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Comment by throwaway283719 | Hacker News Reader

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).

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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

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?

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) it would be longer by kilometers.

Yes, exactly; if you don't start out by doing the math, the result can seem unintuitive.

#2 is 3/Pi.

It's a good puzzle and it helps to know the answer :)

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.

#2 is called this: 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.

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."

I fell for this too. My initial answer, suspect of course, was thereby zero. :-)

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