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 Bird on a Wire (Posted on 2004-06-07)
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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 Two Birds | Comment 9 of 42 |

Working in units of 10 meters (as Charlie did), the two-bird problem is to find the expected value of abs(x-y) over the unit square [0,1]x[0,1].  The probability density is just the constant 1 over that square, and by symmetry the expected value sought is twice the integral from 0 to 1 of the integral from 0 to x of (x-y)dxdy.  This then turns out to be the integral from 0 to 1 of x^2 dx which is 1/3.

Edited on June 7, 2004, 3:08 pm
 Posted by Richard on 2004-06-07 15:02:49

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