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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(2): Some thoughts on the 2 bird problem. (partial solution) | Comment 7 of 42 |
(In reply to re: Some thoughts on the 2 bird problem. (partial solution) by Bob)

If the first bird lands at the edge of the wire and the second one lands in a random spot then the average distance between them would be half the wire.

If the first bird lands in the middle and the second one lands at a random spot then the average distance between them would be 1/4 of the wire.

So, I thought, if the 1st bird landed 1/4 away from one end that the average distance would be 3/8 if it was directly proportional. But it's not. Because if the 1st bird land 1/4 away from the left end then there is a 25% chance that the 2nd bird will land on it's left and a 75% chance on the right. Of all the times that the 2nd bird lands on the left the average distance would be 1/8. And for all the times it lands on the right the average would be .375. But it is 3 times more likely to land on the right so the average would be...

(1/8 + 3(3/8))/4 = 5/16 = .3125

So if you can calculate the average distance between them for all possible landing spots for the 1st bird, that would be the answer. Which I guess is 1/3 since that's what Charlie said.... and he's never wrong! :-)

Edited on June 7, 2004, 2:51 pm
 Posted by Danny on 2004-06-07 14:49:37

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