Comment on anyway, i started blastin'

• errer@lemmy.world ⁨1⁩ ⁨day⁩ ago

  Maybe in your gut biome, but mine is just two or three really, really large bacteria

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  • DannyBoy@sh.itjust.works ⁨14⁩ ⁨hours⁩ ago

    Have you ever been prescribed a wolf?

    xkcd.com/1471/

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  • thejml@lemm.ee ⁨1⁩ ⁨day⁩ ago

    So, what are their names?

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  • HappycamperNZ@lemmy.world ⁨1⁩ ⁨day⁩ ago

    What are their names, and do they bite?

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  • CrazyLikeGollum@lemmy.world ⁨1⁩ ⁨day⁩ ago

    That just sounds like the gut biome version of a spworm.

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• yetAnotherUser@discuss.tchncs.de ⁨1⁩ ⁨day⁩ ago

  To expand a little:

  For a much smaller sample size of just 1 million, the probability to lose just 1% is basically zero.

  WolframAlpha doesn’t even bother to calculate the exact result and just rounds it:

  www.wolframalpha.com/input?i=P[490000+<+X+<+51000…

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  • Enkers@sh.itjust.works ⁨1⁩ ⁨day⁩ ago

    Yeah, I was trying to compute the “ballpark” of thr odds, but it’s actually hard to do because of how astronomically improbable it is. Even computation systems that are designed to compute rather big/small numbers (think 100,000,000^1,000,000 big) fail.

    Here’s another example: If a human only had 1,000 gut microbes, the chance that over 900 of them get snapped is 1 in ~10^162 [[WA(www.wolframalpha.com/input?i=CDF[BinomialDistribu…)]].

    Now if you do that for every human on earth, the probability is still essentially zero. [WA]

    When you consider that humans don’t have 1000 gut microbes, they have over 10 trillion, it’s just mind bogglingly improbable.

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    • yetAnotherUser@discuss.tchncs.de ⁨17⁩ ⁨hours⁩ ago

      I’ve found a proper approximation after some time and some searching.

      Since the binomial distribution has a very large n, we can use the central limit theorem and treat it as a normal distribution. The mean would be obviously 500 billion, the standard deviation is √(n * p * (1-p)) which results in 500,000.

      You still cannot plug that into WA unfortunately so we have to use a workaround.

      Since WA would calculate it through:

      Φ(b) - Φ(a), with
      b = (510 billion - mean) / (standard deviation) = 20,000
      and
      a = (490 billion - mean) / (standard deviation) = -20,000
      and
      Φ(x) = 0.5 * (1 + erf(x/√2))
      

      erf(x) is the error function which has one good property: erf(-x) = -erf(x)

      Therefore:

      Φ(20,000) - Φ(-20,000)
      = 0.5 * [ erf(20,000/√2) - erf(-20,000/√2) ]
      = erf(20,000/√2)
      ≈ erf(14,142)
      

      WolframAlpha will unfortunately not calculate this either.

      However, according to Wikipedia an approximation exists which shows that:

      1 - erf(x) = [(1 - e^(-Ax))e^(-x²)] / (Bx√π)
      

      And apparently A = 1.98 and B = 1.135 give good approximations for all x≥0.

      After failing to get a proper approximation from WA again and having to calculate every part by itself, the result is very roughly around 1 - 10^(-86,857,234).

      So it is very safe to assume you will lose between 49% and 51% of your gut bacteria. For a more realistic 10 trillion you should replace a and b above with around ±63,200 but I don’t want to bother calculating the rest and having WolframAlpha tell me my intermediary steps are equal to zero.

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      • Enkers@sh.itjust.works ⁨16⁩ ⁨hours⁩ ago

        Whoa, good work! I think I’m going to have to go over this a few times to grock how it works, especially the Φ(b) - Φ(a) bit. My stats textbook has a bit too much dust on it. ;)

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