40
C. Just get it over with. Can't wait for the spin and reshoot.
Spoiler:
Tap for spoiler
0/6 chance to live <β actually Iβm probably wrong about this one, I missed the βspin after each pullββ¦ should be (5/6)^6 chance to live (~34%)
2/3 * 2/3 * 2/3 chance to live = 8/27 chance to live (~30%)
2/6 chance to live (~33%)
I'd go with C. Here are my MENTAL maths for odds of survival:
- A: (5/6)^6
- B: (4/6)^3 = (2/3)^3 = 8/27
- C: 2/6 = 1/3 = 9/27
C is clearly better than B. I have no clue how much A is, but it follows the same basic reasoning as B, so it's probably worse than C too.
Spoilers:
spoiler
A > C > B in terms of survival, or you could reverse it if you want to die and end your misery. Choices choices... π
Who wants to live forever anyway? :-PDiscussion
I did the maths after commenting. (5/6)βΆ = 33.49%, not too far from 1/3 = 33.33%.
That's a fun game by the way!
Your misery will be greatly reduced if you live and are released.
Yea but you walk out with zero money. π₯² You still have to find a job, and you might be pardoned, but the missing job history is gonna raise a lot of red flags.
So again... choices choices...
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Attempts to end your own life is illegal
What are they gonna do? Kill me?
Also, I'm gonna say B before doing the math.
I've done the math, I think.
Okay, the probability of living is the amount of unloaded chambers over the amount of chambers in total to the power of the amount of shots taken. Given that, we can conclude that...
- Option A (
(5/6)^6) gives you a 33.49% chance of survival, - Option B (
(4/6)^3) gives you a 29.63% chance of survival, and - Option C (
(2/6)^1) gives you a 33.33% chance of survival
So, turns out, I've chosen the worst odds. Whoops. However, they're so close that it honestly doesn't matter that much.
Also, sorry to those whose clients don't do spoilers. I'm with you on that one, I use Eternity.
load 4 chambers, attempt to take a guard or 4 with me
π€£
Lol you're not loading it, the executioner is.
You're just voicing your choice.
To the guards, "I choose the four bullets... But not for me!"
Off the top of my head, I think the best chances of survival are: C, A, B. I'm not sure about A vs. C, because A's total odds are hard to calculate in my head, while C is exactly 1/3 (33.33%).
The reason A is better than B is that a 1/6 chance of dying, twice, is better than a 2/6 chance of dying, once. They might seem at first like the same, but consider that one of those 36 chances in the A case is where you get shot twice in a row. That's no worse than a regular death. So it comes out to only 11/36 of dying in the first two rounds of A, but 12/36 of dying in the first one round of B.
spoiler
Using a calculator. it turns out A is actually 0.16% better than C. They're really about the same.
!wouldyourather@lemmy.dbzer0.com
I'd choose C and in a cheesy super villain voice say "look me in the eyes when you do it"
Executioner is a robot, feels no emotion
π₯π«π€
Deciding within 30 seconds, C. B is clearly worse and math seems to imply similar odds for A but it takes too long to calculate.
Gut says C feels safer and definitely less torturous so I chose C. Turns out A is better but only by 0.1% so not worth the wait.
I would choose option C.
4/6 loaded and 1 pull.
Segue, what makes you more comfortable for B?
This ππππππ
Or
That ππππππ
My first read I missed that the you were spinning the chamber between shots. I was thinking people pick B assuming its the first one and then get fucked the second way
Anyway, I think I'd go with C. If it's going to happen, it's going to happen, and I didn't get to death row by not believing in instant gratification
A
After 30 seconds: still A
spoiler
Congrats, you found the highest odds of survival at 33.335%
Downside is, you get to be shitting your pants for 6 revolver spins. π
C is 33.333% slightly lower, but its only one and done. No shitting in pants needed.
I died on the 5th round. 3,3,5,5,1
Nice knowing you!
I'm not taking their offer, better certain death than an unknown.
That is a WILD take, man. If youβre serious, youβre a new kind of person Iβve never met before.
This might count as "unusual" punishment, but imo it's far less cruel than what's usually on the table on death row
After quick math, I'll choose B. I believe it has 50% success rate.
Edit: oh no, "spinning after each shot"
A. It's either A or B, something about the fractions, but it'll take me more than 30 seconds to reason it out.
Edit: rip, nevermind, they're all about equal
ATM, I probably shouldn't be left with a gun and a choice. I don't care how many you load or if you spin. I would ask them if they would like to have a real conversation? They seem like an interesting type of person to get talking openly.
Seems like 1:6 Γ 6 is the most survivable as I intuit the problem. The previous shots have no bearing on the statistics of the next.
E: So yeah, the weight of the bullets spinning out of balance likely has more than a 1% bearing on the outcome. Statistics irrelevant IMO... but that is what a loser would say.
The previous shots do matter. Because for you to even reach the 6:th shot, all previous attempts have to be in your favor.
It's (5/6) you'll live each pull. But to reach pull #2 you'll have to survive the first. To reach pull #3 you have to survive the first 2.
You're looking at events that have to take place is a specific order. You have to multiply each pull to work out the probability of this event following one of those orders. It will come out to (5/6)^6.
(5/6) is the probability you survive. And ^6 because you have to survive it 6 times.
You're looking at ~33% of getting empty slots 6 times in a row.
Previous attempts always have a bearing on statistics if things need to happen in a certain order.
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(5 replies)
Guessed A then looked in the comments and that looked to be the best option so now I'm gonna roll some dice to see if I got released or not
Edit: aw beans
I would choose not to take part in this stupid game.
But regarding your math, the option C has the highest chance of surviving
A: (5 * 5 * 5 * 5 * 5 * 5) / (6 * 6 * 6 * 6 * 6 * 6) = terribly low
B: (2 * 2 * 2) / (3 * 3 * 3) = something like 0.3
C: 1/3 = 0.33
Spoilers
spoiler
Chances of survival:
A: (5/6)^6 β 33.335%
B: (4/6)^3 β 30%
C: (2/6) = 33.333%
How are you looking at ~15k / ~45k and deem that terribly low?
Does she spin before the first shot? Otherwise the bullet would always be in position to be fired. I assume so. Then option C, simply by expected value.
With 30 seconds being up, I must admit I can't calculate (5/6)^6 or (2/3)^3 fast enough to make a better comparison between the options.
Also, to be politically correct, what are my executioners pronouns? I just assumed she now, which may of course be offensive and I'm sorry for that.
Yea reworded the post
Edit: Your executioner is a robot, genderless. π
I'll pick B
Can I choose where I pull the trigger?
Cause I can make it the best or the worst funny ever.
C. I'm fine with either outcome. I'll be free.
C.
Because the barrel is spun each time itβs a simple percentage assuming the weight of the bullets donβt affect the spin. A is 16.2% for each trigger pull. B is 33.3% chance each trigger pull. C is 66.7%. The chances donβt stack because the barrel is spun again before each shot.
By this logic, if you roll a die 1000 times, you have only a 1/6 chance to get at teast one 6.
That's not how statistics work.
Leta say you're on your 6:th shot on option A.
For you to even get there, it requires all previous attempts to be in your favor. You're looking at events that all have to happen in a favorable order. And that is as follows
5/6 chance you live after the first time. (5/6)^2 chance you live after the second. (Because you have to survive the first) (5/6)^3 chance you live after the third. (Because you have to survive the first AND the second) .... and so on until (5/6)^6 ~ 33%
Think about flipping a coin. Do you really think getting 6 heads in a row is 50/50? The coin is "reset" between each flip. But it's not a 50/50 chance to get 6 heads in a row. If you don't believe me. Try it and see. According to statistics. It will take you 64 attempts to get 6 heads in a row.
They don't stack, but statistically the probability of a fire goes up each time it is spun again. It's not additive, but it does increase.
statistically the probability of a fire goes up each time it is spun again.
Please show your work here.
Probability = 1 - (1 - 1/6)^6
So the probability that the gun goes off is 66.51%
view more: next βΊ
this post was submitted on 30 Nov 2024
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