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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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 2 birds, 1 stone | Comment 41 of 42 |
(In reply to Strange, I get 2ln1,5 - 3/8 ~ .43593 by FrankM)

Strange to be writing to myself to point out a mistake, but here goes:

If the first bird has landed at x<=,5 then the average length to be painted is the area under the curve for L as described in my earlier comment:

L(x) = x^2/2 + (1-x)^2/2. Averaging this over x=0 to x=,5 gives <L> = 1/3

 Posted by FrankM on 2008-02-03 12:14:38

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