No using search engines or calculator, just imagine you were given these choices in jail and got 30 seconds to choose.

Would you rather have the executioner (Please excuse my use of firearm terminology, I’m not an expert in guns):

[A] Load 1 chamber of 6 and spin before pull the trigger 6 times (spinning the chamber after each pull)

or

[B] Load 2 chamber of 6 and spin before pull the trigger 3 times (spinning the chamber after each pull)

or

[C] Load 4 chamber of 6 and spin before pull the trigger once

If you survive, you are pardoned of all crimes you’ve committed in the past and get to leave prison (alive, obviously), no compensation tho. (Attempts to end your own life is illegal)

Which of these options would you pick?

No search engine, no calculator pls. I’m bored and wanna make this fun. 😉

If you are confused, ask in comments.

Edit: Reworded it to clarify that the chamber is spinned before the first trigger is pulled.

    • subtext@lemmy.world
      4·
      27 days ago

      Spoiler:

      Tap for spoiler

      0/6 chance to live <— actually I’m probably wrong about this one, I missed the “spin after each pull”… 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%)

  • Lvxferre@mander.xyz
    201·
    27 days ago

    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.

  • Slyme@lemm.eeEnglish
    13·
    27 days ago

    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.

  • Limonene@lemmy.world
    6·
    27 days ago

    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.

  • Iceblade@lemmy.world
    5·
    26 days ago

    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.

  • Presi300@lemmy.worldEnglish
    5·
    27 days ago

    I’d choose C and in a cheesy super villain voice say “look me in the eyes when you do it”

  • BougieBirdie@lemmy.blahaj.zoneEnglish
    5·
    27 days ago

    Segue, what makes you more comfortable for B?

    This 🌕🌕🌑🌑🌑🌑
    Or
    That 🌕🌑🌑🌕🌑🌑

    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

  • j4k3@lemmy.worldEnglish
    3·
    27 days ago

    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 × 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.

    • Atomic@sh.itjust.works
      2·
      26 days ago

      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.

      • j4k3@lemmy.worldEnglish
        1·
        25 days ago

        I think there is some philosophical elements that invalidate this , or rather constrain the conclusions to an idealized mathematical irrelevance to the real world due to all the other uncertainties involved in the entropy of the universe all the way down.

        One might argue that the statistics hold true with a large enough sample size. But what is the noise floor of such a sample size. When all the other real world uncertainties are accounted for, I believe these would accrue far greater than such a noise floor.

        Nothing about the situation or mechanism in question is a known or consistent factor in this equation. The orientation, mechanical properties, internal friction, and force applied, should have a significant effect on the outcome of each spin. The distribution of bullets alone is a major factor and not specified. For instance, in the case of 4/6, one must assume the bullets include 2 sequential empty chambers, so how does the imbalance alter the probability. It is likely to have more of a tendency for those bullets to end in a gravitational low point, and therefore a statistically more likely chance for an empty chamber on top in the firing position.

        The noise floor of all additional factors being larger than the statistical difference, IMO, means there is not relevance to the statistics outside of academia or situations where many millions of people are subjected to this method and the inputs are normalized with better constrained input factors. The actual sample size may need to be much larger. I’m just picking a big number from my little brain.

        If the noise of all the unconstrained inputs is larger than the statistical difference, then the statistics don’t matter at this scale for the individual.

        In my experience, mechanical imbalances are an inherent feature of goods produced under capitalism. I likely have a higher probability that the mechanism will have a bad spot where the rotation tends to stop. With a lighter revolving load of a single bullet, this bad spot and behavior should be more prevalent. In all likelihood the single bullet will wind up within 1 chamber of the same point in each spin. The person loading the firearm likely knows this and may invalidate all statistics by simply choosing where to load the chamber. For instance perhaps I am the executioner and am telling you how I load the chamber depending on the size of your nose or how much I like you.

        Ignoring loading bias, if the mechanism is indeed imbalanced, as it likely is, I have a much better probability of an empty chamber in a 1 of 6 scenario and a much higher probability of a similar result with a variance of ±1. So I can largely invalidate the statistics of repetition and conclude that, if I survive the first turn, I am much more likely to survive subsequent turns. The result is that, within the real world factors, I have a much higher probability of survival… IMO.

        • Atomic@sh.itjust.works
          1·
          25 days ago

          For instance, in the case of 4/6, one must assume the bullets include 2 sequential empty chambers

          No. One must not assume that. You are trying to make all the assumptions that benefit you. A certain bald spot somewhere. A bias in the mechanism that they know about, somehow it’s also about capitalism bla bla bla. You put a lot of effort into purposefully misunderstanding how simple statistics work.

          And no. If you survive the first pull. You are LESS likely to survive the second. Not more.

          It’s still the same 5/6 chance. But you having to get that chance multiple times in row makes it less likely the longer you go on. And the math will have it so after 6 times. It comes out to about ~33%

          How you feel about it on a philosophical level doesn’t change the reality around you.

          You can choose which ever option you feel more comfortable with. And that’s ok. But it’s not going to change how statistics work.

          • j4k3@lemmy.worldEnglish
            1·
            25 days ago

            I don’t believe in fate and you do. That is where this abstract overlying philosophical element ends in a stalemate. Your statistics are assuming an oversimplified set of constraints and coming to arbitrary conclusions.

            • Atomic@sh.itjust.works
              1·
              25 days ago

              I do not believe in fate.

              You should stop assuming things because you’re not very accurate with your assumptions.

              Do you want to speculate if the rounds fired are hand loaded or factory made? Maybe this specific batch of rounds had black pepper instead of gunpowder? Maybe the hammer will break just as he’s about to shoot, what are the chances of the executioner having a stroke or an aneurysm? maybe your mom would be your dad if she had balls. And your grandma would be a bike if she had wheels.

              I’m not going to debate elementary school level statistics with you. Option A and C both will give you ~33% chance of survival. But for you specifically. I’d recommend option B, for all of our sake.

  • mittyta@lemmy.world
    3·
    27 days ago

    After quick math, I’ll choose B. I believe it has 50% success rate.

    Edit: oh no, “spinning after each shot”

  • agent_nycto@lemmy.world
    3·
    27 days ago

    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

  • LostXOR@fedia.io
    41·
    27 days ago

    I’m not taking their offer, better certain death than an unknown.