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Bird on a Wire (Posted on 2004-06-07) Difficulty: 5 of 5
A telephone wire stretched tight between two poles placed ten meters apart is a favorite resting spot for a flock of crows.

Suppose one morning two crows land on the wire, each at a random spot (the probability is uniformly distributed). With a bucket of paint and a brush you mark the stretch of wire between them. A certain length of wire will have been painted.

On average, what length of wire would you expect to have painted? Assume that each bird is a single point along the line, and so has no width.

Suppose instead that a dozen crows landed on the wire, each at an independent, random location, and you painted the stretch of wire between each bird and its nearest neighbor. On average, what total length of wire would you expect to have painted now?

And if a thousand crows landed?

A computer-generated solution could be found, but bonus points will be awarded for a formal proof!

No Solution Yet Submitted by Sam    
Rating: 3.7000 (10 votes)

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(probably repetitive) 2 and 3 bird solution | Comment 14 of 42 |

I skimmed below comments.

As Charlie did, I'm going to measure in decameters (dam).

Let's say the first bird landed n dam from the left pole, and the second landed x dam from the left pole.  The length painted is abs(n-x).  The average of the integral from x=0 to 1 is n2/2 + (1-n)2/2.  Simplify to n2 - n + 1/2.  The average integral of that equation from n=0 to 1 is 1/3.  I've just proved that the answer to the first part is indeed 1/3 dam.

3 birds is simple enough because, like the 2 bird case, there are no unpainted spots between the leftmost and rightmost birds.  I'd like to work onto the 2 bird solution.  Lets just say that the first two birds together make one very fat 1/3 dam bird.  Note: I'm making an (probably wrong) assumption here that the fact that the width of the large bird varies doesn't affect the solution. The third bird has a 1/3 chance of landing on the oversized bird and not making a difference.  Therefore, the length of paint added is max(abs(n-x)-1/6,0) (n is the center of the bigger bird).  Integrating as before, we get n2/2 + (1-n)2/2 - 5/36.  Simplify to  n2 - n + 13/36.  Integrating again, this time from n=1/6 to 5/6, we get 8/81.  Then we, of course, add the 1/3 dam of paint under that vast bird for a total of 35/81 dam.  This time, though, I'm not so sure that I left all the rules of probability intact.

Edit: I immediately saw a mistake in 3 bird part, and also added a note.

Edited on June 7, 2004, 7:28 pm

Edited on June 7, 2004, 7:36 pm
  Posted by Tristan on 2004-06-07 19:20:11

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