What's the probability of surviving (n) rounds of Russian roulette?

Question

It's a 6 shot revolver and you get to spin the cylinder on each round. (no missed shots and no misfires... you die if it goes off!)

Survivability = f(n) = ?

It doesn't depend upon how many people are playing... Whatz the probability of me doing it to myself.....

swapnil has the answer! very good! you get points.

Answer 1

without thinking about it too much, I 'm going to go with:

f(n)=(5/6)^n

Answer 2

Two assumptions:
Assume each time anyone pulls the trigger is a round
Second assume that you spin the chamber after each pull: So each time you pull the trigger, you assume that you have one chamber that could contain the bullet, or 1/6.

This means that to survive n successive pulls is: P(n) = (5/6)^n

Meaning zero pulls gives P(0) = 5/6^0 = 1. Or you have 100 percent chance to live pulling zero times.

And 5/6 after that, 25/36 etc.

However, there are a few things that fudge this. First off, most of the time you're using an empty chamber for the dead rounds (not fake bullets), so you can tell (since it's a six shooter) when the side chambers are actually empty. This means that only the top chamber and bottom chamber contain the bullets if it's actually empty.

This chances chance of death to 50/50, which sucks a lot more.

And ofcourse, if you see the bullet in a side chamber, you know it won't kill you so the chance to live would still be 100%. But if you're doing it honorably and pulling the trigger after spinning the chamber regardless of what you see, then it defaults back to the 5/6 rule.

Answer 3

1

Answer 4

Actually, the number of people playing DOES matter, because is someone else loses while it's their turn, you survive. Your survival is also going to depend on to whether you go first, second, or whatever. As an example, if there's 10 people playing and you go last, there is a
[1 - (5/6)^9] = 80.62% chance that you won't even take a turn in the first round.

If, however, you play that you reload after someone dies and everyone still alive has to take the same number of shots, then
f(n) = (5/6)^n

Answer 5

Its a very complex explanation....

Russian Roullete has only a single bullet.......... but the bullet can be inserted in 6 ways... ie in any of the six spaces...

Further, let F(a,b,c) be the probability of a man c losing his life....where a,b is in fourth space..
If we make a triangle with F[a,b,c], then we get a triangle of fractions where its denominator and numerator form a pascal's triangle-like pattern.

Now, lets venture into the properties of Death, Death can cheat you sometimes.... and most interestingly, it comes slowly but steadily....... so probability of a death if hit by bullet in terms of two options are : you die or not die....is 1/2



On the scientific side, You would want to be the 6th player in a game of Russian Roulette : As the revolver has 6 chambers, with a bullet in 1 of them, player number 6 seems guaranteed to die. Player 1 would only have a change of 1/6 to die, player 2 a chance of 1/5, etc. So Playing first seems to be the best bet. But, if you get shot being the first person then you could either survive... ( if you wear a helmet) and might just dodge the death..(see above for details..)

Player 1 has indeed a 5/6 chance (approx. 83 %) to survive. Player 6 also has approx. 83 % to survive : each player before him can get the bullet, so there's a good chance that someone will shoot himself before the gun reaches player 6. That chance is 5/6, or - again - approximately 83 %.

I hope I have you confused !!!!!!!!I believe now.. Russian Roullete is your only option to survive with sanity..

Answer 6

The answer is 5/6. Each round is independent of the one before it and the one after it. Therefore, each round has the same number of chances of living, five, and the same total chances of any outcome, six.
After, let's say 6 rounds, you have a total of 36 possible outcomes (6 x 5 "live" (30) + 6 x 1 (6) "die"), so 30 out of 36, which reduces down to 5 out of 6. No matter how many rounds you do, there's always the same probability of surviving.

Answer 7

This Site Might Help You.

RE:
What's the probability of surviving (n) rounds of Russian roulette?
It's a 6 shot revolver and you get to spin the cylinder on each round. (no missed shots and no misfires... you die if it goes off!)

Survivability = f(n) = ?

Answer 8

I'd go no further than 3 rounds, if I was playing with another.......Your probability would have run out, after 6 pulls. But i'd be more tempted to not do any rounds, thankyou. Watch the Deer Hunter. It messes you up, man. Just as a tip, don't play it with a self loading pistol, unless you've got second turn....O.K.

Answer 9

1/6 chance you'l die and 5/6 chance you'l live. Its note dependent on the number of players if the rule is you have to spin it all over again each round.

Answer 10

f(n) = n/6

In other words, you have 1 chance out of 6.

Might I add that the probability is the same on each spin of the cylinder. This means that, most probably, within 6 turns the trigger puller will die and be ushered directly to hades. Using business probabilty, figure on it happening the first time.