Russian Roulette

[BonJoviJones]

Me and 5 buddys get together over a 6 chamber revolver and a bullet. They offer me the chance to pick where to stand. Assuming I want to live, where do I stand? Assuming I want to die, where do I stand? Does it make a differance?

What if we had 5 bullets? What if we had 1 bullet and 20 chamber gun, but still 6 guys?

It seems to me I'd want to stand last, because the chance that it will get passed to me is low. The problem is that if it does get passed to me, the bullet will have my name on it. I have a sneaky feeling it all evens out for the simple case, and my choice of where to stand is moot.

[brad]

if you go last you only have 6.6% chance of getting shot which is less than 1/6 chance first guy has. assuming no replacement.

[Mike Haven]

if you go first you have a 5 to 1 chance of surviving = 16.7%

if you go second, of the 5 times in 6 that it gets that far
you have a 4 to 1 chance of surving = 83.3% x 20% = 16.7%

etc etc

so it doesn't matter where you stand (same sort of calcs go for the 1 in 20 scenario - don't stand first or second in the 5 out of 6 scenario - stand third to appear to be brave)

except if you are last in the first case and it gets that far you know you have the live bullet in the chamber and you can shoot the twat who thought of the game

[brad]

i got the 6.6% by assuming the chamber thing is spun every time.

[Mike Haven]

if the chamber is spun each time, each person is effectively starting a new set of blow brains out the trying to

so each person has a 5 to 1 chance of "success" = 16.7%

it's the old tossed penny has no memory thing

[BruceZ]

If the chamber is spun every time, then you obviously want to go last if you want to live, and first if you want to die. The way the poster plays /forums/images/icons/grin.gif he doesn't spin it every time, so it doesn't matter where you stand for 1 bullet in 6 chambers. For 5 bullets in 6 chambers, only the first 2 can die, so if you want live just be 3-6, and if you want to die be 1st. For 6 bullets in 20 chambers, even if you don't spin it every time, you don't want to be 1st or 2nd if you want to live since then 4 chambers kill you, but only 3 chambers kill everyone else.

[Bozeman]

Why do you assume the game ends after one person dies?

[brad]

i thought that was the rules too.

[Mike Haven]

"Me and 5 buddys get together over a 6 chamber revolver and a bullet."

[BruceZ]

It seems to me I'd want to stand last, because the chance that it will get passed to me is low.

[Mike Haven]

"If the chamber is spun every time, then you obviously want
to go last if you want to live, and first if you want to die."

thanks, bruce - i was letting mathematics get in the way of
common sense!

IF it reaches you you still have only a 16.7% chance of
shooting yourself at that point, the same as the first guy - but
why TAKE the chance of possibly shooting yourself if you
don't have to, is the point i missed - duh

while you are not pulling the trigger you have a zero chance
of dying - as soon as you pull the trigger, this chance
increases - so when it increases to the same chance
wherever you stand it is best to keep to zero chance for as
long as possible, and stand last if you want to live

***

"The way the poster plays he doesn't spin it every time, so it doesn't matter where you stand for 1 bullet in 6 chambers."

true in percentage terms, but again why increase from zero to 16.7% chance of dying by firing? stand last again if you want to live

***

when bozeman said "Why do you assume the game ends
after one person dies?" he was presumably talking about the
5 bullets in 6 chambers scenario

if the game doesn't end then if you go first here you have a 1
in 6 chance of surviving = 16.7%

if you go second then if the first guy shot himself you have a
1 in 5 chance chance of surving = 20% x 83.3% = 16.7%

plus if he didn't shoot himself then you have 0 in 5 chance of surviving = 0%

etc etc

so in percentage terms it doesn't matter where you stand,
you always have a 16.7% chance of surviving

but, again, as you always have an 83.3% chance of dying,
maybe you should put it off as long as you can?

i am tempted to say that you should go first in this particular case, because of the added 100% chance of dying if one in front of you has survived his shot - is this right?

***

"For 6 bullets in 20 chambers, even if you don't spin it every time, you don't want to be 1st or 2nd if you want to live since then 4 chambers kill you, but only 3 chambers kill everyone else."

i don't follow this - could you explain please? thanks

[BruceZ]

"The way the poster plays he doesn't spin it every time, so it doesn't matter where you stand for 1 bullet in 6 chambers."

true in percentage terms, but again why increase from zero to 16.7% chance of dying by firing? stand last again if you want to live

This case is totally different from the case where it is spun every time. In this case, the person who dies is predetermined by the intial spin. Each person has a 1 in 6 chance no matter where he stands. If you are last, you have a smaller chance of pulling the trigger, but you have the same chance of dying. In fact, if you do pull the trigger you die for sure!


when bozeman said "Why do you assume the game ends
after one person dies?" he was presumably talking about the
5 bullets in 6 chambers scenario

if the game doesn't end then if you go first here you have a 1
in 6 chance of surviving = 16.7%

if you go second then if the first guy shot himself you have a
1 in 5 chance chance of surving = 20% x 83.3% = 16.7%

plus if he didn't shoot himself then you have 0 in 5 chance of surviving = 0%

etc etc

so in percentage terms it doesn't matter where you stand,
you always have a 16.7% chance of surviving

but, again, as you always have an 83.3% chance of dying,
maybe you should put it off as long as you can?

i am tempted to say that you should go first in this particular case, because of the added 100% chance of dying if one in front of you has survived his shot - is this right?

If the game doesn't end until one person is left, then it doesn't matter where you stand. The "winner" is predetermined just as the loser was in the case of one bullet, and each person has a 1 in 6 chance of surviving by getting the blank chamber.


"For 6 bullets in 20 chambers, even if you don't spin it every time, you don't want to be 1st or 2nd if you want to live since then 4 chambers kill you, but only 3 chambers kill everyone else."

i don't follow this - could you explain please? thanks

Actually my remarks apply to 1 bullet in 20 chambers, assuming the gun keeps getting passed around the circle until someone dies. In that case, the first two guys have 4 chambers that can kill them (1,7,13,19 for first person) while the other 4 guys only have 3 chambers that can kill them. For 6 bullets in 20 chambers, I made a spreadsheet to get the probabilites. It is best to be last:

probabilities of dying:

1: 0.33374613
2: 0.241640867
3: 0.171594427
4: 0.119195046
5: 0.080727554
6: 0.053095975