# Russian Roulette OddsJanuary 3, 2008 10:04 PM   Subscribe

Please settle an ill-informed bar argument re. Russian Roulette.

Assuming that there are 6 "players" and one bullet in a six-chambered revolver, and the game is over when one of the six participants is toast:

Would there be any difference in the ultimate statistical odds of surviving Russian Roulette if you were to re-spin the chamber before handing the gun to the next player (as opposed to one bullet, six trigger pulls [or less], game over)?

And, if there are any differing odds of winning or losing based on each of the six players' positions in the circle, what are they and how/would they change based on a re-spinning of the bullets after every turn?

I did google this from a phone at the bar, and again when I got home, but I scored poorly in stats class for a reason and am hoping for the layman-est explanation possible.

Thanks!
posted by macrowave to Sports, Hobbies, & Recreation (25 answers total) 6 users marked this as a favorite

I'm no stats nerd, but if you don't respin the cylinder the odds increase with each trigger pull.
posted by rhizome at 10:08 PM on January 3, 2008

Yes, the odds are different.

If you don't spin the chamber, there is a 100% chance that someone is toast at the end of the first round.

Assume 6 players and a six-shot revolver.

A has a 1 in 6 chance of eating lead; if A survives,
B has a 1 in 5 chance of death; if B survives,
C has a 1 in 4 chance of death; if C survives,
D has a 1 in 3 chance of death; if D survives,
E has a 1 in 2 chance of death; if E survives,

If you re-spin before every trigger pull, each player only has a 1 in 6 chance of getting the bullet, or a 5 in six chance of not getting the bullet. Thus, if you re-spin, it is possible to go all the way through a cycle with no one dying. If you do not re-spin, someone is dead (or at least has a bullet in their head) at the end of the first round.
posted by jewishbuddha at 10:12 PM on January 3, 2008 [2 favorites]

Response by poster: But are the odds 1/6 no matter what position you're in?
posted by macrowave at 10:13 PM on January 3, 2008

Response by poster: jewishbuddha-

I was typing my question as you responded.

But still, overall, are the odds going into a "straight" game not 1/6?
posted by macrowave at 10:16 PM on January 3, 2008

They're going to keep playing till someone dies, right? Then it's a 1/6 chance of dying per player either way; it's just that if you don't spin the chamber, player #6 might get to watch his odds of dying go 1/6, "click", 1/3, "click", 1/2, "click", 2/3, "click", 5/6, "click", 1, "BLAMMO". If you do spin the cylinder, there might be more than one round, and it's becomes a flat 1/6 chance each time the trigger is pulled, but it's still even odds which player gets it.
posted by nicwolff at 10:16 PM on January 3, 2008

(Of course, if you don't spin, you're going to have a hard time keeping player #6 in the game...)
posted by nicwolff at 10:18 PM on January 3, 2008

macrowave, yes it's still 1/6 with no spinning after the game starts; you get more information as the gun goes around, but as long as no-one is allowed to quit the information doesn't change who was already doomed.
posted by nicwolff at 10:20 PM on January 3, 2008

If you dont spin, then the odds are equal (1/6) for everybody.
The easiest way to see this is that everyone has a chamber pre-assigned. And there is a bullet in one of them. Who goes first or whatever is just a ritual.

If you do spin, the odds are better for you if you are the 6th player.
The easiest way to see this is to imagine an extreme case: There is a line of 1000 men. One gun. six chambers. One bullet. The game ends when someone dies. Your odds as the 100th man look pretty good, I'd say. As the first its 1/6th.
posted by vacapinta at 10:22 PM on January 3, 2008

You have better odds of surviving any individual turn if you re-spin, a la the Monty Hall problem. (Except for the first turn, where nobody has gone before you.) But with re-spinning, you may be forced to take more than one turn, so it ends up evening out, kind of.

With no re-spins: the initial spin determines who gets killed, and everybody has a 1/6 chance of getting it.

With re-spins: your odds depend on where you are in the order of players. For example, player A has a 1/6 chance of dying on his first turn. Player B has a 5/36 chance of dying on his first turn: B only dies if A survives (5/6) and the re-spin gives B the bullet (1/6). Player C has a 5/6 * 5/6 * 1/6 chance of dying on his first turn. And so on, all the way through the first round and back to player A again, etc.

I'm too lazy right now to work out the actual probabilities for each position in the order, but if the players choose their order randomly, then it doesn't matter, thanks to the linearity of expectation; everyone has a 1/6 chance.

on preview: crap, I took too long...
posted by equalpants at 10:25 PM on January 3, 2008

If you re-spin after every squeeze, every player has a 1 in 6 chance of getting shot (assuming a 6-chamber revolver), every time. The game could, theoretically, continue forever without anybody getting shot.

If there's no re-spin, somebody is guaranteed to get shot within six squeezes.

The first player has a 1 in 6 chance of getting shot.

If the first player doesn't get shot, the second player has a 1 in 5 chance of getting shot, since there are now only 5 possible chambers for the bullet to be in instead of six.

If neither of the first two players get shot, the third player has a 1 in 4 chance of getting shot.

Fourth player: 1 in 3 chance.

Fifth player: 1 in 2 (50/50) chance.

Sixth player: certain to be shot.

However, these odds are those available during game play. Considering the no-respin game as a whole: it's a deal, made amongst six people, that one of them is definitely going to get shot. The chances of any particular player being the one who gets shot, at the outset of the game, are the same: 1 in 6.
posted by flabdablet at 10:28 PM on January 3, 2008

Oh, duh, right: if you spin, then player 2 has a 1/6 chance of dying on his first pull, but only a 5/6 chance of having to play at all since player 1 may have ended the game by dying. Player 3 has a 1/6 chance of dying if he has to play, but only a 25/36 chance of having to play. And so on around and potentially around, but always in favor of the later players. vacapinta for the win.
posted by nicwolff at 10:34 PM on January 3, 2008

Best answer: Interestingly, only the spinning version is fair. If you don't spin, the later you go the better your odds are.

With no spin:

A has a 1/6 chance of dying on the first shot
B has a 5/6 chance of A not dying, then a 1/5 chance of dying on the 2nd shot
C has a 5/6 chance of A not dying, then a 4/5 chance of B not dying, ...

A: 1/6 = 1/6
B: 5/6 * 1/5 = 1/6
C: 5/6 * 4/5 * 1/4 = 1/6
D: 5/6 * 4/5 * 3/4 * 1/3 = 1/6
E: 5/6 * 4/5 * 3/4 * 2/3 * 1/2 = 1/6
F: 5/6 * 4/5 * 3/4 * 2/3 * 1/2 * 1 = 1/6

With spin:

Every shot fired has a 1/6 chance of killing someone.

A's chance of dying is 1/6 for the first shot + (5/6)^6*1/6 for the second shot (since six shots have to miss to get to A's second shot) + (5/6)^12*1/6 for the third shot + ...

A: Sum of (1/6)*(5/6)^(6*round number) - about .25

B's chance of dying is 5/6*1/6 for the first shot + (5/6)^7*1/6 for the second shot (seven shots have to miss to get to B's second shot) + ...

B: Sum of (1/6)*(5/6)^(6*round number+1) - about .20
C: Sum of (1/6)*(5/6)^(6*round number+2) - about .17
D: Sum of (1/6)*(5/6)^(6*round number+3) - about .14
E: Sum of (1/6)*(5/6)^(6*round number+4) - about .11
F: Sum of (1/6)*(5/6)^(6*round number+5) - about .09
posted by Diz at 10:35 PM on January 3, 2008

And by spinning I mean no spinning
posted by Diz at 10:35 PM on January 3, 2008 [1 favorite]

...So I was too lazy to work out the math, but I ran a simulation. Here are the odds of death for each position in the order, from 10 million games:

0.2506
0.2088
0.1740
0.1450
0.1207
0.1008

...on preview, this agrees with Diz's computation.
posted by equalpants at 10:40 PM on January 3, 2008 [1 favorite]

Slight derail but,
I remember reading somewhere, perhaps in a novel, so this info may be dubious, if your 6-gun is well maintained (clean and well oiled), and there is just 1 bullet in 1 of the 6 chambers, your chances of being shot are much less than 1 in 6, because the weight of the bullet tends to force it to end up low, away from the barrel and out of firing position.
So if you're given choice, respin!
posted by baserunner73 at 11:02 PM on January 3, 2008 [2 favorites]

yeah, it is interesting that the non-spinning version is fair.

makes sense though, cuz the advantage the late-players get going later is perfectly offset by the increased probability of getting the bullet.

The UD stats class I took in school was, easily, the most useful class in my 7 years there.
posted by panamax at 11:48 PM on January 3, 2008

Also, those probabilities so carefully and correctly worked out by Diz are the a-priori probabilities for the with-spin game: they're only correct before the game starts.

Once the game is actually in progress, the player who has just been passed the gun has the same chance of being shot as Diz worked out for player A (ignoring the chamber-weighting thing that baserunner73 mentioned).
posted by flabdablet at 2:14 AM on January 4, 2008

yeah, it is interesting that the non-spinning version is fair.

One way of explaining why the non-spinning version is fair is that the initial spin assigns each player one of the six chambers. Each player dies if and only if their assigned chamber has the bullet in it. Each player has exactly a 1 in 6 chance of that happening at the time that the initial spin is made.

The reason why this logic doesn't apply to the spinning version is that the game stops when one player is dead, so that even if player 6 was going to die based on his random spin, any of the previous players might stop the game ahead of time. Remember that with the non-spinning version, its impossible for one of the previous players to stop the game when player 6 will die, because the bullet cannot be in two chambers at once.

One way to make the spinning version fair would be to only play one round, and not stop if someone gets killed. Each player would have an independent, 1 in 6 chance of dying, and on average only one person would die. There would be about a 1 in 3 chance that no one would die, and a very unlikely but still possible chance that all 6 would die at the same time. Although it seems like nobody would keep playing after someone died, it would be at least a little more realistic than the non-spin version, where player 6 would need to pull the trigger knowing that he has a 100% chance of dying. A way to keep the same odds and remove the chicken-out factor would be to use 6 revolvers and six bullets, and have everyone pull the trigger at the same time.
posted by burnmp3s at 3:59 AM on January 4, 2008

I read the same book as baserunner73. It was one of the Jack Reacher novels by Lee Child. Reacher walked away intact after spinning the cylinder each time before he pulled the trigger. I remember thinking, "I should make note of this if I ever find myself playing Russian Roulette. Always spin the cylinder."
posted by fiery.hogue at 5:29 AM on January 4, 2008

When confronted with these kinds of problems, try to reduce it to a weird small case and see if it still makes sense.

Would there be any difference in the ultimate statistical odds of surviving Russian Roulette if you were to re-spin the chamber before handing the gun to the next player (as opposed to one bullet, six trigger pulls [or less], game over)?

Think about a revolver with exactly two cylinders. If the first person didn't die, and you're next, you had better spin it, right?
posted by cmiller at 6:29 AM on January 4, 2008

camden: True, and I think a Christopher Walken skit about a Math teacher presenting this problem to a class of sixth graders could work very well.
posted by Coventry at 7:07 AM on January 4, 2008

Best answer: Exact probabilities for the spinning version are:
Player 1: 7776/31031
Player 2: 6480/31031
Player 3: 5400/31031
Player 4: 4500/31031
Player 5: 3750/31031
Player 6: 3125/31031

Simple proof: if player 1 isn't shot on the first pull, which happens with 5/6 probability, then player 2's probability of dying has to be the same as what player 1's was before pulling the trigger. Thus, player 2's probability of dying is 5/6 of player 1's, and likewise, player 3's probablility of dying is 5/6 of player 2's, and so forth. Those, combined with the fact that exactly one person must die (i.e., the probabilities for all players must sum to 1) gives the above figures. The numbers also match well with equalpant's simulation.
posted by DevilsAdvocate at 7:29 AM on January 4, 2008

As an interesting aside (and more of a visual illustration of the increased odds against when you don't respin), you might want to watch Tzameti 13, not your typical American feel-good movie.
posted by subajestad at 9:20 AM on January 4, 2008

re: the Lee Child novel referenced by fiery.hogue and baserunner73, Reacher plays Russian roulette in Persuader. Better make sure that the gun is a) well-maintained and well-oiled and b) the load is heavy (i.e., don't try this with a .22 round). the Reacher novels are excellent candy, if you're into that sort of thing.
posted by rtha at 9:54 AM on January 4, 2008

I do remember reading a story in which a suicidally depressed character played Russian roulette with a well oiled revolver, but instead of realising that the heavy bullet was pulling the loaded chamber to the bottom, believed that fate or divine intervention was responsible and cheered up immensely, concluding he had an important task in life (shortly before being eaten by wolves or something).
posted by tomble at 5:40 PM on January 6, 2008

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