• Schmoo@slrpnk.net
    link
    fedilink
    English
    arrow-up
    30
    ·
    28 days ago

    Key word here is “infinitesimally.” Of course if you’re calculating the odds of hitting something infinitesimally small you’re going to get 0. That’s just the nature of infinities. It is impossible to hit an infinitesimally small point, but that’s not what a human considers to be a “perfect bullseye.” There’s no paradox here.

    • Wolf314159@startrek.website
      link
      fedilink
      English
      arrow-up
      13
      ·
      28 days ago

      Another lesson I the importance of significant digits, a concept I’ve had to remind many a young (and sometimes an old) engineer about. An interesting idea along similar lines is that 2 + 2 can equal 5 for significantly large values of 2.

        • skulblaka@sh.itjust.works
          link
          fedilink
          English
          arrow-up
          18
          ·
          27 days ago

          Depending on how you’re rounding, I assume. Standard rounding to whole digits states that 2.4 will round to 2 but 4.8 will round to 5. So 2.4+2.4=4.8 can be reasonably simplified to 2+2=5.

          This is part of why it’s important to know what your significant digits are, because in this case the tenths digit is a bit load bearing. But, as an example, 2.43 the 3 in the hundredths digit has no bearing on our result and can be rounded or truncated.

    • emeralddawn45@discuss.tchncs.de
      link
      fedilink
      English
      arrow-up
      3
      ·
      27 days ago

      Also the circumference of the dart tip is not infinitesimally small, so theres a definite chance of it overlapping the ‘perfect bullseye’ by hitting any number of nearby points.