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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)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Some thoughts on the 2 bird problem. (partial solution) | Comment 3 of 42 |

Bird 1 has even ods of landing at any point on the wire.

The odds of Bird 2 landing to the left or to the right of Bird 1 are the same as the distance between Pole 1 and Bird 1 divided by the total length of wire, ie. the closer Bird 1 is to Pole 1 the less chance there is of Bird 2 landing between them.

Once the odds of Bird 2 landing to the left or right of Bird 1 are calculated, it has even odds of landing on any point between Bird 1 and the pole.

My initial guess was that the span between them would average out to be 1/2, but after running some computer simulations, it seems the real answer approaches 1/3.  I think that has to do with there being a higher probablity that the distance between the birds is smaller than 1/2 that there is of the distance being greater than 1/2.  In other words, there is only 1 chance of the distance betwen the two birds being 10 meters, but the number of infinitesimally small distances approaches infinity. So even though the smaller spans affect the average in a much smaller way, there is a much higher probability of the span being small.

I haven't figured out why the solution for 2 birds approaches 1/3 yet...

 Posted by Erik O. on 2004-06-07 13:56:13
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