[Guest]

So, let's say you're playing a game of Russian Roulette. For those of you who don't know what that is, you're given a gun with 1 bullet in it and 5 blanks. The first person points the gun at their head after spinning it and shoots. If he survives, the next person attempts to shoot himself. The game goes on until someone dies.

The question is, "Where should you sit to have the greatest chance of survival?"

Seat 1(First to shoot): 1 bullet, 5 blanks = about 16.66% chance of death

Seat 2(Second to shoot): 1 bullet, 4 blanks = 5/6 * 1/5 = 1/6 chance of death

Bonus: There's a game on Newgrounds called "Pico vs. the Uberkids" where they play "Rock, Paper, Scissors Roulette. The game is basically "Rock, Paper, Scissors." However, the loser of the game has to play Russian Roulette. So, what are your chances of survival?

[Guest]

I'd want to go last. That way, if the first five got through, I could opt out. But seriously, statistically speaking each spot has the same chance at the beginning of the game. The odds only change as the players exhaust their turns. Those players, however, may not make it so I'm still going last.

[Guest]

Its the odds accumulating. I'm not exactly sure how you'd do that. He's talking about how the first person will have 1/6 chance, of dying, then next 1/5, then 1/4, 1/3, 1/2, 1/1. Not exactly sure how to calculate that. I'm good at math, (I just finished 7th grade algebra at top of my class) but it was just algebra. You don't have oppourtunity to "opt out" of Russian Roulette. So, if there were two people, would you want to be first to pull trigger, or second. That is what your saying right filly?

TOMORROW IS ROOKIE'S BIRTHDAY!

Edited July 14, 2010 by NickFleming

[Guest]

Assuming each time it's your turn you have a 1/6 chance of being shot (if it doesn't matter weather a hole has a blank or it's empty) then I'll go last, of course. I guess it's how it works cause in the example both the first and second persons have a 1/6 chance. But I'm not very good with guns. If it's different, and once a hole is "used" it can't be "chosen" anymore, and therefore the odds are like NickFleming wrote, I guess the chances will be the same for all the people.

[Guest]

I would want to be last. If the gun got to me, I could shoot the person next to me. 100% survival!

[Guest]

Assuming each time it's your turn you have a 1/6 chance of being shot (if it doesn't matter weather a hole has a blank or it's empty) then I'll go last, of course. I guess it's how it works cause in the example both the first and second persons have a 1/6 chance. But I'm not very good with guns. If it's different, and once a hole is "used" it can't be "chosen" anymore, and therefore the odds are like NickFleming wrote, I guess the chances will be the same for all the people.

Ufortunately doesn't apply. Someone will be shot within 6 shots, for it doesn't skip over bullets. The later the turn, the higher the probability of being shot. But if someone's already shot, you won't be. There will be no more than 6 turns. Goes around until someone is shot. Then ends. I believe this riddle is for Russian Roulette played between only two people.

[Guest]

Either the first or second spot

they both have equal probability of dying.

All other are higher:

Seat Probability of death

1 0.167

2 0.167

3 0.208

4 0.264

5 0.368

6 0.632

1/3 odds of winning rock paper scissors, so odd of dying become 1/3*1/6 = 1/18.

[Guest]

I've heard that a blank can kill you so I would say sit in the audience

[Guest]

if 1/6 chance each shot then as late as possible.

1/6 *(5/6)^2 = .115 or 11.5% chance of death for third,

1/6 *(5/6)^3 = .096 or 9.6% chance of death for fourth, etc.

if decreasing odds (1/6, 1/5, 1/4 etc.)

then as early as possible, (second preferably).

with rock paper scissors,

that simply deceases odds by 1/3.

[Guest]

No matter where you sit the probability will be the same:

1: 1/6

2: 1/5 * 5/6 = 1/6

3: 1/4 * 4/5 * 5/6 = 1/6

4: 1/3 * 3/4 * 4/5 * 5/6 = 1/6

5: 1/2 * 2/3 * 3/4 * 4/5 * 5/6 = 1/6

6: 1/1 * 1/2 * 2/3 * 3/4 * 4/5 * 5/6 = 1/6

I'm ready to call B Tremell the winner though. Who's gonna deny the guy with the gun when the next shot is guaranteed to have the bullet!?

[Guest]

The Russian mobsters.

[bonanova]

The question is, "Where should you sit to have the greatest chance of survival?"

Let s = survival probability.

Only if there are 4, 5 or greater than 7 players does any player have an advantage.

Conditional probabilities aside, if the bullet is in the n^th chamber, the n^th player dies.

1 Player -- 1 1 1 1 1 1 - all players: s=0/6

2 Players - 1 2 1 2 1 2 - all players: s=3/6

3 players - 1 2 3 1 2 3 - all players: s=4/6

4 players - 1 2 3 4 1 2 - players 1 2: s=4/6; players 3 4: s=5/6

5 players - 1 2 3 4 5 1 - player 1: s=4/6; players 2 3 4 5: s=5/6

6 players - 1 2 3 4 5 6 - all players: s=5/6

more than 6 - 1 2 3 4 5 6 7 8 ... - players 1-6: s=5/6; others: s=6/6.