# In Russian roulette, is it best to go first?

Assume that we are playing a game of Russian roulette (6 chambers). Assume 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.

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Don't play russian roulette. Just say no. –  lhf Jan 4 '12 at 12:42
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 4 '12 at 16:53
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 4 '12 at 16:56
There's an implicit assumption here that it's an advantage to survive. Apparently what's now called "Russian roulette" originated in circumstances where the players were sometimes weary of life and didn't mind "losing". –  joriki Jan 4 '12 at 18:03
If after the first guy fires a blank, the host offers you to swap turns, should you? –  Larry OBrien Jan 4 '12 at 23:35

For a $2$ Player Game, it's obvious that player one will play, and $1/6$ chance of losing. Player $2$, has a $1/6$ chance of winning on turn one, so there is a $5/6$ 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 - $6/6$ (Chance Turn $1$ happening) * $1/6$ (chance of dying) = $1/6$

Player 2 - $5/6$ (Chance Turn $2$ happening) * $1/5$ (chance of dying) = $1/6$

Player 1 - $4/6$ (Chance Turn $3$ happening) * $1/4$ (chance of dying) = $1/6$

Player 2 - $3/6$ (Chance Turn $4$ happening) * $1/3$ (chance of dying) = $1/6$

Player 1 - $2/6$ (Chance Turn $5$ happening) * $1/2$ (chance of dying) = $1/6$

Player 2 - $1/6$ (Chance Turn $6$ happening) * $1/1$ (chance of dying) = $1/6$

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-$2/6$, P2-$2/6$, P3-$1/6$, P4-$1/6$

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 - $3/6$, P2 - $2/6$, Captor - $1/6$

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+1 for the last paragraph! –  Dilip Sarwate Jan 4 '12 at 17:45
+1, created account to vote on last paragraph! –  vsz Jan 4 '12 at 17:54
+1 for an epic conclusion. –  Samuel Reid Jan 4 '12 at 17:58
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 4 '12 at 21:13
@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. –  XenElement Jan 4 '12 at 21:29

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.

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Reminds me of the baseball manager's strategy of putting the best hitters first in the lineup –  JoelFan Jan 4 '12 at 15:46
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 4 '12 at 17:08
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 4 '12 at 17:20
@slim LOL! That's like the joke about tennis: Winning the last ball is a guarantee to win the game. –  Tormod Jan 4 '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.

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I'd like to see that interviewer. –  Tim Pietzcker Jan 4 '12 at 21:58
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 4 '12 at 22:26
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 5 '12 at 7:35
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 5 '12 at 9:54
@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 5 '12 at 15:21

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)

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As Clive's answer says, this would be true if and only if the number of participants is a factor of $6$. –  Jonas Meyer Jan 5 '12 at 9:48

This is a question on conditional probability. The first player will have a 5/6 chance of surviving. If the first player doesn't die, the second player will have a 4/5 chance of surviving, etc. etc. Think about if there are six players, and the first five dudes don't die. Would you want to be the sixth player?

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I don't think this is an accurate model of the situation, because, for example, the turns are decided before you know whether player 1 survives. Before you start, there is a 1/6 chance that the bullet is in the second chamber. –  Jonas Meyer Jan 5 '12 at 9:45
Yeah, you're right. I made a mistake. Probability of first guy surviving is 5/6. Probability of second guy surviving is the probability of the first guy dying OR them both surviving which equals [1/6 + (5/6)(4/5)] = 5/6 again. Probability of third guy surviving is the probability of first guy dying + the second guy dying + all three surviving. This is = 1/6 + 1/6 + 3/6 = 5/6. You can add because they're mutually exclusive. –  Taylor Jan 5 '12 at 19:43

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.

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## protected by Willie WongJan 5 '12 at 9:50

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