played in the video game Killer7 where the revolver used turned out to hold seven possible bullets.

How to, article

If this isnt a how to article on how to play russian roulette then I don't know what is. I think the Odd's part of the article should be cleaned up to make it less like an article explaining how to play russian roulette. It could possibly give the wrong impression to many. ETod09 06:50, 27 October 2006 (UTC)

Trimmed Deer Hunter summary

The article has an entire summary of The Deer Hunter, including spoilers/conclusions with no warnings tags. I've trimmed it to just the beginning part and am posting the remainder below.

Michael (De Niro) decides to take a risk by demanding that he and Nick (Walken) play with three bullets in the chamber. Their captors agree, believing them to be crazed and desiring to end the game sooner than later. Both convince their captors of their sincerity by taking a shot, and in a one in four chance, both survive. The two POWs now possess six bullets between them, and open fire on their captors, leading to their escape. The three men are forever changed by the experience: Michael (DeNiro), the strongest of the three, returns to his hometown but is plagued by a sense of shame and is no longer able to enjoy the thrill of deer hunting. Steve(Savage) retreats to a psychiatric hospital for veterans. Michael sets himself to finding his one missing mate, Nick. Michael travels to the slums of Saigon, where he finds that Nick has become a heroin user and voluntarily plays Russian roulette for the benefit of gamblers. Michael forces himself to the table across from Nick, and begs him to return to the U.S. and rejoin his friends and former life. Nick shows only the slightest glimmer of recognition of Michael and his life before the war, and insists on playing one more round. Michael shows his loyalty to his friend by participating in the round himself, hoping for a bloodless outcome that would allow him to rescue Nick. However, right before firing, Nick says to Michael "one shot", a reference to their goal of killing a deer with one shot, and confirming that he does recognize and remember Michael. Nick's barrel is loaded and his head is demolished.

--Mithunc 19:43, 30 January 2007 (UTC)

Deer Hunter

This article claims twice that in Deer Hunter the actors paly the variety of russian roulette where one points the gun at another's head. I just finished watching the film, and this just isn't true.

Parachuting

Is there a source for this? Right now it just looks like more clutter. Surely we can come up with all sorts of gruesome variations on the theme, but that isn't the purpose of the article. --Edwin Herdman 08:33, 26 March 2007 (UTC)

Odds

When you spin the chamber once, fire once, then pass the gun on without being spun again, the chances are significantly different. The first guy has 1/6, but the second player has less than 1/5. This is because the bullet placement has already been decided by the second turn, so it becomes a 1/5 chance that the chamber will hold a bullet, but also on top of that is another 1/5 chance that the bullet is already next in line. These odds get worse and worse as the game goes on. JayKeaton 08:56, 15 April 2007 (UTC)

You are confusing two situations: The odds of dying at the beginning of the game (1:6 for everyone) versus the odds of dying after one person has pulled the trigger and not died (1:5 for everyone). In other words, it doesn't matter what order you are sitting in around the table - your odds of dying are exactly the same. However, after one person has pulled the trigger and survived, your odds get worse - but it still doesn't depend on where you are sitting. SteveBaker 15:47, 26 September 2007 (UTC)

Mathematics

Would it be a good idea to write more on the subject of mathematics in the game? As in the chance one has of winning if, say, you are the third player of seven in a game where the gun is passed around with the barrel being spun between every pull of the trigger? What if you are the second player of three and the game is instead being played without the barrel being spun? Point is, I think I can write at least a fair bit on the subject. As I'm new to Wikipedia, I just wanted to know whether moving the mathematics from this article into another article and expanding on it is a good idea or not. Karl-kjeks-Erik 22:38, 24 April 2007 (UTC)

Boogie Nights

The movies section can boogie nights be mentioned as there is a scene..82.24.175.199 14:01, 16 June 2007 (UTC)

Racial overtone

I deleted the line, "This is...known as Polish Roulette" for its obvious racially charged overtone. The "Darwin Award" (hardly encyclopedic in its own right) is given to obviously stupid persons. The following association to Polish persons is unnecessary and materially harmful in that it propagates a ridiculous stereotype. 66.35.34.85 16:45, 14 August 2007 (UTC)Pioneerman

But you're point rests on maintaining that the Poles are a "race". The English, the French, the Germans...are they "races"? Certainly, the Poles are a nation, as are the others I mentioned. —Preceding unsigned comment added by 204.111.112.202 (talk) 01:41, 1 September 2009 (UTC)

Removed mathematics text

I've removed the following:

The terminology for this section:

Player: One participant in the game

P1, P2 ... Pn: Player 1 to Player n respectively

T: The total number of players in the game.

B: The number of bullets in the gun

C: The number of chambers in the gun

Round: A round occurs when a player takes one shot at his head with the gun. For example, the normal game with B = 1 and C = 6 and the cylinder isn't being spun would have a maximum of 6 rounds. Without re-spinning the chamber, subsequent players have an arithmetic disadvantage, as their odds decline from (1/5 to 1/4 to 1/3 to 1/2 to finally 1/1. It is assumed that P1 goes first, then P2 and so on.

R1, R2 ... Rn: Rounds 1 to n respectively

Winning a round: The loaded chamber is not aligned with the barrel; weapon is not discharged.

Losing a round: The loaded chamber is discharged.

The game stops on the first losing round.

A player Pn dies if Rx results in a death and ${(x-n)}\mod {T}$. For example, if there are 2 players (T = 2) then player 1 dies if round 13 is a death : ${(13-1)}\mod {2}$. Put another way, Pn loses if any of the rounds n, n + T, n + 2T... results in a loss (these can be represented by the formula $n+xT$ where x is a positive integer or 0).

The most common Russian roulette game has T = 2; B = 1; C = 6; P1 loses on rounds 1, 3, 5 and P2 loses on rounds 2, 4, 6.

If the cylinder is spun after every shot, the odds of losing a round is ${\frac {B}{C}}$. Alternatively, the odds of winning a round is $1-{\frac {B}{C}}$. However, the odds of making it to round n drop as n gets larger. This is because to make it to round n, rounds n-1, n-2... must have been won. So the odds for the game to stop on round n is $(1-{\frac {B}{C}})^{(n-1)}*{\frac {B}{C}}$. Then, the odds of Px to lose is $\sum _{k=0}^{n}(1-{\frac {B}{C}})^{(x+kT-1)}*{\frac {B}{C}}$ as n approaches infinity. This can be simplified to $A^{x-1}.{\frac {B}{C}}.{\frac {1}{(1-A^{T})}}$ where $A=1-{\frac {B}{C}}$.

For a standard game, P1 has a 6/11 chance of losing, while P2 has a 5/11 chance. Hence it is better to go last. Also, note the $A^{x-1}$ part of the equation. A is always less than 1, so as (x-1) increases, the chance to lose decreases. Hence it is always better to go last independent of number of players, and other parameters.

If the cylinder is not spun after each shot, the probability of losing a game can be determined by looking at each possibility of the bullet configuration in the gun. For example, in a standard game, if the bullet was in position 3, player 1 would lose. There are six possible positions for the bullet to be in a standard game: 1,2,3,4,5 or 6. Player 1 would lose if it is in position 1,3,5 (a 3/6 chance) and player 2 would lose if it is in position 2,4 or 6 (a 3/6 chance). Therefore both have an equal probability of losing (1/2).

Another example is with 6 players and 9 chambers with 1 bullet. There are seven possible positions for this game: 1,2,3,4,5,6,7,8 or 9. Player 1 would lose if it is in position 1,4,9 (a 3/9 chance), player 2: 2,5 (2/9), player 3: 3,6 (2/9), player 4: 4,7 (2/9), player 5: 5,8 (2/9), player 6: 6,9 (2/9) and player 7: 7 (1/9). In this case, it is much better to go last as compared to going first.

I don't think it's right. Drawing it in a tree, where the top branch each time indicates discharge and bottom indicates continuation to the next round:

 Round 1     Round 2     Round 3     Round 4     Round 5      Probability of discharge in each round
+-- 1/6 -->                                                 = 1/6
|
+-- 5/6 --> +-- 1/5 -->                                     = 5/6 * 1/5
            |
            +-- 4/5 --> +-- 1/4 -->                         = 5/6 * 4/5 * 1/4
                        |
                        +-- 3/4 --> +-- 1/3 -->             = 5/6 * 4/5 * 3/4 * 1/3
                                    |
                                    +-- 2/3 --> +-- 1/2 --> = 5/6 * 4/5 * 3/4 * 2/3 * 1/2
                                                |
                                                +-- 1/2 --> = 5/6 * 4/5 * 3/4 * 2/3 * 1/2 * 1/1

The probability of discharge in round 1 is 1/6. In round 2 it's 5/6 (probability of getting to this round) * 1/5 (probability of discharge), which again is 1/6. Likewise each of the others in 1/6. So for two players, the probability of discharge in rounds 1, 3 and 5 is the same as the probability in rounds 2, 4 and 6 -- there is the same risk for both players. --h2g2bob (talk) 00:46, 2 September 2007 (UTC)

I think you misunderstood what I wrote for that section. Your right that it is the same risk for both players when the chamber is not spun between rounds. This was in the original description as "For example, in a standard game, if the bullet was in position 3, player 1 would lose. There are six possible positions for the bullet to be in a standard game: 1,2,3,4,5 or 6. Player 1 would lose if it is in position 1,3,5 (a 3/6 chance) and player 2 would lose if it is in position 2,4 or 6 (a 3/6 chance). Therefore both have an equal probability of losing (1/2). ". If the chamber is spun between rounds however it is always better to go first (see the above description, it is fairly complicated). However I think that the original description was way too complicated. Maybe just put the odds for the no chamber spinning case, that is the most common? --WikiWizard 07:11, 27 October 2007 (UTC)

Stopping the chamber - evening the odds?

In a couple of places in the article, it says that stopping the chamber as it spins eliminates the effect of a heavy bullet tending to settle at the bottom of the gun - which would otherwise reduce the odds of the gun firing below 1:6.

I disagree.

When the cylinder is spinning, it'll slow down as the bullet is going upwards and speed up as it descends. So the cylinder is spinning faster when the bullet is at the bottom than when it is in line with the firing pin. If you stop the cylinder at a random point in time, then because it spends more time at the top of the spin than at the bottom, then the odds of the gun firing when you stop the cylinder spinning manually is actually MORE than 1:6.

SteveBaker 15:42, 26 September 2007 (UTC)

You are wrong, the cylinder is stopped while it is spinning quickly and any difference in the velocity during the rotation due to the rounds location relative to the ground is imperceptible to the person stopping the cylinder. Also you have absolutely no idea or data as to the relative weights of the round and the cylinder or weight distribution in the cylinder relative to the round (hint: the round is MUCH lighter than the solid metal of the cylinder and the cylinder's mass is not distributed evenly). Not only have you not taken these factors OR the variation in revolvers or their ammunition, but you have not quantified even one example. --Deon Steyn (talk) 13:33, 23 November 2007 (UTC)

Odds again

Excuse my boldness, but I think that the word "probablity" should be used where the article uses "odds". The probabilty for a discharge is 1/6, the odds are 1/5, see Odds. 18:20, 25 October 2007 (UTC) de:Benutzer:Marinebanker

Fixed. I also replaced that weird 1/3 1/2 2/3 5/6 sequence with the more intuitive and mathematically correct 1/5 1/4 1/3 1/2 sequence until someone can explain where the original one came from. Sam Hocevar (talk) 09:15, 22 January 2008 (UTC)

Notability of each case

What constitutes a notable incident? This 14-year-old girl unfortunately died today and there's the suspicion that she was playing Russian roulette. [3] I see that most incidents mentioned are of famous people, but there's also the case of Clinton Pope, who isn't famous. Artur (talk) 19:24, 25 December 2007 (UTC)

Drinking Games

The drinking games section has a description of "russian roulette" played with fireworks, but the description given doesn't sound anything like Russian Roulette, it is a game of Chicken instead. Does anyone else agree that it should be removed? amRadioHed (talk) 01:41, 18 January 2008 (UTC)

Variats?

Isn't there a version eyes closed white a knife stabing between your fingers and one with 6 jars with none containing what the label says (one of them poisin)? --Armanalp (talk) 18:53, 8 February 2008 (UTC)

A Hero of Our Time

There is a reference to Russian roulette in that story, but Serbian officer Vulich has nothing to do with that - he tried to shot himself to prove or refute the notion of inevitable fate to Pechorin. —Preceding unsigned comment added by 92.100.48.95 (talk) 19:39, 23 May 2008 (UTC)

This story is actually where the whole notion of "Russian roulette" comes from and I am shocked that I don't see it in the article. -- From a Russian literature major. —Preceding unsigned comment added by 24.19.0.210 (talk) 22:22, 7 December 2009 (UTC)

Why Russian Darts?

I'm having trouble understanding why Russian Darts is part of the article, since the only things it seems to have in common with Russian Roulette are the use of a gun and the word "Russian." I am also having trouble locating other references to this game. —Preceding unsigned comment added by 24.90.235.104 (talk) 18:36, 13 June 2008 (UTC)

Russian Roulette in Spawn comic

In issue 129 of the comic book “Spawn”, Twitch forces Spawn to play Russian Roulette with him (summary on the spawn website here [4]). Is this worth adding into the “in literature” section? Prophaniti (talk) 15:36, 28 September 2008 (UTC)

Odds, yet again

I think the first person to spin the revolver and play has much less than 1/6. Well, there's some gravity in the world, which means that most probably the bullet will be down, while the barrel is up.

G19C (talk) 10:09, 3 October 2008 (UTC)

Russian?

Most of the examples in the article seem to refer to the United States, not Russia. When was the revolver invented anyway? I don't think many Russians would have had access to one, except military officers of course.Steve Dufour (talk) 05:02, 5 March 2009 (UTC)

Polish Roulette?

"Another variation is polish roulette, in which a semi automatic handgun is used instead of a revolver. Mortality rate in this variant of the game is 100%."

Has it ever been verified that this variation has actually been played -- by Poles or anyone else? It bears an uncanny resemblance to a Polish joke I first heard in 1972. --Wwyzzard (talk) 12:37, 12 October 2009 (UTC)Wwyzzard

: Being a Pole whole my life I've read about 'Polish Roulette' for the first time ever here. Mchl (talk) 12:20, 12 January 2010 (UTC)

Odds (for the love of Pete!)

The odds (more accurately probability, but going with the discussion) are never 1/6 unless the gun is re-spun every time. If the gun is spun once (and assuming two alternating players), the odds are always 1/2. Once the spin is complete the lethal chamber is set, and once the order of play is decided the outcome is determined. If the bullet is in an "even" chamber, the "even" player will lose. Likewise for the odd. The number of turns is irrelevant.

--12.106.209.61 (talk) 18:28, 16 December 2009 (UTC)