Lets say you have a revolver with one bullet filled out of eight...

Lets say you have a revolver with one bullet filled out of eight. Suppose you put the gun to your head and shoot (Russian roulette style), you have a 1/8 chance of being struck by the bullet and dying.

If you fire the gun a second time now knowing the first chamber was empty, what is the probability of dying overall? Is it just 2/8 or is it some combination of 1/7 (your chance of eating the second bullet) and 1/8 (original bullet).

I understand this is some conditional probability shit, but how do you calculate the probability for each number of shots. confusing the fuck out of me

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any ideas?

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Nigger, if you have 1/8 chances but the first shot is safe, then you have a total of 7 shots that can kill you, equals 1/7 chances to kill yourself

8/1
7/1
6/1. And so on until One bullet left wich is 1/1

I get that after each bullet you survive the chance of you surviving the next bullet decreases, but is the odds of surviving n bullets really just 1 - n/total number of chambers?

Just noticed that 7/8 * 6/7 * 5/6 is the same as 5/8 -> chances of surviving 3 bullets ( 1 - 5/8 = 3/8)

its just the probability for the first bullet + the probability for the second bullet. So 15/56

It's not 2/8 because there are two instances. Since there are two scenarios it's 1/8+1/7 which equals 15/63. This is assuming you do not reroll for the chambers.

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>It's not 2/8 because there are two instances. Since there are two scenarios it's 1/8+1/7 which equals 15/63. This is assuming you do not reroll for the chambers.

I'm an idiot it's 15/56. Did the math wrong on the bottom lol

Close. You have to multiply not add the two fractions so the odds of killing yourself in the first two shots is 1/8 * 1/7 = 1/56

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The first time you pull the trigger, the chances of you dying are 7/8. If you want to calculate the probability of dying after n trigger pulls, you would use 1 - (7/8)^n. Say you pull the trigger twice; You multiply 7/8 by 7/8 to get 49/64. That's the probability of you living. Now, just subtract that from 1 and you get the probability of you not living.

Rocket science fag here.

Every individual shot will have 1 less on the denominator and denomitor

ie. if you shoot & a bullet doesn't come out, your chances go from 7/8 to 6/7 on the next attempt.

However, if you're asking the odds of pulling twice in a row and still being fine: this is the equation: 7/8 * 6/7

Your odds would be 42/ 56, or simplified, 3/4 to be okay.

It never hurts to gather too much data...so go ahead, try it yourself. Post results.

This is assuming that you randomly roll the chamber each time. I see now that this is not the case in OP's scenario

one less on denominator and numerator*

Sorry, am a high rocket science fag

yep. Think of it this way. If you decide you're going to fire 6 bullets, then out of the 8 possible equally likely positions of the cylinder, there are only two that will allow you to survive.

nope.

If you say "I'm gonna point to my head and fire twice," then you have a 2/8 chance of dying. If you fire once and survive, then say "i think I'll fire the next shot" then the new chance of dying is 1/7.

no this is assuming you are calculating the probability to have the bullet be in the chamber with 8 shots and the chamber with 7 shots. If youre finding the chance to die from each shot then you have to add.

Before pulling the trigger the first time, you have a 2/8 chance of dying if you pull the trigger twice. 1/8 (the probability of dying on the first bullet) + 7/8 (the probability of surviving the first bullet) * 1/7 (the probability of dying to the second trigger pull).

Right. Your chances go DOWN the more times you do it. So, you are practically GUARANTEED to survive if you shoot 8 times. You should totally try it!

Rocket science fag here again, and to be 100% clear on the answer:

Find out the probability of each event.

To find the probability of both happening back-to-back, simply multiply the two values together.

I.e: The chances of flipping a coin and getting heads once is 1/2. The chances of flipping a coin and getting heads twice in a row is 1/4 because 1/2 * 1/2 = 1/4

Nope. Imagine firing all 8. If you're at the last shot your chance to die is 1/1. If you add that to all the previous probabilities you have a probability over 1, meaning you did something wrong.

That's not at all what that means, faggot. learn2math

you do multiply, but not the probability of dying. 1/56 is your chance of a 1/8 event AND an independent 1/7 event both happening.

For the chance that at least one happens, you have to use a complement. You take the probability that both don't happen and subtract it from one. So 1 - (7/8)(6/7) = 1/4.

Alternatively, you could note that out of 8 possible cylinder positions at the start, 2 will kill you. Which gives you the same probability.

Your initial assumption changes the problem. Since it's given that the first shot won't kill you, you end up with a 1 in 7 chance of the bullet killing you. It is conditional but you still end up with 1/7.

If you're asking what are the chances of blowing your brains out on nth shot, that's different.

Both are also different from asking the chances of blowing your brains out within n number of shots.

It's a trick called telescoping. When you multiply fractions and the numerator of one term is equal to the denominator of the next term and so on its the same as the last numerator of the first denominator

Carrying on nth shot.
For the nth shot probability you use the probabilities of every shot before that missing AND the nth containing the bullet. For example, the 4th shot would be ((7/8) * (6/7) * (5/6)) * (1/5)

Carrying on further chances within n shots is the addition of all n-i shots for all i < n. The chances of surviving is then just that number subtracted from 1.