3 ms·

It reminds me of two other neat problems - 1. Imagine a band stretched taught around the diameter of the earth (which, for the purposes of this question, is a

It reminds me of two other neat problems - 1. Imagine a band stretched taught around the diameter of the earth (which, for the purposes of this question, is a smooth sphere). Now imagine that the band is raised one metre from the ground at every single point along its length. How much longer is it? 2. Imagine perfectly parallel lines painted on the floor, exactly one foot apart, and a rigid needle of length one foot. If you throw the needle to the floor at random, what is the probability that it crosses one of the lines? (This one has a nice 'cheat' solution just like the OP article).

KRuchan 12y ago
Maybe I am missing something here, but why is 1 a neat problem? You are increasing the radius by 1 meter so the length of the band is now 2pi(r_e + 1) making the increment 2*pi meters. Is the surprisingly low increment the point of the problem?
- Jun8 12y ago
  Most people when faced with 1 and knowing the huge circumference of the Earth estimate (using Sytem 1 type thinking, http://en.wikipedia.org/wiki/Dual_process_theory#System_1 http://en.wikipedia.org/wiki/Dual_process_theory#System_1) it would be longer by kilometers.
- JoshTriplett 12y ago
  Yes, exactly; if you don't start out by doing the math, the result can seem unintuitive.
stansmith 12y ago
If you read to the bottom of the cylinder article, there's a link to another blog post by the same guy about other Martin Gardner puzzles, and the 'band around the Earth' puzzle is the first one! http://www.datagenetics.com/blog/may12012/index.html http://www.datagenetics.com/blog/may12012/index.html
eps 12y ago
#2 is 3/Pi. It's a good puzzle and it helps to know the answer :)
- Chinjut 12y ago
  Rather, 2/pi. The nice answer (referred to above as a "cheat") is to note that the sought probability is the mean number of crossings made by a 1 foot needle with the lines on the floor, where we are implicitly supposing our throw-distribution to be uniform with respect to both translation and rotation. This uniformity, along with "linearity of expectation", is such that the mean number of line-crossings from throwing any shape is simply proportional to the length of the shape (imagine breaking the shape up into many tiny straight lines, all identical except for location and orientation, the number of which is proportional to the length). Note that a circle of 1 foot diameter always makes exactly 2 crossings. Thus, the constant of proportion is 2/pi per foot, and accordingly the sought probability is 2/pi.
  - mdkras 12y ago
    #2 is called this: http://en.wikipedia.org/wiki/Buffon%27s_needle http://en.wikipedia.org/wiki/Buffon%27s_needle. This is also mentioned in Jordan Ellenberg's book How Not to Be Wrong, and has a similar passage with your explanation above. I mention the book as much because it's a great read.
- baddox 12y ago
  Maybe it's just me, but that one could use better wording. Because of the "exactly one foot apart" phrase, I interpreted it to mean that there are only two parallel lines, which obviously makes it a poorly defined problem. I probably would have understood it if the phrase was replaced by "at one foot intervals."
  - alok-g 12y ago
    I fell for this too. My initial answer, suspect of course, was thereby zero. :-)

KRuchan 12y ago

Maybe I am missing something here, but why is 1 a neat problem? You are increasing the radius by 1 meter so the length of the band is now 2pi(r_e + 1) making the increment 2*pi meters. Is the surprisingly low increment the point of the problem?

Jun8 12y ago
Most people when faced with 1 and knowing the huge circumference of the Earth estimate (using Sytem 1 type thinking, http://en.wikipedia.org/wiki/Dual_process_theory#System_1 http://en.wikipedia.org/wiki/Dual_process_theory#System_1) it would be longer by kilometers.
JoshTriplett 12y ago
Yes, exactly; if you don't start out by doing the math, the result can seem unintuitive.

stansmith 12y ago

If you read to the bottom of the cylinder article, there's a link to another blog post by the same guy about other Martin Gardner puzzles, and the 'band around the Earth' puzzle is the first one! http://www.datagenetics.com/blog/may12012/index.html http://www.datagenetics.com/blog/may12012/index.html

eps 12y ago

#2 is 3/Pi. It's a good puzzle and it helps to know the answer :)

Chinjut 12y ago
Rather, 2/pi. The nice answer (referred to above as a "cheat") is to note that the sought probability is the mean number of crossings made by a 1 foot needle with the lines on the floor, where we are implicitly supposing our throw-distribution to be uniform with respect to both translation and rotation. This uniformity, along with "linearity of expectation", is such that the mean number of line-crossings from throwing any shape is simply proportional to the length of the shape (imagine breaking the shape up into many tiny straight lines, all identical except for location and orientation, the number of which is proportional to the length). Note that a circle of 1 foot diameter always makes exactly 2 crossings. Thus, the constant of proportion is 2/pi per foot, and accordingly the sought probability is 2/pi.
- mdkras 12y ago
  #2 is called this: http://en.wikipedia.org/wiki/Buffon%27s_needle http://en.wikipedia.org/wiki/Buffon%27s_needle. This is also mentioned in Jordan Ellenberg's book How Not to Be Wrong, and has a similar passage with your explanation above. I mention the book as much because it's a great read.
baddox 12y ago
Maybe it's just me, but that one could use better wording. Because of the "exactly one foot apart" phrase, I interpreted it to mean that there are only two parallel lines, which obviously makes it a poorly defined problem. I probably would have understood it if the phrase was replaced by "at one foot intervals."
- alok-g 12y ago
  I fell for this too. My initial answer, suspect of course, was thereby zero. :-)