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So I was too lazy to work out the math, but I ran a simulation. If neither of the first two players get shot, the third player has a 1 in 4 chance of getting shot. When confronted with these kinds of problems, try to reduce it to a weird small case and see if it still makes sense. I read the same book as baserunner It was one of the Jack Reacher novels by Lee Child. If you don't spin, the later you go the better your odds are. I remember thinking, "I should make note of this if I ever find myself playing Russian Roulette. Each player would have an independent, 1 in 6 chance of dying, and on average only one person would die. Yes, the odds are different. The chances of any particular player being the one who gets shot, at the outset of the game , are the same: 1 in 6. 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. Thus, if you re-spin, it is possible to go all the way through a cycle with no one dying. Fourth player: 1 in 3 chance. Here are the odds of death for each position in the order, from 10 million games: 0. If there's no re-spin, somebody is guaranteed to get shot within six squeezes. And there is a bullet in one of them. I'm no stats nerd, but if you don't respin the cylinder the odds increase with each trigger pull. Each player dies if and only if their assigned chamber has the bullet in it. Assume 6 players and a six-shot revolver. Your odds as the th man look pretty good, I'd say. Each player has exactly a 1 in 6 chance of that happening at the time that the initial spin is made. 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. 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. 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. Gifts for a loved one with an interest in This thread is closed to new comments. The game ends when someone dies. One gun. Interestingly, only the spinning version is fair. Reacher walked away intact after spinning the cylinder each time before he pulled the trigger. And so on, all the way through the first round and back to player A again, etc. 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. 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. 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. Who goes first or whatever is just a ritual. Of course, if you don't spin, you're going to have a hard time keeping player 6 in the game The easiest way to see this is that everyone has a chamber pre-assigned. 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, F is dead. Those, combined with the fact that exactly one person must die i. Sixth player: certain to be shot. 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. 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. Tags math.{/INSERTKEYS}{/PARAGRAPH} 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. With re-spins: your odds depend on where you are in the order of players. The UD stats class I took in school was, easily, the most useful class in my 7 years there. 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. 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? Better make sure that the gun is a well-maintained and well-oiled and b the load is heavy i. 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. So if you're given choice, respin! You have better odds of surviving any individual turn if you re-spin, a la the Monty Hall problem. The numbers also match well with equalpant's simulation. Always spin the cylinder. And so on around and potentially around, but always in favor of the later players. {PARAGRAPH}{INSERTKEYS}Russian Roulette. 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. But with re-spinning, you may be forced to take more than one turn, so it ends up evening out, kind of. One way of explaining why the non-spinning version is fair is that the initial spin assigns each player one of the six chambers. The game could, theoretically, continue forever without anybody getting shot. Think about a revolver with exactly two cylinders. However, these odds are those available during game play. One way to make the spinning version fair would be to only play one round, and not stop if someone gets killed. They're going to keep playing till someone dies, right? The easiest way to see this is to imagine an extreme case: There is a line of men. If you do spin, the odds are better for you if you are the 6th player. 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. 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? 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. One bullet. If the first person didn't die, and you're next, you had better spin it, right? The first player has a 1 in 6 chance of getting shot. Except for the first turn, where nobody has gone before you.