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/sci/ - Science & Math

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5015783 No.5015783 [Reply] [Original]

In a game of Russian roulette, there are six chambers in a gun and a bullet in one of the chambers.

You're playing against an opponent. Your turn is first, and your probability of shooting a bullet is 1/6. You shoot a blank.

What is probability that your opponent, in the next turn, shoots a bullet?

Protip: it's not 1/5.

>> No.5015784

1/5

>> No.5015785

1/6, idiot

>> No.5015789

>>5015785

The chamber is not spun again, you continue with the same gun, you fucking child.

>> No.5015792

Would... it be 1/2?

>> No.5015795

>>5015789
>itt OP is a butthurt faggot because he forgot how Russian roulette is normally played

>> No.5015796

0.
Blanks would still kill you at such close range, so your opponent walks away victorious.

>> No.5015797

>You shoot a blank.

if you shot a blank at your temple you have a very very high chance of killing yourself, and a 99.999999999999% chance of serious injury.

so you are wrong.

>> No.5015799

>>5015795

It's just a scenario, faggot, if we assume it's reconfigured at every turn of course it's 1/6 why the fuck would I even ask?

>> No.5015806

>>5015789

For that to happen, all needed is that the the blank shot was the blank before the bullet.
As there is only 1 blank right before the one bullet (in the 6 possible spaces), probability remains 1/6.

Also, stop sperging out, you are the child here. Should've stated before it wasn't actually Russian roulette, but some other shit you invented and decided to call Russian roulette even though.

>> No.5015807

>>5015799
>reconfigured at every turn

what do you mean by this?

>> No.5015809

>>5015799

We don't assume it's reconfigured every turn. It's not a dice. You pull once. He pulls once. You don't keep spinning it every turn like a dice. Russian roullete =/= dice game.

The chance is 1/5, then 1/4, then 1/3, then 1/2, then 1/1.

>> No.5015811
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5015811

>>5015806

>NT detected

>> No.5015812

>>5015809
what the fuck is reconfiguring a firearm mean? your not making any sense.

>> No.5015814

>>5015812

RECONFIGURRING means you take the gun apart, clean it, talk to the other guy about what you did last Friday, put it back together, spin the chamber, and then hand it to him to fire. you just don't fucking do it.

>> No.5015817

Let <span class="math">B_i[/spoiler] denote bullet in the i'th chamber. What you want is <span class="math">P(B_2|B_1^c)=\frac{P(B_2\cap B_1^c}{P(B_1^c}[/spoiler]
<span class="math">B_2[/spoiler] is a subset of <span class="math">B_1^c[/spoiler] so it becomes <span class="math">\frac{P(B_2)}{P(B_1^c)}=\frac{1/6}{5/6}=1/5 brah.[/spoiler]

>> No.5015818

>>5015789

Shooting a blank then shooting a bullet ->

P(blank)*P(bullet)=(5/6)*(1/5)=1/6

>> No.5015820

>>5015818
The events are not independent, for example, if the bullet was in the first chamber there would be 0 chance of shooting the second time.

>> No.5015821

Wait is it 1/2 because he either shoots a bullet or he shoots a blank. Think about it....

>> No.5015822

>>5015820

That's why you need the blank that's right before the bullet and nothing else. See >>5015806

>> No.5015824

>>5015817
Fun fact, without spinning the chamber after pulling the trigger there is an equal chance of losing if there are 2 playing.

>> No.5015826

>>5015820

Thank you master can I lick your genitals please.

>> No.5015829

>>5015821
There's actually people that think like this.

>> No.5015831

>>5015820

He didn't. Read OP post again. He didn't shoot a bullet the first time. Suck it!~

>> No.5015833

>>5015822
You can only multiply probabilities of the intersecting sets if they are independent.

>> No.5015836

1/3

>> No.5015835

>>5015831
That was an example of a different situation which displays that there is an error in the method. What's with all the trolls at the moment?

>> No.5015837

22/7

>> No.5015838

>>5015814
>you just dont do it
>cleaning your firearm

holy fuck dont listen to this kid.

what does reconfiguring mean in this instance?

>> No.5015839

>>5015833

You have 6 chambers.
Let's say:
C1 - blank
C2 - blank
C3 - blank
C4 - bullet
C5 - blank
C6 - blank

Person 1 must get C3 so that when the gun passes to person 2 and rotates 1 chamber, person 2 gets C4.

So, ALL needed is that Person 1 gets C3. (C3 is 1 in the 6 chambers, hence 1/6)

I thought /sci/ knew high school probability...

This doesn't even need any calculations.

>> No.5015841

Why do you faggots choose to troll with one of my favorite mathematical subjects? fuck

is it because probability is so easily misinterpreted?

>> No.5015842

>>5015839
This is the chance before we know that player 1 pulled a blank which takes C4 off the table.

>> No.5015846

>>5015841
But that makes the troll claims hard to address sometimes which can be interesting. You know it's wrong because the math says so but your "intuition" is fucking with you.

>> No.5015849

> /sci/ can't into conditional probability

>> No.5015852

>>5015783

Troll. The P is 1/5. You already shot an empty chamber, so it's no longer considered as a probability. So you have 5 chambers left, 1 of which has a bullet.

Once the 1st shot is taken, the 2nd shot becomes an independent event.

>> No.5015853

>>5015817

The events are mutually exclusive so there is no intersection.

>> No.5015855

I'm from /x/, is this what always happens here?
Like, no one really understands high-school probability here?

>> No.5015857

>>5015806

bat DOOD
now YOU KNOW THAT CHAMBER IS EMPTY AND YOUR MOVING ONTO THE NEXT CHAMBER. UNLESS THE CHAMBERS RANDOMLY SWITCH PLACES FOR SOME QUALIA PHYSICS REASONS IDK DOOD.

so ya that chambers empty and you watch that chamber go to the back of the line of chambers.

and shit who knows if youre going to die, all you know is that the next chamber you try definitely wont be that empty because once again to reiterate, you watched it cycle to the back of the cycle.

>> No.5015865

guys is russian roulette just a game of chance?
I mean you have a 83.33% chance of getting the chamber right unless you happen to know the answer, and lifelines can only get you so far into the game.

>> No.5015871

>>5015853

They are not mutually exclusive events. The chance depends on the past event very much so.

>> No.5015875

>>5015853
No, it's B_2(bullet in 2) and B_1^c(no bullet in 1). Bullet in 2 means there is no bullet in 1.

>> No.5015885

all you dipshits have to do is solve schroedingers eQuation for the bullet and the answer will become fucking clear.

The answer is 0. Quantum Suicide states that we will live forever despite trying to commit suicide.

>> No.5015901

1/5. Can't into conditional probability. Another way to look at it is to notice that you are altering the sample space when you fire a blank.

>> No.5015914 [DELETED] 

the answer is 1/4.

>>5015901

would you rather play russian roulette with a gun that has 5 chambers and 1 bullet, and it's your turn, or a normal revolver with six chambers that has rotated by one in the first turn and now it's your turn?

i'd pick the first, which has a 1 in 5 probability of shooting the bullet, the second it's slightly higher (1 in 4) because the bullet has moved towards your chamber, so there's zero probability that it's in the 5th chamber.

>> No.5015917

It seems there is a lot of confusion in this thread because OP didn't say if the players must "reroll" te revolver's magazine

>> No.5015920

There are 6 chambers, 1 of which has a bullet so for the first turn the probability of shooting a bullet if 1/6.

Then there are now 5 chambers left to go, 1 of which has a bullet so the probability of firing a bullet on the second go is 1/5 .

>> No.5015927

>shoot
>not bullet
>reroll
>1/6 again
>shoot
>not bullet
>reroll
>1/6 again

and so on
sci can't into russian roulette

>> No.5015947

>>5015927
Every time you shoot, there is a higher risk of chance of getting the bullet chamber.

>> No.5015961

>>5015947
no, because you roll the revolver's magazine and that leaves you at 1/6 chance of getting a bullet once more

you may aswell shoot the same blank that you just shot because you rolled the magazine

>> No.5015965

Nice ambiguous question there...

Probability you shoot a blank AND he shoots the bullet:
5/6 x 1/6 = 5/36

Probability he shoots the bullet is just 1/6

>> No.5015966

>>5015961
Oh, so it rolls again? Glad you cleared that up.

So it would be 1/6, I think a lot of confusion arose when we thought that you just hand the gun to the other guy without rolling the magazine.

>> No.5015968

>>5015965
haha fucked that up...

5/6 * 1/5 = 5/30 = 1/6

>> No.5015976

>>5015966
glad to help

"The term "Russian Roulette" dates back to a 1937 short story of the same title:
‘Did you ever hear of Russian Roulette?’ [...] With the Russian army in Romania, around 1917, some officer would suddenly pull out his revolver, put a single cartridge in the cylinder, spin the cylinder, put it to his head and pull the trigger."

I am from Romania. My grandfather used to play Russian Roulette. He was always so lucky. He lived for 93 years.

>> No.5015988

>>5015965

The events are dependent not independent the AND rule doesn't apply here.

Meaning the probability of shooting a bullet is affected if a blank has been fired before (more likely)

The probability is 1/5 as there are 5 chambers left with 1 having a bullet in it.

>conditional probability

>> No.5015986

On a somewhat related note, I'll give 100 internets to whoever links to a video of people playing russian roulette

>> No.5016016

>>5015986
http://www.youtube.com/watch?v=lqakCa-MysE

>> No.5016019

>>5015988
He's working under the boring assumption that you roll the cartridge between trigger pulls.

>> No.5016024

>>5016019

OP is a master troll

>> No.5016060

>>5015986
http://www.youtube.com/watch?v=mJneySUo8V4

movie trailer

>crossed russian roulette
>they point the guns at each other and shoot
>more intriguing

Also, what is the probability that one of them gets shot? What about both of them?

>> No.5016085

>>5016060
The only thing I got from that was that everyone in Romania shoots at everyone else with revolvers

>> No.5016370

It's 1/5, OP is trolling.

>> No.5016383
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5016383

the answer is 1/3

3rd shot would be 1/2

4th 2/3

5th 5/6

6th 1

>> No.5016470

If it was the first empty chamber then the next is also empty
if second empty chamber then next empty
if third empty chamber next empty
if fourth empty chamber then empty
if fifth empty chamber then boom, so 1/5

>> No.5016479

Zero.

Sure, throw a bullet in the chamber. Won't hurt me at all without the rest of the round in there.

>> No.5016866

This whole thing becomes really easy if you rephrase the question. What is the chance that the bullet is in the second slot of the pistol?

In this case we just need the chance the shot is in any given slot. Hence 1/6

>> No.5016935

>>5016479
XD

>> No.5016968

The probability that the bullet is shot in the next turn is 1/5. After shooting once with the bullet not getting shot, the problem is reduced to the bullet residing in one of the five remaining chambers. If you ask before starting the game what the probability is that the bullet appears on the 2nd turn, then that would be 1/6 because there are only six possible outcomes in Russian Roulette if played until the bullet gets shot (all of this presumes that the bullet has an equal chance of being in any of the six chambers).

>> No.5016989
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5016989

>>5015821
yew stoopid
every single imaginable event has a 1/2 "probability" of occuring. It's not a matter of whether it happens or not, it's a matter of what the chances are that it will happen given the data provided to you.

>> No.5016992

The answer is 1/5. OP's just being a dumbass.

If the first shot was a dud, then 1 out of the remaining 5 is going to be the loaded shot. 1/5. gogdam

>> No.5017069

It's 1/2.
You're all fucking idiots.
It's 50%. The bullet is either in the next chamber, or it isn't.

The probability of shooting yourself in a game of russian roulette is 1/6. The probability of shooting yourself IN ONE SINGLE TURN of russian roulette is 1/2.

/thread.

>> No.5017076

>>5017069
Op here. We have a winrar

>> No.5017082

>>5017069
If you think this guy is wrong, it's because you don't know that in the Official rules of Russian Roulette, there only 3 turns per match.

>> No.5017089

Yeah it's 1/5.

>> No.5017111

>>5017069
No dumbass, read the following
>>5017089

>> No.5017152

>>5017069
If you and a friend chip in and buy two lottery tickets, you scratch yours and find that you didn't win, you pass the other lottery ticket to your friend, he has a 50% chance of winning because its either win or loose.

^This is what you look like to people who aren't retarded.

>> No.5017187

YOU ARE ALL FUCKING IDIOTS.
The reason it isn't 1/5 is simple.
How does Russian Roulette work?
You grab the revolver, you open the cylinder, you put a round in one chamber, then you spin it and click the cylinder back in place. The probability the round will be in the chamber you hammer is 1/6.
Then you give it to the next person. He also spins the cylinder. For this reason it will never be 1/5, it will always be 1/6.

This is how Russian Roulette is played. Otherwise people would fight over who gets to go first don't you thing?

>> No.5017190

<div class="math">P(A)=Bullet\;\; P(B)=Blank \;\;P(A\mid B)= \frac{P(A\cap B)}{P(B)}\:=\frac{P(B\mid A)\cdot P(A)}{P(B)}\;=\;\frac{1\cdot \frac{1}{6}}{\frac{5}{6}}\:=\:\frac{1}{6}\cdot\frac{6}{5}=\frac{6}{30}=\frac{1}{5}<div class="math"></div></div>

>> No.5017192

You're all forgetting that you spin the cylinder before each turn. Thereby randomizing the results once more and making sure it's always 1/6

>> No.5017201
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5017201

noone realizes that it is 1/4
u wot m8

>> No.5017203

>>5015838
it's pretty fucking obvious he was referring to spinning the gun again

>> No.5017205

>>5015826
I am going to say this sentence to my master on Sunday :3

>> No.5017213

>>5017201
wait a sec
i misread the question
i suck cocks

>> No.5017236

>>5015796
No, unless maybe you held the gun touching your temple. Otherwise it would shoot very high velocity very hot gas into your face and you would scream to the bitch and run to the hospital with your retarded drunk russian friend laughing the whole way.

>> No.5017240

http://en.wikipedia.org/wiki/Monty_Hall_Problem

Is this what you were referring to, OP?

Unfortunately, there is no control of the information whatsoever (no host to tell us that the first round is a blank). As well, the Monty Hall problem was never about the exact location of the car, only about whether or not switching did anything.

I would still say 1/5.

>> No.5017276

>>5017187
As opposed to fighting who gets to go second?>>5015806
We're given that the first shot is a blank though. Conditional probability. (1/6) / (5/6) = 1/5.

>> No.5017306

This.
>>5017240
In the monty hall problem the kicker is the host always opens the door to a goat. In russian roulette you aren't given that option.

>> No.5017308
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5017308

Errrbody gettin' rustled up in here

>> No.5017347

It is 1/5 though.