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.
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.
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.
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.
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.
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.
What do you mean by significantly large
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.
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.
Oh. That’s what they mean. That’s dumb lol.