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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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 re(5): Full solution | Comment 37 of 42 |
(In reply to re(4): Full solution by Bon)

"between EACH bird" isolated makes no sense.  "Between" has to relate two things.  In this instance what two things? Between (1) each bird and (2) its nearest neighbor.

In ThoughtProvoker's example of birds at locations 1, 2, 8 and 9:

Bird 1's nearest neighbor is at 2.
Bird 2's nearest neighbor is at 1.
Bird 8's nearest neighbor is at 9.
Bird 9's nearest neighbor is at 8.

Each bird is accounted for.  The stretch between 2 and 8 is unpainted, because 8 is not 2's nearest neighbor and 2 is not 8's nearest neighbor.  You might consider the stretches between 1 and 2 and between 8 and 9 to be doubly painted, but that's just two coats of paint--the extra painting does not add to the total of painted length.

 Posted by Charlie on 2004-08-04 21:59:52

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